$$ \mathrm{Ext}-\int_X f_n(x) d^\# \mu^\# \to_{\#} \mathrm{Ext}-\int_X f(x) d^\# \mu^\#. \quad (3.19) $$

**Theorem 3.2 (Generalized Fatou's lemma)** If $(f_n)_{n=1}^\infty$ is any hyper infinite sequence in $L_+^1(X)$ then

$$ \mathrm{Ext}-\int_X \#\text{-lim } \inf_{n \to \infty} (f_n(x)) d^\# \mu^\# \le \#\text{-lim } \inf_{n \to \infty} \left( \mathrm{Ext}-\int_X f_n(x) d^\# \mu^\# \right). \quad (3.20) $$

**Theorem 3.3 (The dominated #-convergence theorem)** Let $f$ and $g$ be #-measurable, let $f_n$ be #-measurable for any $n \in \mathbb{N}$ such that $|f_n(x)| \le s$ $g(x)$ #-a.e., and $f_n \to_{\#} f$ #-a.e. If $g$ is #-integrable then $f$ and $f_n$ are also #-integrable and

$$ \mathrm{Ext}-\int_X f(x) d^\# \mu^\# = \#\text{-lim}_{n \to \infty} \mathrm{Ext}-\int_X f_n(x) d^\# \mu^\#. \quad (3.21) $$

**Proof:** $f$ is #-measurable and, since $|f| \le s$ $g$ $\mu^\#$-a.e., we have $f \in L_+^1(X)$. We have that $g+f_n \ge 0$ $\mu^\#$-a.e. and $g-f_n \ge 0$ so, by Fatou's lemma,

$$ \begin{aligned} \mathrm{Ext}-\int_X g d^\# \mu^\# + \mathrm{Ext}-\int_X f d^\# \mu^\# &\le \#\text{-lim } \inf_{n \to \infty} \left( \mathrm{Ext}-\int_X [g+f_n] d^\# \mu^\# \right) = \\ \mathrm{Ext}-\int_X g d^\# \mu^\# &+ \#\text{-lim } \inf_{n \to \infty} \left( \mathrm{Ext}-\int_X f_n d^\# \mu^\# \right), \end{aligned} \quad (3.22) \\ \begin{aligned} \mathrm{Ext}-\int_X g d^\# \mu^\# - \mathrm{Ext}-\int_X f d^\# \mu^\# &\le \#\text{-lim } \inf_{n \to \infty} \left( \mathrm{Ext}-\int_X [g-f_n] d^\# \mu^\# \right) = \\ \mathrm{Ext}-\int_X g d^\# \mu^\# &- \#\text{-lim } \sup_{n \to \infty} \left( \mathrm{Ext}-\int_X f_n d^\# \mu^\# \right) \end{aligned} $$

Therefore

$$ \#\text{-lim } \inf_{n \to \infty} \left( \mathrm{Ext}-\int_X f_n d^\# \mu^\# \right) \ge \mathrm{Ext}-\int_X f d^\# \mu^\# \ge \#\text{-lim } \sup_{n \to \infty} \left( \mathrm{Ext}-\int_X f_n d^\# \mu^\# \right) \quad (3.23) $$

and the required result follows from (3.23).

## § 4. #-Convergence in #-measure.

**Definition 4.1.** We say that a hyper infinite sequence $(f_n)_{n=1}^\infty$ of #-measurable functions on $(X,M,\mu^\#)$ is Cauchy in #-measure if, for every $\varepsilon \approx 0, \varepsilon > 0$,

$$ \mu^\#(\{x : |f_n(x) - f_m(x)| \ge \varepsilon\}) \to_{\#} 0 \quad \text{as } m, n \to \infty, \quad (4.1) $$

and that $(f_n)_{n=1}^\infty$ #-converges in #-measure to $f$ if, for every $\varepsilon \approx 0, \varepsilon > 0$,

$$ \mu^\#(\{x : |f_n(x) - f(x)| \ge \varepsilon\}) \to_{\#} 0 \quad \text{as } n \to \infty. \quad (4.2) $$

**Proposition 4.1.** If $f_n \to_{\#} f$ in $L^1$ then $f_n \to_{\#} f$ in #-measure.

**Proof.** Let $E_{n,\varepsilon} = \{x : |f_n(x) - f(x)| \ge \varepsilon\}$. Then

$$ \mathrm{Ext}-\int_X |f_n - f| d\mu^\# \ge \mathrm{Ext}-\int_{E_{n,\varepsilon}} |f_n - f| d\mu^\# \ge \varepsilon \mu^\#(E_{n,\varepsilon}), \\ \text{so } \mu(E_{n,\varepsilon}) \le \varepsilon^{-1} \mathrm{Ext}-\int_X |f_n - f| d\mu^\# \to_{\#} 0. $$

**Theorem 3.1.** Suppose that $(f_n)_{n=1}^\infty$ is Cauchy in #-measure. Then there is a #-measurable function $f$ such that $f_n \to_{\#} f$ in #-measure, and there is a hyper infinite subsequence $(f_{n_j})_{j \in \mathbb{N}}$ that #-converges to $f$ #-a.e. Moreover, if $f_n \to_{\#} g$ in #-measure then $g = f$ #-a.e.

**Proof.** We can choose a hyper infinite subsequence $(g_j)_j = (f_{n_j})_j$ of $(f_n)_{n=1}^\infty$ such that if $E_j = \{x : |g_j(x) - g_{j+1}(x)| \ge 2^{-j}\}$ then $\mu^\#(E_j) \le 2^{-j}$. If $F_k = \bigcup_{j=k}^\infty E_j$ then
an $\varepsilon > 0, \varepsilon \approx 0$ such that $f(x) < \mu$ for $c \le x < c - \varepsilon$, so $c$ is not an upper bound for $S$.
This is also a contradiction. Therefore, $f(c) = \mu$. The proof for the case where
$f(b) < \mu < f(a)$ can be obtained by applying this result to $-f(x)$.

**Lemma.15.1.** If $f$ is #-continuous at $x_0$ and $f(x_0) > \mu$, then $f(x) > \mu$ for all $x$ in some #-neighborhood of $x_0$.

15.9.Uniform #-Continuity.

**Definition 15.15.** A function $f$ is uniformly #-continuous on a subset $S$ of its domain if, for every $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $|f(x) - f(x')| < \varepsilon$ whenever $|x - x'| < \delta$ and $x, x' \in S$.

We emphasize that in this definition $\delta$ depends only on and $S$ and not on the
particular choice of $x$ and $x'$, provided that they are both in $S$.

**Theorem 15.14.** If $f$ is #-continuous on a #-closed and bounded or hyperbounded interval $[a, b]$, then $f$ is uniformly #-continuous on $[a, b]$.

**Proof.** Suppose that $\varepsilon > 0, \varepsilon \approx 0$. Since $f$ is #-continuous on $[a, b]$, for each $t \in [a, b]$ there is a positive number $\delta_t$ such that

$$
|f(x) - f(t)| < \varepsilon/2 \qquad (15.18)
$$

if $|x - t| < \delta_t$ and $x \in [a, b]$. If $I_t = (t - \delta_t, t + \delta_t)$, the collection $H = \{I_t | t \in [a, b]\}$ is an open covering of $[a, b]$. Since $[a, b]$ is #-compact, the generalized Heine–Borel theorem implies that there are hyper finitely many points $t_1, t_2, \dots, t_n$ in $[a, b]$ such that $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$. Now define

$$
\delta = \min \{ \delta_{t_1}, \delta_{t_2}, \dots, \delta_{t_n} \}. \qquad (15.19)
$$

We will show that if

$$
|x - x'| < \delta \text{ and } x, x' \in [a, b] \qquad (15.20)
$$

then $|f(x) - f(x')| < \varepsilon$. From the triangle inequality one obtains:

$$
|f(x) - f(x')| = |(f(x) - f(t_r)) + (f(t_r) - f(x'))| \le |f(x) - f(t_r)| + |f(t_r) - f(x')| \quad (15.21)
$$

Since $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$, $x$ must be in one of these intervals. Suppose that $x \in I_{t_r}$ that is,

$$
|x - t_r| \le \delta_{t_r}. \tag{15.22}
$$

From (11.18) with $t = t_r$,

$$
|f(x) - f(t_r)| \le \frac{\varepsilon}{2}. \qquad (15.23)
$$

From (11.20), (11.22), and the triangle inequality,

$$
|x' - t_r| = |(x' - x) + (x - t_r)| \le |x' - x| + |x - t_r| < \delta + \delta_{t_r} \le 2\delta_{t_r}. \quad (15.24)
$$

Therefore, (11.18) with $t = t_r$ and $x$ replaced by $x'$ implies that

$$
|f(x') - f(t_r)| \le \frac{\varepsilon}{2}. \qquad (15.25)
$$

Thus (11.25), (11.21) and (11.23) imply that $|f(x') - f(t_r)| \le \varepsilon/2$.

15.10. Monotonic External Functions $f : \mathbb{R}^\#_c \to \mathbb{R}^\#_c$.

**Theorem 15.15.** If $f$ is monotonic and nonconstant on $[a, b]$, then $f$ is #-continuous on $[a, b]$ if and only if its range **range** ($f$) = {$f(x)|x \in [a, b]$} is the #-closed interval with endpoints $f(a)$ and $f(b)$.
$$ \mu^{\#}(F_k) \leq \mathrm{Ext} - \sum_{j=k}^{\infty} 2^{-j} = 2^{1-k}, \text{ and if } x \notin F_k \text{ we have for } i \geq j \geq k $$

$$ |g_j(x) - g_i(x)| \leq \mathrm{Ext} - \sum_{l=j}^{i-1} |g_{l+1}(x) - g_l(x)| \leq \mathrm{Ext} - \sum_{l=j}^{i-1} 2^{1-j}. \quad (4.3) $$

Thus $(g_j)_j$ is pointwise Cauchy on $F_k^c$. Let $F = \bigcap_{k=1}^{\infty} F_k = \lim \sup_j E_j$. Then $\mu^\#(F) = 0$, and if we set $f(x) = \lim_{j \to \infty} g_j(x)$ for $x \notin F$, and $f(x) = 0$ for $x \in F$, then $f$ is #-measurable and $g_j \to_{\#} f$ a.e. By (4.3), we have that $|g_j(x) - f(x)| \le 2^{1-j}$ for $x \notin F_k$ and $j \ge k$. Since $\mu^\#(F_k) \to_{\#} 0$ as $k \to \infty$, it follows that $g_j \to_{\#} f$ in #-measure, because

$$ \{x : |f_n(x) - f(x)| \ge \varepsilon\} \subseteq \{x : |f_n(x) - g_j(x)| \ge (1/2)\varepsilon\} \cup \{x : |g_j(x) - f(x)| \ge (1/2)\varepsilon\}, \quad (4.4) $$

and the sets on the right both have infinite small #-measure when $n$ and $j$ are infinite large. Likewise, if $f_n \to_{\#} g$ in #-measure

$$ \{x : |f(x) - g(x)| \ge \varepsilon\} \subseteq \{x : |f(x) - f_n(x)| \ge (1/2)\varepsilon\} \cup \{x : |f_n(x) - g(x)| \ge (1/2)\varepsilon\} \quad (4.5) $$

for all $n \in \mathbb{N}$, hence $\mu^\#(\{x : |f(x) - g(x)| \ge \varepsilon\}) = 0$ for all $\varepsilon > 0$, and $f = g$ #-a.e.

**Theorem 3.2** Let $f_n \to_{\#} f$ in $L_1^\#$ then there is a hyper infinite subsequence $(f_{n_k})_k$ such that $f_{n_k} \to_{\#} f$ #-a.e.

**Proof.** Let $E_{n, \varepsilon} = \{x : |f_n(x) - f(x)| \ge \varepsilon\}$. Then

$$ \mathrm{Ext} - \int_X |f_n - f| d^\# \mu \geq \mathrm{Ext} - \int_{E_{n,\varepsilon}} |f_n - f| d^\# \mu \geq \varepsilon \mu^\#(E_{n,\varepsilon}), $$

so $\mu^\#(E_{n,\varepsilon}) \to_{\#} 0$. Then, by Theorem 3.1, there is a hyper infinite subsequence $(f_{n_k})_k$ such that $f_{n_k} \to_{\#} f$ #-a.e.

# 5. The Extension of #-Measure

## § 5.1. Outer #-measures.

**Definition 5.1.1.** Let $X$ be a nonempty set. An outer #-measure (or #-submeasure) on $X$ is a function $\zeta^\# : \tilde{P}(X) \to [0, \infty], \tilde{P}(X) \subset P(X)$ that satisfies:

(a) $\zeta^\#(\emptyset) = 0$;

(b) $\zeta^\#(A) \le \zeta(B)$ if $A \subseteq B$;

(c) $\zeta^\#\left(\bigcup_{j=1}^{\infty} A_j\right) \le \mathrm{Ext} - \sum_{j=1}^{\infty} \zeta^\#(A_j)$ for all hyper infinite sequences $(A_j)_{j=1}^{\infty}$ in $\tilde{P}(X)$.

The common way to obtain an outer #-measure is to start with a family $G$ of “elementary sets” on which a notion of measure is defined (such as rectangles or cubes in $\mathbb{R}_c^{\#n}$ and then approximate arbitrary sets from the outside by hyper infinite unions of members of $G$.

**Proposition 5.1.1** Let $G \subseteq \tilde{P}(X)$ be a set such that $\emptyset \in G, X \in G$ and let $\rho : G \to [0, \infty]$ be a function such that $\rho(\emptyset) = 0$. For any $A \subseteq X$, define

$$ \zeta^\#(A) = \rho^*(A) = \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{\infty} \rho(G_j) : G_j \in G \text{ and } A \subseteq \bigcup_{j=1}^{\infty} G_j \right\}. \quad (5.1.1) $$

if $\rho^*(A)$ exists. Then $\zeta^\#$ is an outer #-measure.

**Definition 5.1.2.** We will say that $A \subseteq X$ is admissible if $\rho^*(A)$ exists.
**Theorem 15.16.** Suppose that $f$ is increasing and #-continuous on $[a, b]$ and let $f(a) = c$ and $f(b) = d$. Then there is a unique function $g$ defined on $[c, d]$ such that

$$g(f(x)) = x, a \le x \le b, \qquad (15.26)$$

and

$$f(g(y)) = y, c \le y \le d. \qquad (15.27)$$

Moreover, $g$ is #-continuous and increasing on $[c, d]$:

The function $g$ of Theorem 15.16 is the inverse of $f$, denoted by $f^{-1}$. Since (15.26) and (15.27) are symmetric in $f$ and $g$, we can also regard $f$ as the inverse of $g$, and denote it by $g^{-1}$.

## 15.11. The #-derivative of a $\mathbb{R}^\#_c$-valued function $f: D \to \mathbb{R}^\#_c$.

A function $f: D \to \mathbb{R}^\#_c, D \subset \mathbb{R}^\#_c$ is differentiable at an #-interior point $x_0 \in D$ of its domain

$D \subset \mathbb{R}^\#_c$ if the difference quotient

$$\frac{f(x) - f(x_0)}{x - x_0}, x \neq x_0 \qquad (15.28)$$

approaches a #-limit as $x$ approaches $x_0$, in which case the #-limit is called the #-derivative

of $f$ at $x_0$, and is denoted by $f^{#'}(x_0)$ or by $f^{#\prime}(x_0)$ or by $d^{\#}f(x_0)/d^{\#}x$ i.e.,

$$d^{\#}f(x_0)/d^{\#}x \triangleq f^{#\prime}(x_0) = \#\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} \qquad (15.29)$$

# Chapter IV. $\mathbb{R}^\#_c$-Valued abstract measures

## 1. $\sigma^\#$-algebras

**Definition 1.1** ($\sigma^\#$-algebra). Let $X$ be any set. We denote by $2^X = P(X) = \{A : A \subset X\}$ the set of all subsets of $X$. A family $\mathcal{F} \subset 2^X$ is called a $\sigma^\#$-algebra (on $X$) if:

(i) $\emptyset \in \mathcal{F}$;

(ii) $\mathcal{F}$ is closed under complements, i.e. $A \in \mathcal{F}$ implies $X \setminus A \in \mathcal{F}$;

(iii) $\mathcal{F}$ is closed under hypercountable unions, i.e. if $(A_n)_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence in $\mathcal{F}$ then $\bigcup_{n \in \mathbb{N}^\#} A_n \in \mathcal{F}$.

**Proposition 1.1.** If $\mathcal{F}$ is a $\sigma^\#$-algebra on $X$ then:

1. $\mathcal{F}$ is closed under hypercountable intersections, i.e. if $(A_n)_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence in $\mathcal{F}$ then $\bigcap_{n \in \mathbb{N}^\#} A_n \in \mathcal{F}$.

2. $X \in \mathcal{F}$.

3. $\mathcal{F}$ is closed under hyperfinite unions and hyperfinite intersections.

4. $\mathcal{F}$ is closed under set differences.

5. $\mathcal{F}$ is closed under symmetric differences.

**Proposition 1.2.** Suppose $\mathcal{F} \subset 2^X$ is a family of subsets satisfying the following:

1. $\emptyset \in \mathcal{F}$;

2. $\mathcal{F}$ is closed under complements;

3. $\mathcal{F}$ is closed under hyperinfinite intersections.
**Proof.** For any admissible $A \subseteq X$, $\zeta^{\#}(A)$ is well defined. Obviously $\zeta^{\#}(\emptyset) = 0$. To prove $\*-\text{countable subadditivity}$, suppose $\{A_j\}_{j=1}^{*\infty} \subseteq \tilde{P}(X)$ and $\varepsilon \approx, \varepsilon > 0$.

For each $j \in *\mathbb{N}$, there exists $\{G_k^j\}_{k=1}^{*\infty} \subseteq G$ such that $A_j \subseteq \bigcup_{k=1}^{*\infty} G_k^j$ and
$$ \mathrm{Ext}-\sum_{k=1}^{*\infty} \rho(G_k^j) \le \zeta^{\#}(A_j) + \varepsilon 2^{-j}. $$
Then if $A = \bigcup_{j=1}^{*\infty} A_j$, we have $A \subseteq \bigcup_{j,k=1}^{*\infty} G_k^j$ and
$$ \mathrm{Ext}-\sum_{j,k=1}^{*\infty} \rho(G_k^j) \le \sum_{j=1}^{*\infty} \zeta(A_j) + \varepsilon, $$
whence $\zeta^{\#}(A) \le \mathrm{Ext}-\sum_{j=1}^{*\infty} \zeta^{\#}(A_j) + \varepsilon$. Since $\varepsilon > 0$ is arbitrary, we have done.

**Definition 5.1.3.** A set $A \subseteq X$ is called $\zeta^{\#}$-measurable if $\rho^*(A)$ exists and $\forall B \subseteq X$ such that $\rho^*(B)$ exists the equality (5.1.2) holds

$$ \zeta^{\#}(B) = \zeta^{\#}(B \cap A) + \zeta^{\#}(B \cap (X \setminus A)). \qquad (5.1.2) $$

Of course, the inequality $\zeta^{\#}(B) \le \zeta^{\#}(B \cap A) + \zeta^{\#}(B \cap (X \setminus A))$ holds for any (admissible) set $A$ and $B$.

So, to prove that $A$ is $\zeta^{\#}$-measurable, it suffices to prove the reverse inequality, which is trivial if $\zeta^{\#}(B) = *\infty$. Thus, we see that $A$ is $\zeta^{\#}$-measurable iff for any admissible $B \subseteq X$, $\zeta^{\#}(B) < *\infty$

$$ \zeta^{\#}(B) \ge \zeta^{\#}(B \cap A) + \zeta^{\#}(B \cap (X \setminus A)). \qquad (5.1.3) $$

**Theorem 5.1.1 (Generalized Caratheodory's theorem)** Let $\zeta^\#$ be an outer $\#$-measure on $X$. Then the family $\Sigma$ of all $\zeta^\#$-measurable sets is a $\sigma^\#$-algebra, and the restriction of $\zeta^\#$ to $\Sigma$ is a complete $\#$-measure.

**Proof:** First, we observe that $\Sigma$ is closed under complements, since the definition of $\zeta^\#$-measurability of $A$ is symmetric in $A$ and $A^c \triangleq X \setminus A$. Next, if $A, B \in \Sigma$ and $E \subseteq X$,

$$ \zeta^{\#}(E) = \zeta^{\#}(E \cap A) + \zeta^{\#}(E \cap A^c) = \zeta^{\#}(E \cap A \cap B) + \zeta^{\#}(E \cap A \cap B^c) + \zeta^{\#}(E \cap A^c \cap B) + \\ + \zeta^{\#}(E \cap A^c \cap B^c). $$

But $(A \cup B) = (A \cap B) \cup (A \cap B^c) \cup (A^c \cap B)$ so, by subadditivity, $\zeta^{\#}(E \cap A \cap B) + \zeta^{\#}(E \cap A \cap B^c) + \zeta^{\#}(E \cap A^c \cap B) \ge \zeta^{\#}(E \cap (A \cup B))$, and hence $\zeta^{\#}(E) \ge \zeta^{\#}(E \cap (A \cup B)) + \zeta^{\#}(E \cap (A \cup B)^c)$.

It follows that $A \cup B \in \Sigma$, so $\Sigma$ is an algebra. Moreover, if $A, B \in \Sigma$ and $A \cap B = \emptyset$, $\zeta^{\#}(A \cup B) = \zeta^{\#}((A \cup B) \cap A) + \zeta^{\#}((A \cup B) \cap A^c) = \zeta^{\#}(A) + \zeta^{\#}(B)$, so $\zeta^{\#}$ is hyperfinitely additive on $\Sigma$.

To show that $\Sigma$ is a $\sigma^\#$-algebra, it suffices to show that $\Sigma$ is closed under $\*-\text{countable disjoint unions}$. If $(A_j)_{j=1}^{*\infty}$ is a sequence of disjoint sets in $\Sigma$, set

$$ B_n = \bigcup_{j=1}^{n} A_j \wedge B = \bigcup_{j=1}^{*\infty} A_j. $$
Then, for any admissible $E \subseteq X$,

$$ \zeta^{\#}(E \cap B_n) = \zeta^{\#}(E \cap B_n \cap A_n) + \zeta^{\#}(E \cap B_n \cap A_n^c) = \zeta^{\#}(E \cap A_n) + \zeta^{\#}(E \cap B_{n-1}), $$
so a hyperfinite induction shows that $\zeta^{\#}(E \cap B_n) = \mathrm{Ext}-\sum_{j=1}^{n} \zeta^{\#}(E \cap A_j)$. Therefore

$$ \zeta^{\#}(E) = \zeta^{\#}(E \cap B_n) + \zeta^{\#}(E \cap B_n^c) \ge \mathrm{Ext}-\sum_{j=1}^{n} \zeta^{\#}(E \cap A_j) + \zeta^{\#}(E \cap B^c) $$

and, letting $n \to *\infty$, we obtain

$$ \zeta^{\#}(E) \ge \mathrm{Ext}-\sum_{j=1}^{*\infty} \zeta^{\#}(E \cap A_j) + \zeta^{\#}(E \cap B^c) \ge \zeta^{\#}\left(\bigcup_{j=1}^{*\infty} E \cap A_j\right) + \zeta^{\#}(E \cap B^c) = \zeta^{\#}(E \cap B) + $$
Then $\mathcal{F}$ is a $\sigma^\#$-algebra.

**Proposition 1.3.** If $(\mathcal{F}_\alpha)_{\alpha \in I}$ is a collection of $\sigma^\#$-algebras on $X$, then $\bigcap_\alpha \mathcal{F}_\alpha$ is also a $\sigma^\#$-algebra on $X$.

**Proposition 1.4.** ($\sigma^\#$-algebra generated by subsets). Let $K$ be a collection of subsets of $X$. There exists a $\sigma^\#$-algebra, denoted $\sigma^\#(K)$ such that $K \subset \sigma^\#(K)$ and for every other $\sigma^\#$ algebra $\mathcal{F}$ such that $K \subset \mathcal{F}$ we have that $\sigma^\#(K) \subset \mathcal{F}$
We call $\sigma^\#(K)$ the $\sigma^\#$-algebra generated by $K$.

**Proof.** Define $\sigma^\#(K) \triangleq \bigcap \{\mathcal{F} | \mathcal{F}$ is a $\sigma^\#$-algebra on $X, K \subset \mathcal{F}\}$.
This is a $\sigma^\#$-algebra with the required properties.

**Proposition 1.5.** If $K \subset \mathcal{L}$ then $\sigma^\#(K) \subset \sigma^\#(\mathcal{L})$. Also, if $K \subset \mathcal{F}$ and $\mathcal{F}$ is a $\sigma^\#$-algebra, then $\sigma^\#(K) \subset \mathcal{F}$.

**Definition 1.2.** (Borel $\sigma^\#$-algebra). Given a topological space $X$, the Borel $\sigma^\#$-algebra is the $\sigma^\#$-algebra generated by the open sets. It is denoted $B^\#(X)$.
Specifically in the case $X = *\mathbb{R}_c^{\#d}, d \in \mathbb{N}^\#$ we have that
$$B_d^\# \triangleq B^#(*\mathbb{R}_c^{\#d}) = \sigma^#(U|U \text{ is an } \#\text{-open set}).$$

A Borel-#-measurable set, i.e. a set in $B^\#(X)$, is called a #-Borel set.

**Measurable functions.** Let $f$ be a $*\mathbb{R}_c^\#$-valued function defined on a set $X$. We suppose that some $\sigma^\#$-algebra $\Omega \subseteq P(X)$ is fixed.

**Definition 1.3.** We say that $f$ is #-measurable, if $f^{-1}([a,b]) \in \Omega$ for any hyperreals $a, b \in *\mathbb{R}_c^\#$ such that $a < b$.

The following three propositions are obvious.

**Proposition 1.7.** Let $f : X \to *\mathbb{R}_c^\#$ be a function. Then the following conditions are equivalent:

(a) $f$ is #-measurable;

(b) $f^{-1}([0, b)) \in \Omega$ for any hyperreal $b \in *\mathbb{R}_c^\#$;

(c) $f^{-1}((b, \infty)) \in \Omega$ for any hyperreal $b \in *\mathbb{R}_c^\#$;

(d) $f^{-1}(B) \in \Omega$ for any $B \in B(\mathcal{R})$.

**Proposition 1.8** Let $f$ and $g$ be #-measurable functions, then

(a) $\alpha \times f + \beta \times g$ is #-measurable for any $\alpha, \beta \in *\mathbb{R}_c^\#$;

(b) functions $\max\{f, g\}$ and $f \times g$ are #-measurable.

In particular, functions $f^+ := \max\{f, 0\}, f^- := (-f)^+, \text{ and } |f| := f^+ + f^-$ are
\#-measurable.

## §2.\#-Measures and measure \#-space

**Definition 2.1.** A pair $(X, \mathcal{F})$ where $\mathcal{F}$ is a $\sigma^\#$-algebra on $X$ is called a #-measurable space. Elements of $\mathcal{F}$ are called #-measurable sets.

Given a #-measurable space $(X, \mathcal{F})$, a function $\mu^\# : \mathcal{F} \to [0, \infty^\#]$ is called a #-measure on $(X, \mathcal{F})$ if

1. $\mu^\#(\emptyset) = 0$;

2. (Hyper infinite additivity) For all hyper infinite sequences $(A_n)_{n \in \mathbb{N}^\#} \subset \mathcal{F}$ of pairwise disjoint sets in $\mathcal{F}$, we have that $\mu^\#\left(\bigcup_{n \in \mathbb{N}^\#} A_n\right) = \text{Ext}-\sum_{n \in \mathbb{N}^\#} \mu^\#(A_n)$.
$(X, \mathcal{F}, \mu^\#)$ is called a \#-measure space.

**Definition 2.2.** A measure space $(X, \mathcal{F}, \mu^\#)$ is called: (a) hyperfinite if $\mu^\#(X) < \infty^\#$.
+$\zeta^{\#}(E \cap B^c) \geq \zeta^{\#}(E)$.

Thus the inequalities in this last calculation become equalities. It follows $B \in \Sigma$.

Taking $E = B$ we have $\zeta^{\#}(B) = \text{Ext}-\sum_{j=1}^{*\infty} \zeta^{\#}(A_j)$, so $\zeta^{\#}$ is $\sigma^{\#}$-additive on $\Sigma$. Finally, if

$\zeta^{\#}(A) = 0$ then we have for any admissible set $E \subseteq X$

$\zeta^{\#}(E) \leq \zeta^{\#}(E \cap A) + \zeta^{\#}(E \cap A^c) = \zeta^{\#}(E \cap A^c) \leq \zeta^{\#}(E)$, so $A \in \Sigma$.

Therefore $\zeta^{\#}(E \cap A) = 0$ and $\zeta^{\#}|_{\Sigma}$ is a complete $\#$-measure.

Combination of Proposition 5.1.1 and Theorem 5.1.1 gives the following corollary which is also called generalized Caratheodory's theorem.

**Corollary 5.1.1** Let $G \subseteq P(X)$ be a set such that $\emptyset \in G, X \in G$, and let $\rho: G \to [0, *\infty]$ satisfy $\rho(\emptyset) = 0$. Then the family $\Sigma$ of all $\rho^* \#$-measurable sets (where $\rho^*$ is given by (5.1.1)) is a $\sigma^* \#$-algebra, and the restriction $\rho^*|_{\Sigma}$ of $\rho^*$ to $\Sigma$ is a complete $\#$-measure.

**Definition 5.1.4** Let $\bar{A}$ be an algebra of subsets of $X$, i.e. $\bar{A}$ contains $\emptyset$ and $X$, and $\bar{A}$ is closed under hyperfinite intersections and complements. A function

$\zeta: \bar{A} \to [0, *\infty]$ is called a $\#$-premeasure if $\zeta(\emptyset) = 0$ and $\zeta(\bigcup_{j=1}^{*\infty} A_j) = \text{Ext}-\sum_{j=1}^{*\infty} \zeta(A_j)$ for any disjoint sequence $(A_j)_{j \in *\mathbb{N}}$ of elements of $\bar{A}$ such that $\bigcup_{j=1}^{*\infty} A_j \in \bar{A}$.

**Theorem 5.1.2** If $\zeta$ is a $\#$-premeasure on an algebra $\bar{A} \subseteq P(X)$ and $\zeta^*: P(X) \to [0, *\infty]$ is given by (5.1.1) then $\zeta^*|_A = \zeta$ and every $A \in \bar{A}$ is $\zeta^* \#$-measurable.

# § 5.2. The Lebesgue and Lebesgue – Stieltjes $\#$-measure on $*\mathbb{R}_c^\#$.

The most important application of generalized Carathedory's theorem is the construction of the Lebesgue $\#$-measure on $*\mathbb{R}_c^\#$. Take $G$ as the set of all intervals $[a, b]$, where $a, b \in *\mathbb{R}_c^\# \cup \{-*\infty, +*\infty\}$ and $[a, b] = \emptyset$ if $a > b$. Define the function $\rho: G \to *\mathbb{R}_c^\# \cup \{*\infty\}$ by

$$ \forall a \forall b (a \le b) [\rho([a, b]) = b - a] \text{ and } \forall a \forall b (a > b) [\rho([a, b]) = 0]. \quad (5.2.1) $$

The function $\rho$ has the obvious extension (which we denote also by $\rho$) to the algebra $A$ generated by all intervals, and this extension is a $\#$-premeasure on $A$. The $\sigma^*$-algebra $\Sigma$ given by Corollary 5.1.1 is called the the Lebesgue $\sigma^*$-algebra in $R$, and the restriction of $\rho^*$ to $\Sigma = \Sigma(*\mathbb{R}_c^\#)$ is called the Lebesgue $\#$-measure on $*\mathbb{R}_c^\#$ and is denoted by $\mu^*$. By Theorem 5.1.2, $\mu^*$ is the unique extension of $\rho$. By the construction, $B^*(*\mathbb{R}_c^\#) \subseteq \Sigma(*\mathbb{R}_c^\#)$. Hence the Lebesgue $\#$-measure is a Borel $\#$-measure. It can be shown that $B^*(*\mathbb{R}_c^\#) \neq \Sigma(*\mathbb{R}_c^\#)$ and that the Lebesgue $\#$-measure can be obtained also as the completion of any Borel $\#$-measure $\omega^*$ such that $\omega^*([a, b]) = b - a(\forall a \le b)$.

The notion of the Lebesgue measure on $*\mathbb{R}_c^\#$ has the following generalization. Suppose that $\mu^*$ is a $\sigma^*$-finite Borel measure on $*\mathbb{R}_c^\#$, and let $\forall x \in *\mathbb{R}_c^\#$

$$ F(x) = \mu^*((-*\infty, x]) \quad (5.2.2) $$

Then $F$ is increasing and right $\#$-continuous. Moreover, if $b > a$, $(-*\infty, b] = (-*\infty, a] \cup (a, b]$, so $\mu^*((a, b]) = F(b) - F(a)$.
(b) It is called $\sigma^\#$-hyperfinite if $X = \bigcup_{n \in \mathbb{N}^\#} A_n$ where $A_n \in \mathcal{F}$ and $\mu^\#(A_n) < \infty^\#$ for all $n \in \mathbb{N}^\#$.

**Definition 2.3.** Let $\Sigma$ be a $\sigma^\#$-algebra of subsets of a set $X$, and let $E = (E, \|\cdot\|_\#)$ be a non-Archimedean Banach space. A function $\mu^\# : \Sigma \to E \cup \{\infty\}$ is called a vector-valued #-measure (or $E$-valued measure) if

(a) $\mu^\#(\emptyset) = 0$;

(b) $\mu^\#\left(\bigcup_{n \in \mathbb{N}^\#} A_n\right) = \mathrm{Ext}-\sum_{n \in \mathbb{N}^\#} \mu^\#(A_n)$ for any pairwise disjoint sequence $A_n, n \in \mathbb{N}^\#$,
$A_n \subseteq \Sigma$;

(c) for any $S \in \Sigma$, $\mu^\#(S) = \infty$, there exists $B \in \Sigma$ such that $B \subseteq S$ and
$0 < \|\mu^\#(B)\|_\# < \infty$.

**Definition 2.4.** (a) A function $\mu^\# : \mathcal{F} \to \mathbb{R}_c^\# \cup \{\infty\}$ is called a complex #-measure if

1. $\mu^\#(\emptyset) = 0$,

2. $\mu^\#\left(\bigcup_{n \in \mathbb{N}^\#} A_n\right) = \mathrm{Ext}-\sum_{n \in \mathbb{N}^\#} \mu^\#(A_n)$ for any sequence $A_n, n \in \mathbb{N}^\#$ of pairwise disjoint sets from $\mathcal{F}$, and, for any $A \in \mathcal{F}$, $\mu^\#(A) = \infty$, there exists $B \in \mathcal{F}$ such that $B \subseteq A$ and $0 < |\mu^\#(B)|_\# < \infty$.

(b) A function $\mu^\# : \mathcal{F} \to \mathbb{R}_c^\# \cup \{\infty\}$ is called a signed #-measure if
$\mu^\#(\emptyset) = 0$

$\mu^\#\left(\bigcup_{n \in \mathbb{N}^\#} A_n\right) = \mathrm{Ext}-\sum_{n \in \mathbb{N}^\#} \mu^\#(A_n)$ for any sequence $A_n, n \in \mathbb{N}^\#$ of pairwise disjoint sets from $\mathcal{F}$, and, for any $A \in \mathcal{F}$, $\mu^\#(A) = \infty$, there exists $B \in \mathcal{F}$ such that $B \subseteq A$ and $0 < |\mu^\#(B)| < \infty$.

**Definition 2.5.** If a certain property involving the points of #-measure space is true, except a subset having #-measure zero, then we say that this property is true #-almost everywhere (abbreviated as #-a.e.).

**Proposition 2.5.** Let $\mu^\#$ be a #-measure on a $\sigma^\#$-algebra $\mathcal{F}$, $A_n \in \mathcal{F}$, and $A_n \to A$. Then $A \in \mathcal{F}$ and $\mu^\#(A) = \lim_{n \to \infty} \mu^\#(A_n)$. In particular, if $(B_n)_{n=1}^\infty$ is a decreasing hyper infinite sequence of elements of $\mathcal{F}$ such that $\bigcap_{n=1}^\infty B_n = \emptyset$, then $\mu^\#(B_n) \to 0$.

**Definition 2.6.** If $\mathcal{F}$ is a $\sigma^\#$-algebra of subsets of $X$ and $\mu^\#$ is a #-measure on $\mathcal{F}$, then the triple $(X, \mathcal{F}, \mu)$ is called a #-measure space. The sets belonging to $\mathcal{F}$ are called #-measurable sets because the #-measure is defined for them.

## §2.1. #-Convergence of functions and the generalized Egoroff theorem.

**Definition 2.1.1.** Let $f_n, n \in \mathbb{N}^\#$ be a hyper infinite sequence of $\mathbb{R}_c^\#$-valued functions defined on $X$. We say that:

1. $f_n \to_{\#} f$ pointwise, if $f_n(x) \to_{\#} f(x)$ for all $x \in X$;

2. $f_n \to_{\#} f$ almost #-everywhere (#-a.e.), if $f_n(x) \to_{\#} f(x)$ for all $x \in X$ except a set of #-measure 0;

3. $f_n \to_{\#} f$ uniformly, if for any $\varepsilon > 0$, $\varepsilon \approx 0$ there is $n(\varepsilon)$ such that $\sup\{|f_n(x) - f(x)|: x \in X\} \le \varepsilon$ for all $n \ge n(\varepsilon)$.

**Theorem 2.1.1.** (generalized Egoroff's theorem) Suppose that $\mu^\#(X) < \infty$,
Our procedure used above can be to turn this process around and construct a measure $\mu$ starting from an increasing, right-continuous function $F$. The special case $F(x) = x$ will yield the usual Lebesgue $\#$-measure. As building blocks we can use the left-$\#$-open, right-$\#$-closed intervals in $*\mathbb{R}_c^\#$ i.e. sets of the form $(a, b]$ or $(a, *\infty)$ or $\emptyset$, where $-*\infty \le a < b < *\infty$. We call such sets $h$-intervals. The family $A$ of all finite disjoint unions of $h$-intervals is an algebra, moreover, the $\sigma^\#$-algebra generated by $A$ is the $\#$-Borel algebra $B^\#(*\mathbb{R}_c^\#)$.

**Lemma 5.2.1.** Given an increasing and right $\#$-continuous function $F : *\mathbb{R}_c^\# \to *\mathbb{R}_c^\#$, if $(a_j, b_j](j = 1, \dots, n), n \in *\mathbb{N}$ are disjoint $h$-intervals, let

$$ \mu_0^\# \left( \bigcup_{j=1}^n (a_j, b_j] \right) = \mathrm{Ext} - \sum_{j=1}^n [F(b_j) - F(a_j)], \qquad (5.2.3) $$

and let $\mu_0^\#(\emptyset) = 0$. Then $\mu_0^\#$ is a $\#$-premeasure.

**Lemma 5.2.2.** Assume that $\{(a_\alpha, b_\alpha)|\alpha \in G\}$ is a hyperfinite or $\#$-countable family of intervals in $*\mathbb{R}_c^\#$ such that $[0, 1] \subseteq \cup_{\alpha \in G} (a_\alpha, b_\alpha)$ then $\mathrm{Ext} - \sum_{\alpha \in G} |a_\alpha - b_\alpha| > 1$.

**Theorem 5.2.1** If $F : *\mathbb{R}_c^\# \to *\mathbb{R}_c^\#$ is any increasing, right $\#$-continuous function, there is a unique Borel $\#$-measure $\mu_F^\#$ on $*\mathbb{R}_c^\#$ such that $\forall a \forall b (a, b \in *\mathbb{R}_c^\#)$

$$ \mu_F^\#((a, b]) = F(b) - F(a). $$

If $G$ is another such function then $\mu_F^\# = \mu_G^\#$ iff $F - G$ is constant.

Conversely, if $\mu^\#$ is a Borel $\#$-measure on $*\mathbb{R}_c^\#$ that is hyperfinite on all $\#$-bounded $\#$-Borel sets, and we define $F(x) = \mu^\#((0, x])$ if $x > 0$, $F(x) = 0$ if $x = 0$, $F(x) = -\mu^\#((x, 0])$ if $x < 0$,

then $F$ is increasing and right $\#$-continuous function, and $\mu^\# = \mu_F^\#$.

**Proof:** Each $F$ induces a $\#$-premeasure on $B^\#(*\mathbb{R}_c^\#)$ by Lemma 5.1.1. It is clear that $F$ and $G$ induce the same $\#$-premeasure iff $F - G$ is constant, and that these

$\#$-premeasures are $\sigma^\#$-finite (since $*\mathbb{R}_c^\# = \bigcup_{j=-\infty}^{\infty} (j, j+1]$). The first two assertions

follow now from Lemma 5.2.2. As for the last one, the monotonicity of $\mu^\#$ implies the monotonicity of $F$, and the $\#$-continuity of $\mu^\#$ from above and from below implies the right $\#$-continuity of $F$ for $x \ge 0$ and $x < 0$. It is evident that $\mu^\# = \mu_F^\#$ on algebra $A$, and hence $\mu^\# = \mu_F^\#$ on $B^\#(*\mathbb{R}_c^\#)$ (accordingly to Lemma 5.2.4). Lebesgue – Stieltjes $\#$-measures possess some important and useful regularity properties.

Let us fix a complete Lebesgue – Stieltjes $\#$-measure $\mu^\#$ on $*\mathbb{R}_c^\#$ associated to an increasing, right $\#$-continuous function $F$. We denote by $\Sigma_{\mu^\#}$ the Lebesgue algebra correspondent to $\mu^\#$. Thus, for any $E \in \Sigma_{\mu^\#}$,

$$ \begin{aligned} \mu^\#(E) &= \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{*\infty} [F(b_j) - F(a_j)] \middle| E \subseteq \bigcup_{j=1}^{*\infty} (a_j, b_j] \right\} = \\ &= \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{*\infty} \mu_F^\#((a_j, b_j]) \middle| E \subseteq \bigcup_{j=1}^{*\infty} (a_j, b_j] \right\} \end{aligned} \qquad (5.2.4) $$

if infinum in RHS of (5.2.4) exists. Since $B^\#(*\mathbb{R}_c^\#) \subseteq \Sigma_{\mu^\#}$, we may replace in the second formula for $\mu^\#(E)$ $h$-intervals by $\#$-open intervals, namely

**Lemma 5.2.3** For any $E \in \Sigma_{\mu^\#}$,
$\{f_n\}_{n=1}^{\infty}$ and $f$ are #-measurable functions on $X$ such that $f_n \to_{\#} f$ #-a.e. Then, for every $\varepsilon \approx 0, \varepsilon > 0$, there exists $E \subseteq X$ such that $\mu^#(E) < \varepsilon$ and $f_n \to_{\#} f$ uniformly on $E^c = X\setminus E$.

**Proof:** Without loss of generality, we may assume that $f_n \to_{\#} f$ everywhere on $X$ and (by replacing $f_n$ with $f_n - f$) that $f \equiv 0$. For $k, n \in *\mathbb{N}$, let

$E_n(k) := \bigcup_{m=n}^{*\infty} \{x : |f_m(x)| \ge k-1\}$. Then, for a fixed $k$, $E_n(k)$ decreases as $n$ increases, and $\bigcap_{n=1}^{*\infty} E_n(k) = \emptyset$. Since $\mu^#(X) < *\infty$, we conclude that $\mu^#(E_n(k)) \to_{\#} 0$ as $n \to *\infty$.

Given $\varepsilon \approx 0, \varepsilon > 0$ and $*\mathbb{N}$, choose $n_k$ such that $\mu^#(E_n(k)) < \varepsilon \times 2^{-k}$, and set

$E = \bigcup_{n=1}^{*\infty} E_n(k)$. Then $\mu^#(E) < \varepsilon$, and we have $|f_n(x)| < k^{-1}(\forall n > n_k, x \notin E)$.

Thus $f_n \to_{\#} 0$ uniformly on $X\setminus E$.

## Generalized exhaustion argument.

Let $(X, \Sigma, \mu^#)$ be a $\sigma^#$-finite #-measure space. Given a hyper infinite sequence $(U_n)_{n=1}^{*\infty} \subseteq \Sigma$, a set $A \in \Sigma$ is called $(U_n)_n$-bounded if there exists $n \in *\mathbb{N}$ such that $A \subseteq U_n$ $\mu^#$-almost everywhere.

**Theorem 2.1.2.** (Generalized Exhaustion theorem) Let $(Y_n)_{n=1}^{*\infty} \subseteq \Sigma$ be a hyper infinite sequence satisfying $Y_n \uparrow X$ and $\mu^#(Y_n) < *\infty$ for all $n \in *\mathbb{N}$.

Let $P$ be some property of $(Y_n)_n$-bounded

#-measurable sets, such that $A \in P$ iff $B \in P$ for all $B, \mu^#(A \Delta B) = 0$. Suppose that any $(Y_n)_n$-bounded set $A$, $\mu^#(A) > 0$, has a subset $B \in \Sigma, \mu^#(B) > 0$ with the property $P$. Moreover, assume that either

(a) $A_1 \cup A_2 \in P$ for every $A_1, A_2 \in P$, or

(b) $\cup_{n \in *\mathbb{N}} B_n \in P$ for every at most hyper infinite family $(B_n)_n$ of pairwise disjoint sets possessing the property $P$.

Then there exists hyper infinite sequence $(X_n)_{n=1}^{*\infty} \subseteq \Sigma$ such that $X_n \uparrow X$, and
$P \ni X_n \subseteq Y_n$

for all $n \in *\mathbb{N}$. Moreover, there exists a pairwise disjoint sequence $(A_n)_{n=1}^{*\infty} \subseteq \Sigma$ such that $\cup_{n \in *\mathbb{N}} A_n = X$ and $A_n \in P$ for all $n \in *\mathbb{N}$.

**Proof:** Let $A$ be a $(Y_n)_n$-bounded set with $\mu^#(A) > 0$. Denote

$P_A := \{B \in P : B \subseteq A\} \cap m(A) := \sup\{\mu^#(B) : B \in P_A\}$.

I(a) Suppose $P$ satisfies (a). Then there exists a sequence $(F_n)_{n=1}^{*\infty} \subseteq P_A$ such that $m(A) = \#\lim_{n \to *\infty} \mu^#(F_n)$, We may assume, that $F_n \uparrow$. By Proposition 2.5 the set $F = \cup_{n=1}^{*\infty} F_n$ satisfies $\mu^#(F) = m(A)$. We show that $\mu^#(A) = m(A)$. If not then $\mu^#(AF) > 0$. The set $AF$ has a subset of positive #-measure $F_0 \in P$.

Then $F_n \cup F_0 \in P_A$ and $\mu^#(F_n \cup F_0) > m(A)$ for a sufficiently large $n \in *\mathbb{N}$, which contradicts to the definition of $m(A)$. Therefore, $\mu^#(A) = m(A)$.

Now we apply this for $A = Y_n$. Thus, there exists hyper infinite sequence $(X'_n)_n \subseteq \Sigma$ such that $X'_n \subseteq Y_n$, $X'_n, n \in P$, and $\mu^#(Y_n \setminus X'_n) < n^{-1}$ for all $n \in *\mathbb{N}$. By (a), we may assume that $X'_n \uparrow$. The set $X'_0 = \cup_{n=1}^{*\infty} X'_n$ satisfies $Y_n \setminus X'_0 \subseteq Y_n \setminus X'_n$, so $\mu^#(Y_n \setminus X'_0) < n^{-1}$ for all $n \in *\mathbb{N}$. Then $\mu^#(Y_n \setminus X'_0) = 0$, and $\mu^#((\cup_{n=1}^{*\infty} Y_n) \setminus X'_0) = 0$, or $\mu^#(X_n \setminus X'_0) = 0$.

Let $X_n = (X'_n \cup (X \setminus X'_0)) \cap Y_n$, then the hyper infinite sequence $(X_n)_n$ has the required properties. The desired pairwise disjoint sequence $(A_n)_{n=1}^{*\infty}$ is given recurrently by
$$
\mu^{\#}(E) = \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{\infty} \mu_{F}^{\#}((a_j, b_j)) \middle| E \subseteq \bigcup_{j=1}^{\infty} (a_j, b_j) \right\}. \qquad (5.2.5)
$$

**Theorem 5.2.2** If $E \in \Sigma_{\mu^{\#}}$ then

$$
\begin{align}
E \in \Sigma_{\mu^{\#}} &= \inf\{\mu^{\#}(U) : U \supseteq E \text{ and } U \text{ is #-open}\} = \nonumber \\
&= \sup\{\mu^{\#}(K) : K \subseteq E \text{ and } K \text{ is #-compact}\}. \tag{5.2.6}
\end{align}
$$

**Proof.** By Lemma 5.2.2, for any $\varepsilon \approx \varepsilon > 0$, there exist intervals $(a_j, b_j)$ such that

$E \subseteq \bigcup_{j=1}^{\infty} (a_j, b_j)$ and $\mu^#(E) \le \mathrm{Ext} - \sum_{j=1}^{\infty} \mu^#((a_j, b_j)) + \varepsilon$. If $U = \bigcup_{j=1}^{\infty} (a_j, b_j)$ then $U$ is #-open, $E \subseteq U$, and $\mu^#(U) \le \mu^#(E) + \varepsilon$. On the other hand, $\mu^#(U) \ge \mu^#(E)$ whenever $E \subseteq U$ so the first equality is valid.

For the second one, suppose first that $E$ is bounded in $\ast\mathbb{R}_c^\#$. If $E$ is #-closed then $E$
is #-compact and the equality is obvious. Otherwise, given $\varepsilon \approx \varepsilon > 0$, we can
choose an #-open $U$, $(\#\bar{E})\setminus E \subseteq U$, such that $\mu^#(U) \le \mu^#((\#\bar{E})\setminus E) + \varepsilon$.

Let $K = (\#\bar{E}) \setminus U$. Then $K$ is #-compact, $K \subseteq E$, and

$$
\begin{align*}
\mu^{\#}(K) &= \mu^{\#}(E) - \mu^{\#}(E \cap U) = \mu^{\#}(E) - [\mu^{\#}(U) - \mu^{\#}(U \setminus E)] \ge \\
&\ge \mu^{\#}(E) - \mu^{\#}(U) + \mu^{\#}((\#\bar{E})\setminus E) \ge \mu^{\#}(E) - \varepsilon.
\end{align*}
$$

If $E$ is unbounded in $\ast\mathbb{R}_c^\#$, let $E_j = E \cap (j,j+1]$. By the preceding argument, for
any $\varepsilon \approx \varepsilon > 0$, there exist a #-compact $K_j \subseteq E_j$ with $\mu^\#(K_j) \ge \mu^\#(E_j) - \varepsilon 2^{-j}$. Let

$$
H_n = \bigcup_{j=n}^{j=n} K_j. \text{ Then } H_n \text{ is } \#\text{-compact, } H_n \subseteq E, \text{ and } \mu^\#(H_n) \ge \mu^\# \bigcup_{j=-n}^{j=n} (E_j) - \varepsilon.
$$

Since $\mu^#(E) = \#\text{-lim}_{n \to \infty} \mu^#\left(\bigcup_{j=-n}^{j=n} E_j\right)$, the result follows.

**Theorem 5.2.3.** If $E \subseteq \ast\mathbb{R}_c^\#$, the following are equivalent:

(a) $E \in \Sigma_{\mu^{\#}}$;

(b) $E = V \cap W_1$, where $V$ is a $G_{\delta^{\#}}$-set and $\mu^{\#}(N_1) = 0$;

(c) $E = H \cup N_2$, where $H$ is an $F_{\sigma^{\#}}$-set and $\mu^{\#}(N_2) = 0$.

**Theorem 5.2.4.** If $E \in \Sigma_{\mu^{\#}}$ and $\mu^{\#}(E) < \infty$ then, for every $\varepsilon \approx \varepsilon > 0$, there is a set $A$ that is a hyperfinite union of #-open intervals such that $\mu^{\#}(E \Delta A) < \varepsilon$.

**Lemma 5.2.4** Let $A \subseteq P(X)$ be an algebra, let $\mu_0^\#$ be a $\sigma^\#$-finite #-premeasure on $A$, and let $\Omega$ be the $\sigma^\#$-algebra generated by $A$. Then there exists a unique extension of $\mu_0^\#$ to a #-measure $\mu^\#$ on $\Omega$.

§ 5.3. Product #-measures.

**Definition 5.3.1.** Let $\{(X_\alpha, \mathcal{F}_\alpha, \mu_\alpha^\#)\}_{\alpha \in \Delta}$ be a nonempty family of #-measure spaces. We define the family $\Omega$ of blocks:

$$
\begin{align}
A(A_{a_1}, A_{a_2}, \dots, A_{a_n}) &:= \nonumber \\
&= A_{a_1} \times A_{a_2} \times \dots \times A_{a_n} \times \mathrm{Ext} - \prod_{\substack{a \in \Delta, a \neq a_k, 1 \le k \le n \\ \bigcup a = A}} X_a, \tag{5.3.1}
\end{align}
$$

where $A_{a_k} \in \mathcal{F}_{a_k}$ and define a function
$A_1 = X_1$ and $A_{k+1} = X_{k+1} \setminus \bigcup_{i=1}^k A_i$.

I(b) Suppose $P$ satisfies (b). Let $F_A$ be the family of all pairwise disjoint families of elements of $P_A$ of nonzero measure. Then $F_A$ is ordered by inclusion and, obviously, satisfies the conditions of the Zorn lemma. Therefore, we have a maximal element in $F_A$, say $\Delta$. Then $\Delta$ is at most hyper infinite family, say $\Delta = \{D_n\}_n$. By (b), its union $D = \cup_n D_n$ is an element of $P_A$ as well. If $D$ is a proper subset of $A$, then $\mu^#(AD) > 0$. The set $AD$ has a subset $F \in P$ of the positive measure. Then $\Delta_1 := \Delta \cup \{F\}$ is an element of $F_A$ which is strictly greater then $\Delta$. The obtained contradiction, shows that $A \in P$ for every $(Y_n)_n$-bounded set $A$. So, we may take $X_n = Y_n$ for each $n \in *\mathbb{N}$.

Now we apply this for $A = Z_m = Y_m \cup \bigcup_{k=1}^{m-1} Y_k$ be a pairwise disjoint union, where $D_n^m \in P$ for all $n, m \in *\mathbb{N}$. The family $\{D_n^m\}_{n,m}$ is an at most hyper infinite disjoint decomposition of $X$, say $\{D_n^m\}_{n,m} = (A_n)^*_{n=1}$. The sequence $(A_n)^*_{n=1}$ satisfies the required properties.

**Theorem 2.1.3.** (The generalized Borel-Cantelli lemma) Let $(X, \Sigma, \mu^#)$ be a #-measure space. Assume that $\{A_n\}_n \subseteq \Sigma$ and $\text{Ext}-\sum_{n=1}^{\infty} \mu(A_n) < *\infty$ then $\lim_{n \to *\infty} \mu^#(A_n) = 0$.

## §2.2. Vector-valued #-measures

In this section, we extend the notion of a measure. Then we study the basic operations with signed measures and present the Jordan decomposition theorem.

### 2.2.1. Vector-valued, signed and complex #-measures.

Let $\Sigma^#$ be a $\sigma^$-algebra of subsets of a set $X$, and let $E^# = (E^\#, \|\cdot\|_\#)$ be a non-Archimedean Banach space.

**Definition 2.2.1** A function $\mu^# : \Sigma^# \to E^# \cup \{*\infty\}$ is called a vector-valued #-measure (or $E^$-valued measure) if

(a) $\mu^#(\emptyset) = 0$;

(b) $\mu^#(\cup_{k=1}^{*\infty} A_k) = \text{Ext}-\sum_{k=1}^{*\infty} \mu^#(A_k)$ for any pairwise disjoint sequence $(A_k)_k \subseteq \Sigma^$;

(c) for any $A \in \Sigma^\#$, $\mu^#(A) = *\infty$, there exists $B \in \Sigma^\#$ such that $B \subseteq A$ and $0 < \|\mu^#(B)\|_\# < *\infty$.

**Example 2.2.1** Take $\Sigma^# = P(*\mathbb{N})$, and $c_0^#$ is the non-Archimedean Banach space of all #-convergent $\mathbb{C}_c^$-valued hyper infinite sequences with a fixed element $(\alpha_n)_n \in c_0^\#$. Define now for any $A \subseteq *\mathbb{N}\psi(A) := (\beta_n)_n$, where $\beta_n = \alpha_n$ if $n \in A$ and $\beta_n = 0$ if $n \notin A$. Then $\psi$ is a $c_0^$-valued #-measure on $P(*\mathbb{N})$.

**Example 2.2.2** Let $X$ be a set and let $\Omega$ be a $\sigma^$-algebra in $P(X)$. Then for any family $\{\mu_k\}_{k=1}^m$ of finite #-measures on $\Omega$ and for any family $\{w_k\}_{k=1}^m$ of vectors of $\mathbb{R}_c^{*n}$, the $\mathbb{R}_c^{*n}$-valued #-measure $\Psi$ on $\Omega$ is defined by the formula
$$ \Psi(E) = \text{Ext}-\sum_{k=1}^m \mu_k(E) \times w_k, (E \in \Omega). $$

**Example 2.2.3** Let $X$ be a set and let $\Omega$ be a $\sigma^$-algebra in $P(X)$. Then for any family $\{\mu_k\}_{k=1}^m$ of finite #-measures on $\Omega$, for any family $\{A_k\}_{k=1}^m$ of pairwise disjoint sets in $\Omega$, and for any family $\{w_k\}_{k=1}^m$ of $\mathbb{R}_c^{*n}$, $n \in *\mathbb{N}$, the $\mathbb{R}_c^{*n}$-valued #-measure $\Phi$ on $\Omega$ is defined by the formula
$$ \Phi(E) = \text{Ext}-\sum_{k=1}^m \mu_k(E \cap A_k) \times w_k, (E \in \Omega). $$
$$ \mu_{\Omega}^{\#} : \Omega \to *\mathbb{R}_{c}^{\#} \cup \{^{*\infty}\} := $$

$$ \mu^{\#}(A_{a_1}) \times \mu^{\#}(A_{a_2}) \times \cdots \times \mu^{\#}(A_{a_n}) \times \left[ \mathrm{Ext} - \prod_{\substack{a \in \Delta, a \neq a_k, 1 \le k \le n \\ \alpha \in \Delta}} \mu^{\#}(X_a) \right]. \quad (5.3.2) $$

This function possesses an extension (by #-additivity) on the #-algebra A generated by Ω. It is easily to show that $\mu_{\Omega}^{\#}$ is a #-premeasure on A.

**Definition 5.3.2** The #-measure $\hat{\mu}^{\#}$ on the $\sigma^{\#}$-algebra $\Sigma$ generated by A accordingly to **Theorem 2.1.3** is called the product #-measure of $\{\mu_{\alpha}^{\#}\}_{\alpha \in \Delta}$, and the triple

$(\prod_{\alpha \in \Delta} X_{\alpha}, \Sigma, \hat{\mu}^{\#})$ is called the product of #-measure spaces $(X_{\alpha}, \Sigma_{\alpha}, \mu_{\alpha}^{\#})$.

We denote the $\sigma^{\#}$-algebra $\Sigma$ by $\bigotimes_{\alpha \in \Delta} \Sigma_{\alpha}$, and the #-measure $\hat{\mu}^{\#}$ by $\bigotimes_{\alpha \in \Delta} \mu_{\alpha}^{\#}$.

**Definition 5.3.3.** If $E \subseteq X_1 \times X_2$ and $x_1 \in X_1, x_2 \in X_2$, we define

$$ E_{x_1} = \{x \in X_2 : (x_1, x) \in E\} \text{ and } E^{x_2} = \{x \in X_1 : (x, x_2) \in E\}. $$

If $f: X_1 \times X_2 \to *\mathbb{R}_c^{\#}$ is a function, we define $f_{X_1}: X_2 \to *\mathbb{R}_c^{\#}$ and $f^{X_2}: X_1 \to *\mathbb{R}_c^{\#}$ by $f_{X_1}(x) = f(x_1, x)$ and $f^{X_2}(x) = f(x, x_2)$.

**Theorem 5.3.1.** (The generalized Fubini's theorem) Let $\mu_1^{\#}, \mu_2^{\#}$ be $\sigma^{\#}$-hyperfinite #-measures on $(X_1, \mathcal{F}_1)$ and $(X_2, \mathcal{F}_2)$,

$$ (X_1 \times X_2, \mathcal{F}_1 \otimes \mathcal{F}_2, \mu_1^{\#} \otimes \mu_2^{\#}) = (X_1, \mathcal{F}_1, \mu_1^{\#}) \times (X_2, \mathcal{F}_2, \mu_2^{\#}), \quad (5.3.3) $$

and let $f \in L_1^{\#}(X_1 \times X_2, \mathcal{F}_1 \otimes \mathcal{F}_2, \mu_1^{\#} \otimes \mu_2^{\#})$. Then $f_{X_1} \in L_1^{\#}(X_2, \mathcal{F}_2, \mu_2^{\#})$, $\mu_1^{\#}$-#-a.e., and $f^{X_2} \in L_1^{\#}(X_1, \mathcal{F}_1, \mu_1^{\#})$, $\mu_2^{\#}$-#-a.e., and

$$ \begin{aligned} \mathrm{Ext} - \int_{X_1 \times X_2} f d^{\#}(\mu_1^{\#} \otimes \mu_2^{\#}) &= \mathrm{Ext} - \int_{X_2} \left[ \mathrm{Ext} - \int_{X_1} f^{X_2} d^{\#} \mu_1^{\#} \right] d^{\#} \mu_2^{\#} = \\ &= \mathrm{Ext} - \int_{X_1} \left[ \mathrm{Ext} - \int_{X_2} f_{X_1} d^{\#} \mu_2^{\#} \right] \end{aligned} \quad (5.3.4) $$

**Lemma 5.3.1.** Let $(X_1, \Sigma_1, \mu_1^{\#})$ and $(X_2, \Sigma_2, \mu_2^{\#})$ be #-measure spaces, $E \in \Sigma_1 \otimes \Sigma_2$, and let $f$ be a $\Sigma_1 \otimes \Sigma_2$-measurable function on $X_1 \times X_2$, then:

(a) $E_{X_1} \in \Sigma_2$ for all $x_1 \in X_1$ and $E_{X_2} \in \Sigma_1$ for all $x_2 \in X_2$;

(b) $f_{X_1}$ is $\Sigma_2$-measurable and $f_{X_2}$ is $\Sigma_1$-measurable for all $x_1 \in X_1$ and $x_2 \in X_2$.

**Proof.** Denote by A the collection of all $A \subseteq X_1 \times X_2$ such that $A_{X_1} \in \Sigma_2$ and $A^{X_2} \in \Sigma_1(\forall x_1 \in X_1, x_2 \in X_2)$.

The family A contains all rectangles. Thus, since

$$ \left[ \bigcup_{n=1}^{\infty} A_n \right]_{x_1} = \bigcup_{n=1}^{\infty} [A_n]_{x_1}, [B_n]^{x_2} = [B_n]^{x_2} \quad (5.3.5) $$

and

$$ [X_1 \times X_2 \setminus A]_{x_1} = X_2 \setminus A_{x_1}, [X_1 \times X_2 \setminus A]^{x_2} = X_1 \setminus A^{x_2}, \quad (5.3.6) $$

A is a $\sigma^{\#}$-algebra. So $\Sigma_1 \otimes \Sigma_2 \subseteq A$, and (a) is proved. Now the part (b) follows from

(a) due to $f_{X_1}^{-1}(A) = [f^{-1}(A)]_{X_1}$ and $[f^{X_2}]^{-1}(A) = [f^{-1}(A)]^{X_2}(\forall A \subseteq *\mathbb{R}_c^{\#})$.

**Definition 5.3.4** A family $M \subseteq P(X)$ is called a monotone class if M is closed under $*$-countable increasing unions and $*$-countable decreasing intersections.

**Lemma 5.3.2.** If $A \subseteq P(X)$ is an algebra then the monotone class generated by A coincides with the $\sigma^{\#}$-algebra generated by A.
§3. The *$\mathbb{R}_c^\#$-Valued Lebesgue #-Integral

In the following consideration, we fix a $\sigma^\#$-finite #-measure space $(X, \mathcal{F}, \mu^\#)$.

**Definition 3.1.** Let $A_i \in \mathcal{F}, i = 1, \dots, n \in *\mathbb{N}$, be such that $\mu^\#(A_i) < *\infty$ for all $i$, and $A_i \cap A_j = \emptyset$ for all $i \neq j$. The external function

$$f(x) = \mathrm{Ext} - \sum_{i=1}^{n} \lambda_i \chi_{A_i}(x), \qquad (3.1)$$

$\lambda_i \in *\mathbb{R}_c^\#$, is called a simple external function. The Lebesgue external integral (Lebesgue #-integral) of a simple external function $f(x)$ is defined as

$$\mathrm{Ext} - \int_X f(x) d^\# \mu^\# = \mathrm{Ext} - \sum_{i=1}^{n} \lambda_i \mu^\#(A_i). \qquad (3.2)$$

The Lebesgue external integral of a simple function is well defined.

**Notation 3.1.** Let $A_i \in \mathcal{F}, i = 1, \dots, n \in *\mathbb{N}$, be such that $\mu^\#(A_i) < *\infty$ for all $i$, and $A_i \cap A_j = \emptyset$ for all $i \neq j$. Let $f_1(x), f_2(x)$ be a simple external function such that

(i) $0 \le f_1(x) \le f_2(x)$ and (ii) $f_1(x) = \mathrm{Ext} - \sum_{i=1}^{n} \lambda_{1,i} \chi_{A_i}(x), f_2(x) = \mathrm{Ext} - \sum_{i=1}^{n} \lambda_{2,i} \chi_{A_i}(x)$.

$$\mathrm{Ext} - \sum_{i=1}^{n} \lambda_{1,i} \le \mathrm{Ext} - \sum_{i=1}^{n} \lambda_{2,i}, \qquad (3.3)$$

then we will write $f_1(x) \le_s f_2(x)$.

**Definition 3.2.** Suppose that $\mu^\#$ is hyperfinite. Let $f: X \to *\mathbb{R}_c^\#$ be an arbitrary nonnegative bounded in $*\mathbb{R}_c^\#$ #-measurable external function and let $(f_n)_{n \in *\mathbb{N}}$, be a hyper infinite sequence of simple external functions which #-converges uniformly to $f$. Then the Lebesgue #-integral of $f$ is

$$\mathrm{Ext} - \int_X f(x) d^\# \mu^\# = \#\lim_{n \to *\infty} \left( \mathrm{Ext} - \int_X f_n(x) d^\# \mu^\# \right). \qquad (3.4)$$

**Remark 3.1.** It can be easily shown that the #-limit in Definition 3.2 exists and does not depend on the choice of a hyper infinite sequence $(f_n)_{n \in *\mathbb{N}}$, and moreover, the hyper infinite sequence $(f_n)_{n \in *\mathbb{N}}$ can be chosen such that $0 \le f_n \le f$ for all $n \in *\mathbb{N}$.

**Notation 3.2.** Let $f_1: X \to *\mathbb{R}_c^\#$ and $f_2: X \to *\mathbb{R}_c^\#$ be an arbitrary nonnegative bounded in $*\mathbb{R}_c^\#$ #-measurable external functions and let $(f_{1,n})_{n \in *\mathbb{N}}$ and $(f_{2,n})_{n \in *\mathbb{N}}$ be a hyper infinite sequences of simple external functions which #-converges uniformly to $f_1$ and to $f_2$ correspondingly. We assume that for all $n \in *\mathbb{N}$ the inequality (3.3) is satisfied, then we will write $f_1(x) \le_s f_2(x)$.

**Definition 3.3.** Let $f: X \to *\mathbb{R}_c^\#$ be a #-measurable function. Then the Lebesgue #-integral of $f$ is defined by

$$\mathrm{Ext} - \int_X f(x) d^\# \mu^\# = \mathrm{Ext} - \int_X f^+(x) d^\# \mu^\# - \mathrm{Ext} - \int_X f^-(x) d^\# \mu^\#. \qquad (3.5)$$

If both of these terms are finite or hyperfinite then the function $f$ is called #-integrable. In this case we write $f \in L_1^\# = L_1^\#(X, \mathcal{F}, \mu^\#)$.

**Notation 3.3.** We will use the following notation. For any $A \in \mathcal{F}$:

$$\mathrm{Ext} - \int_A f(x) d^\# \mu^\# = \mathrm{Ext} - \int_X f(x) \chi_A(x) d^\# \mu^\#. \qquad (3.6)$$

**Lemma 3.1.** (1) Let $f: X \to *\mathbb{R}_c^\#$ be an arbitrary nonnegative #-measurable function then
**Lemma 5.3.3.** Let $(X_1, \Sigma_1, \mu_1^{\#})$ and $(X_2, \Sigma_2, \mu_2^{\#})$ be #-measure spaces, $E \in \Sigma_1 \otimes \Sigma_2$. Then the functions $x_1 \to \mu_2^{\#}(E_{x_1})$ and $x_2 \to \mu_1^{\#}(E^{x_2})$ are #-measurable on $(X_1, \Sigma_1)$ and $(X_2, \Sigma_2)$, and

$$ \mu_1^{\#} \otimes \mu_2^{\#}(E) = \mathrm{Ext} - \int_{X_2} \mu_1^{\#}(E^{x_2}) d^{\#} \mu_2^{\#} = \mathrm{Ext} - \int_{X_1} \mu_2^{\#}(E_{x_1}) d^{\#} \mu_1^{\#}. \quad (5.3.7) $$

**Proof.** First we consider the case when $\mu_1^{\#}$ and $\mu_2^{\#}$ are finite. Let $A$ be the family of all $E \in \Sigma_1 \otimes \Sigma_2$ for which (5.3.7) is true. If $E = A \times B$, then $\mu_1^{\#}(E^{x_2}) = \mu_1^{\#}(A)\chi_B(x_2)$ and $\mu_2^{\#}(E_{x_1}) = \mu_2^{\#}(B)\chi_A(x_1)$, so $E \in A$. By additivity, it follows that hyperfinite disjoint unions of rectangles are in $A$ so, by Lemma 5.3.2, bit will suffice to show that $A$ is a monotone class. If $(E_n)_{n=1}^{*\infty}$ is an increasing hyper infinite sequence in $A$ and $E = \bigcup_{n=1}^{*\infty} E_n$, then the function $f_n(x_2) = \mu_1^{\#}((E_n)^{x_2})$

are #-measurable and increase pointwise to $f(y) = \mu_1^{\#}(E^{x_2})$. Hence $f$ is #-measurable and, by the monotone convergence theorem,

$$ \begin{aligned} \mathrm{Ext} - \int_{X_2} \mu_1^{\#}(E^{x_2}) d\mu_2^{\#} &= \#\lim_{n \to *\infty} \mathrm{Ext} X_1 \int_{X_2} \mu_1^{\#}((E_n)^{x_2}) d\mu_2^{\#} = \\ &\#\lim_{n \to *\infty} \mu_1^{\#} \times \mu_2^{\#}(E_n) = \mu_1^{\#} \times \mu_2^{\#}(E). \end{aligned} \quad (5.3.8) $$

Likewise $\mu_1^{\#} \times \mu_2^{\#}(E) = \mathrm{Ext} - \int_{X_1} \mu_2^{\#}(E_x) d\mu_1^{\#}$, so $E \in A$. Similarly, if $(E_n)_{n=1}^{*\infty}$ is a decreasing hyper infinite sequence in $A$ and $E = \bigcap_{n=1}^{*\infty} E_n$, the function $x_2 \to \mu_1^{\#}((E_1)^{x_2})$ is in $L_1^{\#}(\mu_2^{\#})$ because $\mu_1^{\#}((E_1)x_2) \le \mu_1^{\#}(X_1) < *\infty$ and $\mu_2^{\#}(X_2) < *\infty$, so the dominated convergence theorem can be applied to show that $E \in A$. Thus, $A$ is a monotone class, and the proof is complete for the case of finite #-measure spaces. Finally, if $\mu_1^{\#}$ and $\mu_2^{\#}$ are $\sigma^*$-finite, we can write $X_1 \times X_2$ as the union of an increasing hyper infinite sequence $(X_1^j \times X_2^j)_{j=1}^{*\infty}$ of rectangles of finite or hyperfinite #-measure. If $E \in \Sigma_1 \otimes \Sigma_2$, the preceding argument applies to $E \cap (X_1^j \times X_2^j)$ for each $j$ gives us

$$ \mu_1^{\#} \times \mu_2^{\#}(E \cap (X_1^j \times X_2^j)) = \mathrm{Ext} - \int_{X_2} \mu_1^{\#}(E^{x_2} \cap X_1^j) \mu_2^{\#} = \mathrm{Ext} - \int_{X_1} \mu_2^{\#}(E_{x_1} \cap X_2^j) \mu_1^{\#}. \quad (5.3.9) $$

The application of the monotone convergence theorem then yields the desired result.

**Lemma 5.3.3.** (Generalized Tonelli's theorem) Let $(X_1, \Sigma_1, \mu_1^{\#})$ and $(X_2, \Sigma_2, \mu_2^{\#})$ be #-measure spaces, and $f: X_1 \times X_2 \to *\mathbb{R}_{c+}^\#$ be a $\Sigma_1 \otimes \Sigma_2$-#-measurable function. Then the functions

$$ f_{\mu_2^{\#}}(x_1) = \mathrm{Ext} - \int_{X_2} f_{x_1} d^{\#} \mu_2^{\#} \quad \text{and} \quad f_{\mu_1^{\#}}(x_2) = \mathrm{Ext} - \int_{X_1} f^{x_2} d^{\#} \mu_1^{\#} \quad (5.3.10) $$

are $\Sigma_1$-#-measurable and $\Sigma_2$-#-measurable, respectively, and
$$ \mathrm{Ext}-\int_X f(x)d^\#\mu^\# = \sup\{\mathrm{Ext}-\int_X \varphi(x)d^\#\mu^\# \mid \varphi \text{ is a simple function such that } 0 \le \varphi(x) \le s f(x)\}. \quad (3.7) $$

(2) If $f,g: X \to *\mathbb{R}_c^\#$ are #-measurable, $g$ is #-integrable, and $|f(x)|\le s g(x)$, then $f$ is #-integrable and

$$ \left| \mathrm{Ext}-\int_X f(x)d^\#\mu^\# \right| \le \mathrm{Ext}-\int_X g(x)d^\#\mu^\#. \quad (3.8) $$

(3) $\mathrm{Ext}-\int_X |f(x)|d^\#\mu^\# = 0$ if and only if $f(x) = 0$ #-a.e.

(4) If $f_1, f_2, \dots, f_n: X \to *\mathbb{R}_c^\#$, $n \in *\mathbb{N}$ are integrable then, for $\lambda_1, \lambda_2, \dots, \lambda_n \in *\mathbb{R}_c^\#$, the linear combination $\mathrm{Ext}-\sum_{i=1}^n \lambda_i f_i$ is #-integrable and

$$ \mathrm{Ext}-\int_X (\mathrm{Ext}-\sum_{i=1}^n \lambda_i f_i) d^\#\mu^\# = \mathrm{Ext}-\sum_{i=1}^n \left( \mathrm{Ext}-\int_X \lambda_i f_i d^\#\mu^\# \right). \quad (3.9) $$

(5) Let $f \in L_1^\#(X, \mathcal{F}, \mu^\#)$, then the formula

$$ v^\#(A) = \mathrm{Ext}-\int_A f(x)d^\#\mu^\# = \mathrm{Ext}-\int_X f(x)\chi_A(x)d^\#\mu^\# \quad (3.10) $$

defines a signed #-measure on the $\sigma^\#$-algebra $\mathcal{F}$.

**Remark 3.2.** Assume that $f,g: X \to *\mathbb{R}_c^\#$$ are #-integrable functions and such that $0 \le f \le s g$ #-a.e., then

$$ \mathrm{Ext}-\int_X f(x)d^\#\mu^\# \le \mathrm{Ext}-\int_X g(x)d^\#\mu^\#. $$

#-Convergence theorem

**Definition 3.4.** We say that a hyper infinite sequence $\{f_n\}_{n=1}^{*\infty}$ of #-integrable functions $L_1^#$-#-converges to $f$ (or #-converges in $L_1^\#(X, \mathcal{F}, \mu^\#)$) if

$$ \mathrm{Ext}-\int_X |f_n - f|d^\#\mu^\# \to_\# 0 \text{ as } n \to *\infty. \quad (3.11) $$

**Theorem 3.1** (The monotone #-convergence theorem) If $\{f_n\}_{n=1}^{*\infty}$ is a hyper infinite sequence in $L_1^{\#+}(X, \mathcal{F}, \mu^\#)$ such that $f_j \le s f_{j+1}$ for all $j$ and $f(x) = \sup_{n \in *\mathbb{N}} f_n(x)$ then

$$ \mathrm{Ext}-\int_X f(x)d^\#\mu^\# = \#\text{-lim}_{n\to*\infty} \mathrm{Ext}-\int_X f_n(x)d^\#\mu^\#. \quad (3.12) $$

**Proof:** The #-limit of the increasing sequence

$$ (\mathrm{Ext}-\int_X f_n(x)d^\#\mu^\#)^{*\infty}_{n=1} $$

($*-\text{finite or }*-\text{infinite}$) exists. Moreover by (3.2),

$$ \mathrm{Ext}-\int_X f_n(x)d^\#\mu^\# \le \mathrm{Ext}-\int_X f(x)d^\#\mu^\# $$

for all $n \in *\mathbb{N}$, so

$$ \#\text{-lim}_{n\to*\infty} (\mathrm{Ext}-\int_X f_n(x)d^\#\mu^\#) \le \mathrm{Ext}-\int_X f(x)d^\#\mu^\#. $$

To establish the reverse inequality, fix $\alpha \in (0,1)$, let $\varphi$ be a simple function with $0 \le \varphi \le f$, and let $E_n = \{x : f_n(x) \ge \alpha\varphi(x)\}$. Then $(E_n)^{*\infty}_{n=1}$ is an increasing hyper infinite sequence of #-measurable sets whose union is $X$, and we have

$$ \mathrm{Ext}-\int_X f_n(x)d^\#\mu^\# \ge \mathrm{Ext}-\int_{E_n} f_n(x)d^\#\mu^\# \ge \alpha \left( \mathrm{Ext}-\int_{E_n} \varphi(x)d^\#\mu^\# \right) \quad (3.13) $$
$$
\begin{align}
Ext- \int_{X_1 \times X_2} f d^\# \mu_1^\# \otimes \mu_2^\# &= Ext- \int_{X_2} \left[ Ext- \int_{X_1} f^{x_2} d^\# \mu_1^\# \right] d^\# \mu_2^\# = \tag{5.3.11} \\
&= Ext- \int_{X_1} \left[ Ext- \int_{X_2} f_{x_1} d^\# \mu_2^\# \right] d^\# \mu_1^\#. \nonumber
\end{align}
$$

**Proof:** In the case when $f$ is a characteristic function, the statement of this lemma follows from Lemma 5.3.3. Therefore, by linearity, it holds also for nonnegative simple functions. If a nonnegative #-measurable function $f$ is arbitrary, there exists a sequence of nonnegative simple functions which increase pointwise to $f$, say $(f_n)_{n=1}^\infty$. By the monotone convergence theorem,

$$
\begin{align}
Ext- \int_{X_1} f_{\mu_2^\#} d^\# \mu_1^\# &= \#\lim_{n \to \infty} \left[ Ext- \int_{X_1} f_{\mu_2^\#}^n d^\# \mu_1^\# \right] = \tag{5.3.12} \\
&= \#\lim_{n \to \infty} \left[ Ext- \int_{X_1 \times X_2} f_n d^\# \mu_1^\# \otimes \mu_2^\# \right] \nonumber
\end{align}
$$

and

$$
\begin{align}
Ext- \int_{X_2} f_{\mu_1^\#} d^\# \mu_2^\# &= \#\lim_{n \to \infty} \left[ Ext- \int_{X_2} f_{\mu_1^\#}^n d^\# \mu_2^\# \right] = \tag{5.3.13} \\
&= \#\lim_{n \to \infty} \left[ Ext- \int_{X_1 \times X_2} f_n d^\# \mu_1^\# \otimes \mu_2^\# \right], \nonumber
\end{align}
$$

where

$$
f_{\mu_2^\#}^n(x_1) = Ext- \int_{X_1} [f_n]_{x_1} d^\# \mu_2^\#, f_{\mu_1^\#}^n(x_2) = Ext- \int_{X_2} [f_n]_{x_2} d^\# \mu_1^\#. \quad (5.3.14)
$$

This proves (5.3.11) and the lemma.

**Proof** of Theorem 5.3.1. Since an $*\mathbb{R}_c^{\#n}$-valued function $f$ is Lebesgue #-integrable iff its positive $f^+$ and negative $f^-$ parts are #-integrable, it is sufficient to prove the theorem only for nonnegative function $f \in L_1^\#(X_1 \times X_2, \Sigma, \mu_1^\# \otimes \mu_2^\#)$. But this was exactly done in Lemma 5.3.3.

§ 5.4. Lebesgue #-measure and integral in $*\mathbb{R}_c^{\#n}$.

In this section, we study $*\mathbb{R}_c^{\#n}$, $n \in *\mathbb{N}$ and functions from $*\mathbb{R}_c^{\#n}$ to $*\mathbb{R}_c^\#$ from the point of view of the Lebesgue #-measure and Lebesgue integration. All results presented below possess obvious $*C_c^{\#n}$-valued analogs. Then we define and study generalized Cantor sets which are interesting from the point of view of the set topology and the #-measure theory. Cantor sets are #-closed #-Borel nowhere #-dense subsets of the interval $[0, 1]$ or, more generally, of a Hausdorff #-space.

**Definition 5.4.1.** The Lebesgue #-measure $\mu^{\#n}$ on $*\mathbb{R}_c^{\#n}$ is the #-completion of the product of the Lebesgue #-measure on $*\mathbb{R}_c^{\#n}$ according to Definition 5.3.1. The
By (3.10) and by Proposition 2.5,

$$ \#\lim_{n \to *\infty} \left( \mathrm{Ext} - \int_{E_n} \varphi(x) d^\# \mu^\# \right) = \mathrm{Ext} - \int_X \varphi(x) d^\# \mu^\#, \quad (3.14) $$

and hence

$$ \#\lim_{n \to *\infty} \left( \mathrm{Ext} - \int_{E_n} f_n(x) d^\# \mu^\# \right) \ge \alpha \left( \mathrm{Ext} - \int_X \varphi(x) d^\# \mu^\# \right). \quad (3.15) $$

Since this is true for all $\alpha, 0 < \alpha < 1$, it remains true for $\alpha = 1$:

$$ \#\lim_{n \to *\infty} \left( \mathrm{Ext} - \int_{E_n} f_n(x) d^\# \mu^\# \right) \ge \mathrm{Ext} - \int_X \varphi(x) d^\# \mu^\#. \quad (3.16) $$

Using Lemma 3.1.(1), we may take the supremum over all simple functions $\varphi$, $0 \le \varphi \le s$. Thus

$$ \#\lim_{n \to *\infty} \left( \mathrm{Ext} - \int_{E_n} f_n(x) d^\# \mu^\# \right) \ge \mathrm{Ext} - \int_X f(x) d^\# \mu^\#. \quad (3.17) $$

Proofs of the following two corollaries of Theorem 3.1 are straightforward.

**Corollary 3.1** If $(f_n)_{n=1}^{*\infty}$ is a hyper infinite sequence in $L_+^1(X)$ and $f = \mathrm{Ext} - \sum_{n=1}^{*\infty} f_n$ pointwise then

$$ \mathrm{Ext} - \int_X f(x) d^\# \mu^\# = \mathrm{Ext} - \sum_{n=1}^{*\infty} \left( \mathrm{Ext} - \int_X f_n(x) d^\# \mu^\# \right). \quad (3.18) $$

**Corollary 3.2** If $(f_n)_{n=1}^{*\infty}$ is a hyper infinite sequence in $L_+^1(X)$, $f \in L_+^1(X)$, and $f_n \to_* f$ $\mu^\#$-a.e., then

$$ \mathrm{Ext} - \int_X f_n(x) d^\# \mu^\# \to_* \mathrm{Ext} - \int_X f(x) d^\# \mu^\#. \quad (3.19) $$

**Theorem 3.2** (Generalized Fatou's lemma) If $(f_n)_{n=1}^{*\infty}$ is any hyper infinite sequence in $L_+^1(X)$ then

$$ \mathrm{Ext} - \int_X \#\lim \inf_{n \to *\infty} (f_n(x)) d^\# \mu^\# \le \#\lim \inf_{n \to *\infty} \left( \mathrm{Ext} - \int_X f_n(x) d^\# \mu^\# \right). \quad (3.20) $$

**Theorem 3.3** (The dominated #-convergence theorem) Let $f$ and $g$ be #-measurable, let $f_n$ be #-measurable for any $n \in *\mathbb{N}$ such that $|f_n(x)| \le s$ $g(x)$ #-a.e., and $f_n \to_* f$ #-a.e. If $g$ is #-integrable then $f$ and $f_n$ are also #-integrable and

$$ \mathrm{Ext} - \int_X f(x) d^\# \mu^\# = \#\lim_{n \to *\infty} \mathrm{Ext} - \int_X f_n(x) d^\# \mu^\#. \quad (3.21) $$

**Proof:** $f$ is #-measurable and, since $|f| \le s$ $g$ $\mu^\#$-a.e., we have $f \in L_+^1(X)$. We have that $g + f_n \ge 0$ $\mu^\#$-a.e. and $g - f_n \ge 0$ so, by Fatou's lemma,

$$ \begin{align} \mathrm{Ext} - \int_X g d^\# \mu^\# + \mathrm{Ext} - \int_X f d^\# \mu^\# &\le \#\lim \inf_{n \to *\infty} \left( \mathrm{Ext} - \int_X [g+f_n] d^\# \mu^\# \right) = \nonumber \\ \mathrm{Ext} - \int_X g d^\# \mu^\# &+ \#\lim \inf_{n \to *\infty} \left( \mathrm{Ext} - \int_X f_n d^\# \mu^\# \right), \tag{3.22} \\ \mathrm{Ext} - \int_X g d^\# \mu^\# - \mathrm{Ext} - \int_X f d^\# \mu^\# &\le \#\lim \inf_{n \to *\infty} \left( \mathrm{Ext} - \int_X [g-f_n] d^\# \mu^\# \right) = \nonumber \\ &\quad = \mathrm{Ext} - \int_X g d^\# \mu^\# - \#\lim \sup_{n \to *\infty} \left( \mathrm{Ext} - \int_X f_n d^\# \mu^\# \right) \nonumber \end{align} $$

Therefore

$$ \#\lim \inf_{n \to *\infty} \left( \mathrm{Ext} - \int_X f_n d^\# \mu^\# \right) \ge \mathrm{Ext} - \int_X f d^\# \mu^\# \ge \#\lim \sup_{n \to *\infty} \left( \mathrm{Ext} - \int_X f_n d^\# \mu^\# \right) \quad (3.23) $$
domain $\Sigma^n$ of $\mu^\#$ (of course, $B^{\#(*\mathbb{R}_c^{\#n})} \subseteq \Sigma^n$) is the class of Lebesgue #-measurable sets in $*\mathbb{R}_c^{\#n}$. We write $d^\#x^n$ for $d^\#\mu^{\#n}$ and

$$ \mathrm{Ext}-\int f(x)d^\#x^n = \mathrm{Ext}-\int f d^\#\mu^{\#n}. $$

We extend some of the results of previous section to the $n$-dimensional case with $n \in *\mathbb{N}$. If $E = \mathrm{Ext}-\prod_{j=1}^n E_j$ is a block in $*\mathbb{R}_c^{\#n}$, we call sets $E_j \subseteq *\mathbb{R}_c^{\#n}$ the sides of the block $E$.

**Theorem 5.4.1.** Let $E \in \Sigma^n$. Then

(a) $\mu^{\#n}(E) = \inf\{\mu^{\#n}(U) : E \subseteq U, U \text{ #-open}\} = \sup\{\mu^{\#n}(K) : K \subseteq E, K \text{ #-compact}\}$;

(b) $E = A_1 \cup N_1 = A_2 \cup W_2$, where $A_1$ is an $F_{\sigma}^{\#n}$ set, $A_2$ is a $G_{\delta}^{\#n}$ set, and

$\mu^{\#n}(N_1) = \mu^{\#n}(N_2) = 0$;

(c) If $\mu^{\#n}(E) < *\infty$ then, for any $\varepsilon \approx 0, \varepsilon > 0$, there is a hyperfinite family $\{R_j\}_{j=1}^N$ of disjoint blocks, whose sides are intervals such that $\mu^{\#n}(E\Delta \cup_{j=1}^N R_j) < \varepsilon$.

**Proof:** By the definition of product #-measures, if $E \in \Sigma^n$ and $\varepsilon \approx 0, \varepsilon > 0$, there is a $*$-countable family $\{T_j\}_{j=1}^{*\infty}$ of blocks such that $E \subseteq \cup_{j=1}^{*\infty} T_j$ and

$\mathrm{Ext}-\sum_{j=1}^{*\infty} \mu^{\#n}(T_j) \le \mu^{\#n}(E) + \varepsilon$.

For each $j$, by applying Theorem 5.2.3 to the sides of $R_j$, we can find blocks $U_j \supseteq F_j$ whose sides are #-open sets such that $\mu^{\#n}(U_j) < \mu(T_j) + \varepsilon 2^{-j}$.

If $U = \cup_{j=1}^{*\infty} U_j$ then $U$ is #-open and

$$ \mu^{\#n}(U) \le \mathrm{Ext}-\sum_{j=1}^{*\infty} \mu^{\#n}(U_j) \le \mu^{\#n}(E) + 2\varepsilon. $$

This proves the first equation in part (a). The second equation and part (b) follow as in the proofs of Theorems 2.1.6 and 2.1.7.

Next, if $\mu^{\#n}(E) < *\infty$ then $\mu^{\#n}(U_j) < *\infty$ for all $j$. Since the sides of $U_j$ are $*$-countable unions of #-open intervals, by taking suitable hyperfinite subunions, we obtain blocks $V_j \subseteq U_j$ whose sides are hyperfinite unions of intervals such that $\mu^{\#n}(V_j) \ge \mu^{\#n}(U_j) - \varepsilon 2^{-j}$. If $N \in *\mathbb{N}$ is sufficiently hyperfinite large, we have

$$ \mu^{\#n} \left( E \setminus \bigcup_{j=1}^{N} V_j \right) \le \mu^{\#n} \left( \bigcup_{j=1}^{N} U_j \setminus V_j \right) + \mu^{\#n} \left( \bigcup_{j=N+1}^{*\infty} U_j \right) < 2\varepsilon $$

and

$$ \mu^{\#n} \left( \bigcup_{j=1}^{N} V_j \setminus E \right) \le \mu^{\#n} \left( \bigcup_{j=1}^{*\infty} U_j \setminus E \right) < \varepsilon, $$

so $\mu^{\#n}(E\Delta \cup_{j=1}^N V_j) < 3\varepsilon$. Since $\cup_{j=1}^N V_j$ can be expressed as a hyperfinite disjoint union of rectangles whose sides are intervals, we have proved (c).

## § 5.5. Lebesgue #-integrable functions on $*\mathbb{R}_c^{\#n}$

Let $\mu^{\#n}$ be the Lebesgue #-measure in $*\mathbb{R}_c^{\#n}$. The set $M(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ of all $*\mathbb{R}_c^{\#n}$-valued $\mu^{\#n}$-measurable functions on $*\mathbb{R}_c^{\#n}$ is a vector space (addition and scalar multiplication are pointwise). By $L_1^1(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ we denote its subspace of all Lebesgue #-integrable functions (with finite in $*\mathbb{R}_c^{\#n}$-integral). Now write $f \approx g$ for $f$ and $g$ in $M(*\mathbb{R}_c^{\#n}, \mu^{\#n})$, whenever $f$ and $g$ differ only on a $\mu^{\#n}$-null set (a set of $\mu^{\#n}$-measure zero). It is easily seen that $\approx$ is an equivalencerelation. Let $L_0 = L_0(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ be the set of equivalence classes of functions in $M(*\mathbb{R}_c^{\#n}, \mu^{\#n})$. We denote the equivalence classes of $f, g, \dots$ by $[f], [g], \dots$. The set $L_0$ becomes a vector space over field $*\mathbb{R}_c^{\#n}$ by
and the required result follows from (3.23).

## § 4. #-Convergence in *$\mathbb{R}_c^\#$-Valued #-measure.

**Definition 4.1.** We say that a hyper infinite sequence $(f_n)_{n=1}^{\infty}$ of #-measurable functions on $(X, M, \mu^#)$ is Cauchy in #-measure if, for every $\varepsilon \approx 0, \varepsilon > 0$,

$$ \mu^#(\{x : |f_n(x) - f_m(x)| \ge \varepsilon\}) \to_{\#} 0 \text{ as } m, n \to^{*\infty}, \quad (4.1) $$

and that $(f_n)_{n=1}^{\infty}$ #-converges in #-measure to $f$ if, for every $\varepsilon \approx 0, \varepsilon > 0$,

$$ \mu^#(\{x : |f_n(x) - f(x)| \ge \varepsilon\}) \to_{\#} 0 \text{ as } n \to^{*\infty}. \quad (4.2) $$

**Proposition 4.1.** If $f_n \to_{\#} f$ in $L^1$ then $f_n \to_{\#} f$ in #-measure.

**Proof.** Let $E_{n,\varepsilon} = \{x : |f_n(x) - f(x)| \ge \varepsilon\}$. Then

$$ \mathrm{Ext}-\int_X |f_n - f| d\mu^\# \ge \mathrm{Ext}-\int_{E_{n,\varepsilon}} |f_n - f| d\mu^\# \ge \varepsilon \mu^\#(E_{n,\varepsilon}), $$

$$ \text{so } \mu(E_{n,\varepsilon}) \le \varepsilon^{-1} \mathrm{Ext}-\int_X |f_n - f| d\mu^\# \to_{\#} 0. $$

**Theorem 3.1.** Suppose that $(f_n)_{n=1}^{\infty}$ is Cauchy in #-measure. Then there is a #-measurable function $f$ such that $f_n \to_{\#} f$ in #-measure, and there is a hyper infinite subsequence $(f_{n_j})_{j \in \mathbb{N}}$ that #-converges to $f$ #-a.e. Moreover, if $f_n \to_{\#} g$ in #-measure then $g = f$ #-a.e.

**Proof.** We can choose a hyper infinite subsequence $(g_j)_{j=1}^{\infty}$ of $(f_n)_{n=1}^{\infty}$ such that if $E_j = \{x : |g_j(x) - g_{j+1}(x)| \ge 2^{-j}\}$ then $\mu^#(E_j) \le 2^{-j}$. If $F_k = \bigcup_{j=k}^{\infty} E_j$ then

$$ \mu^\#(F_k) \le \mathrm{Ext}-\sum_{j=k}^{\infty} 2^{-j} = 2^{1-k}, \text{ and if } x \notin F_k \text{ we have for } i \ge j \ge k $$

$$ |g_j(x) - g_i(x)| \le \mathrm{Ext}-\sum_{l=j}^{i-1} |g_{l+1}(x) - g_l(x)| \le \mathrm{Ext}-\sum_{l=j}^{i-1} 2^{1-j}. \quad (4.3) $$

Thus $(g_j)_{j=1}^{\infty}$ is pointwise Cauchy on $F_k^c$. Let $F = \bigcap_{k=1}^{\infty} F_k = \limsup_j E_j$. Then $\mu^#(F) = 0$, and if we set $f(x) = \lim_{j \to \infty} g_j(x)$ for $x \notin F$, and $f(x) = 0$ for $x \in F$, then $f$ is #-measurable and $g_j \to_{\#} f$ #-a.e. By (4.3), we have that $|g_j(x) - f(x)| \le 2^{1-j}$ for $x \notin F_k$ and $j \ge k$. Since $\mu^#(F_k) \to_{\#} 0$ as $k \to \infty$, it follows that $g_j \to_{\#} f$ in #-measure, because

$$ \{x : |f_n(x) - f(x)| \ge \varepsilon\} \subseteq \{x : |f_n(x) - g_j(x)| \ge (1/2)\varepsilon\} \cup \{x : |g_j(x) - f(x)| \ge (1/2)\varepsilon\}, \quad (4.4) $$

and the sets on the right both have infinite small #-measure when $n$ and $j$ are infinite large. Likewise, if $f_n \to_{\#} g$ in #-measure

$$ \{x : |f(x) - g(x)| \ge \varepsilon\} \subseteq \{x : |f(x) - f_n(x)| \ge (1/2)\varepsilon\} \cup \{x : |f_n(x) - g(x)| \ge (1/2)\varepsilon\} \quad (4.5) $$

for all $n \in \mathbb{N}$, hence $\mu^#(\{x : |f(x) - g(x)| \ge \varepsilon\}) = 0$ for all $\varepsilon > 0$, and $f = g$ #-a.e.

**Theorem 3.2** Let $f_n \to_{\#} f$ in $L_1^\#$ then there is a hyper infinite subsequence $(f_{n_k})_k$ such that $f_{n_k} \to_{\#} f$ #-a.e.

**Proof.** Let $E_{n,\varepsilon} = \{x : |f_n(x) - f(x)| \ge \varepsilon\}$. Then

$$ \mathrm{Ext}-\int_X |f_n - f| d^\# \mu \ge \mathrm{Ext}-\int_{E_{n,\varepsilon}} |f_n - f| d^\# \mu \ge \varepsilon \mu^\#(E_{n,\varepsilon}), $$

so $\mu^\#(E_{n,\varepsilon}) \to_{\#} 0$. Then, by Theorem 3.1, there is a hyper infinite subsequence
defining $[f] + [g] = [f + g]$ and $\alpha[f] = [\alpha f]$ for a real $\alpha \in *\mathbb{R}_c^\#$. Observe that these definitions do not depend on the choice of $f$ and $g$ in their equivalence classes. The same is true for the partial order in $L_0$, if we define $[f] \le [g]$ to mean $f(x) \le g(x)$ for all $x \in *\mathbb{R}_c^\#n$ except a null set. In practice, the elements of $L_0 = L_0(*\mathbb{R}_c^\#n, \mu^{\#n})$ are usually denoted by $f, g, \dots$ and treated as if they were functions instead of equivalence classes of functions.

**Definition 5.5.1.**

**Theorem 5.5.1.** If $f \in L_1^\#(\mu^{\#n})$ and $\varepsilon \approx 0, \varepsilon > 0$, there is a simple function $\varphi = \text{Ext}-\sum_{j=1}^N \alpha_j \chi_{R_j}$, where each $R_j$ is a product of intervals such that $\text{Ext}-\int|f-\varphi|d^\#\mu^{\#n} < \varepsilon$, and there is a #-continuous function $g$ vanishing outside of a bounded in $*\mathbb{R}_c^\#n$ set such that $\text{Ext}-\int|f-g|d^\#\mu^{\#n} < \varepsilon < \varepsilon$.

**Proof.** By the definition of Lebesgue #-integrable functions, we can approximate $f$ by simple functions in $L_1^#$-#-norm. Then use Theorem 5.4.1 to approximate a simple function by a function $\varphi$ of the desired form. Finally, use the generalized Urysohn Lemma to approximate such $\varphi$ by a #-continuous function.

**Theorem 5.5.2.** The Lebesgue #-measure on $*\mathbb{R}_c^\#n$ is translation-invariant. Namely, let $a \in *\mathbb{R}_c^\#n$. Define the shift $\tau_a : *\mathbb{R}_c^\#n \to *\mathbb{R}_c^\#n$ by $\tau_a(x) = x + a$.

(a) If $E \in \mathcal{L}^{\#n}$ then $\tau_a(E) \in \mathcal{L}^{\#n}$ and $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$;

(b) If $f : *\mathbb{R}_c^\#n \to *\mathbb{R}_c^\#n$ is Lebesgue #-measurable then so is $f \circ \tau_a$. Moreover, if either $f \ge 0$ or $f \in L_1^\#(\mu^{\#n})$ then

$$ \text{Ext}-\int (f \circ \tau_a) d^\# \mu^{\#n} = \text{Ext}-\int f d^\# \mu^{\#n}. \qquad (5.5.1) $$

**Proof.** Since $\tau_a$ and its inverse $\tau_{-a}$ are #-continuous, they preserve the class of #-Borel sets. The formula $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$ follows easily from the trivial one dimensional variant of this result if $E$ is a block. For a general #-Borel set $E$, the formula $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$ follows from the previous step, since $\mu^{\#n}$ is determined by its action on blocks. Assertion (a) now follows immediately.

If $f$ is Lebesgue #-measurable and $B$ is a #-Borel set in $*\mathbb{R}_c^\#$, we have $f^{-1}(B) = E \cup N$, where $E$ is #-Borel and $\mu^\#(N) = 0$. But $\tau_a^{-1}(E)$ is Borel and $\mu^\#(\tau_a^{-1}(N)) = 0$, so $(f \circ \tau_a)^{-1}(B) \in \Sigma^n$ and $f$ is Lebesgue #-measurable. The equality (5.5.1) reduces to the equality $\mu^{\#n}(\tau_{-a}(E)) = \mu^{\#n}(E)$ when $f = \chi_E$. It is true for simple functions by linearity, and hence for nonnegative #-measurable functions by the definition of #-integral. Taking positive and negative parts of real and imaginary parts, we obtain the result for $f \in L_1^\#(\mu^{\#n})$.

**Theorem 5.5.3.** (Generalized Lusin's theorem) If $f$ is a Lebesgue #-measurable function on $*\mathbb{R}_c^\#n$ and $\varepsilon \approx 0, \varepsilon > 0$ then there exist a #-measurable set $A \subseteq *\mathbb{R}_c^\#n$ such that $\mu^{\#n}(*\mathbb{R}_c^\#n \setminus A) \le \varepsilon$ and the restriction of $f$ onto $A$ is #-continuous.

# Chapter IV. $*\mathbb{R}_c^\#$-valued distributions.

## §1. $*\mathbb{R}_c^\#$-valued test functions and distributions

Definitions and theorems appropriate to analysis on non-Archemedean field $*\mathbb{R}_c^\#$ and on complex field $*\mathbb{C}_c^\# = *\mathbb{R}_c^\# + i *\mathbb{R}_c^\#$ are given in [1]-[2].

**Definition 1.1.** [3].(i) Let $U$ be a free ultrafilters on $\mathbb{N}$ and introduce an equivalence relation on sequences in $\mathbb{R}^\mathbb{N}$ as $f_1 \sim_U f_2$ iff $\{i \in \mathbb{N} | f_1(i) = f_2(i)\} \in U$.
$(f_{n_k})_k$ such that $f_{n_k} \to f$ #-a.e.

## § 5. The Extension of #-Measure

### § 5.1. Outer #-measures.

**Definition 5.1.1.** Let $X$ be a nonempty set. An outer #-measure (or #-submeasure) on $X$ is a function $\zeta^# : \tilde{P}(X) \to [0, *^\infty], \tilde{P}(X) \subset P(X)$ that satisfies:

(a) $\zeta^#(\emptyset) = 0$;

(b) $\zeta^#(A) \le \zeta(B)$ if $A \subseteq B$;

(c) $\zeta^#\left(\bigcup_{j=1}^{*^\infty} A_j\right) \le \text{Ext}-\sum_{j=1}^{*^\infty} \zeta^#(A_j)$ for all hyper infinite sequences $(A_j)_{j=1}^{*^\infty}$ in $\tilde{P}(X)$.

The common way to obtain an outer #-measure is to start with a family $G$ of “elementary sets” on which a notion of measure is defined (such as rectangles or cubes in $*\mathbb{R}_c^{\#n}$ and then approximate arbitrary sets from the outside by hyper infinite unions of members of $G$.

**Proposition 5.1.1.** Let $G \subseteq \tilde{P}(X)$ be a set such that $\emptyset \in G, X \in G$ and let $\rho : G \to [0, *^\infty]$ be a function such that $\rho(\emptyset) = 0$. For any $A \subseteq X$, define

$$ \zeta^#(A) = \rho^*(A) = \inf\left\{\text{Ext}-\sum_{j=1}^{*^\infty} \rho(G_j) : G_j \in G \text{ and } A \subseteq \bigcup_{j=1}^{*^\infty} G_j\right\}. \quad (5.1.1) $$

if $\rho^*(A)$ exists. Then $\zeta^#$ is an outer #-measure.

**Definition 5.1.2.** We will say that $A \subseteq X$ is admissible if $\rho^*(A)$ exists.

**Proof.** For any admissible $A \subseteq X$, $\zeta^#(A)$ is well defined. Obviously $\zeta^#(\emptyset) = 0$. To prove #-countable subadditivity, suppose $\{A_j\}_{j=1}^{*^\infty} \subseteq \tilde{P}(X)$ and $\varepsilon \approx, \varepsilon > 0$.

For each $j \in *\mathbb{N}$, there exists $\{G_j^i\}_{i=1}^{*^\infty} \subseteq G$ such that $A_j \subseteq \bigcup_{k=1}^{*^\infty} G_k^i$ and

$\text{Ext}-\sum_{k=1}^{*^\infty} \rho(G_k^j) \le \zeta^#(A_j) + \varepsilon 2^{-j}$. Then if $A = \bigcup_{j=1}^{*^\infty} A_j$, we have $A \subseteq \bigcup_{j,k=1}^{*^\infty} G_k^j$ and

$\text{Ext}-\sum_{j,k=1}^{*^\infty} \rho(G_k^j) \le \sum_{j=1}^{*^\infty} \zeta(A_j) + \varepsilon$, whence $\zeta^#(A) \le \text{Ext}-\sum_{j=1}^{*^\infty} \zeta^#(A_j) + \varepsilon$. Since $\varepsilon > 0$ is

arbitrary, we have done.

**Definition 5.1.3.** A set $A \subseteq X$ is called $\zeta^#$-measurable if $\rho^*(A)$ exists and $\forall B \subseteq X$ such that $\rho^*(B)$ exists the equality (5.1.2) holds

$$ \zeta^#(B) = \zeta^#(B \cap A) + \zeta^#(B \cap (X \setminus A)). \quad (5.1.2) $$

Of course, the inequality $\zeta^#(B) \le \zeta^#(B \cap A) + \zeta^#(B \cap (X \setminus A))$ holds for any (admissible) set $A$ and $B$.

So, to prove that $A$ is $\zeta^#$-measurable, it suffices to prove the reverse inequality, which is trivial if $\zeta^#(B) = *^\infty$. Thus, we see that $A$ is $\zeta^#$-measurable iff for any admissible $B \subseteq X, \zeta^#(B) < *^\infty$

$$ \zeta^#(B) \ge \zeta^#(B \cap A) + \zeta^#(B \cap (X \setminus A)). \quad (5.1.3) $$

**Theorem 5.1.1 (Generalized Caratheodory's theorem)** Let $\zeta^#$ be an outer #-measure on $X$. Then the family $\Sigma$ of all $\zeta^#$-measurable sets is a $\sigma^#$-algebra, and the restriction of $\zeta^#$ to $\Sigma$ is a complete #-measure.

**Proof:** First, we observe that $\Sigma$ is closed under complements, since the definition
(ii) $\mathbb{R}^N$ divided out by the equivalence relation $\sim_U$ gives us the nonstandard extension $*\mathbb{R}$, the hyperreals; in symbols, $*\mathbb{R} = \mathbb{R}^N/\sim_U$ and similarly $\mathbb{N}^N$ divided out by the equivalence relation $\sim_U$ gives us the nonstandard extension $*\mathbb{N}$, the hyperintegers; in symbols, $*\mathbb{N} = \mathbb{N}^N/\sim_U$.

**Abbreviation 1.1.** If $f \in \mathbb{R}^N$, we denote its image in $*\mathbb{R}$ by $[f]$, i.e., $[f'] = \{g \in \mathbb{R}^N | g \sim_U f\}$.

**Remark 1.1.** For any real number $r \in \mathbb{R}$ let $\mathbf{r}$ denote the constant function $\mathbf{r} : \mathbb{N} \to \mathbb{R}$ with value $r$, i.e., $\mathbf{r}(n) = r$, for all $n \in \mathbb{N}$. We then have a natural embedding

$$ * (\cdot) : \mathbb{R} \hookrightarrow *\mathbb{R} $$

by setting $*r = [\mathbf{r}(n)]$ for all $r \in \mathbb{R}$. We denote its image $*(\mathbb{R})$ in $*\mathbb{R}$ by $*\mathbb{R}_{st}$.

**Definition 1.2.**[3]. An element $x \in *\mathbb{R}$ is called finite if $|x| < r$ for some $r \in \mathbb{Q}, r > 0$.

**Abbreviation 1.2.** For $x \in *\mathbb{R}$ we abbreviate $x \in *\mathbb{R}_{fin}$ if $x$ is finite.

**Remark 1.2.**[3]. Let $x \in *\mathbb{R}_{fin}$ be finite. Let $D_1$, be the set of $r \in \mathbb{Q}$ such that $r < x$ and $D_2$ the set of $r' \in \mathbb{Q}$ such that $x < r'$. The pair $(D_1, D_2)$ forms a Dedekind cut in $\mathbb{R}$, hence determines a unique $r_0 \in \mathbb{R}$. A simple argument shows that $|x - r_0|$ is infinitesimal, i.e., $|x - r_0| \approx 0$.

**Definition 1.3.**[1]. This unique $r_0$ is called the standard part of $x$ and is denoted by $\mathring{x}$ or $st(x)$.

The following notation will be used throughout this paper.

$n \in \mathbb{N}^\#$ is a fixed positive integer and $U \subset *\mathbb{R}_c^{\#n}$ is a fixed non-empty #-open subset of linear space $*\mathbb{R}_c^{\#n}$ over non Archemedan field $*\mathbb{R}_c^\#$.

$\mathbb{N} = \{0, 1, 2, \dots\}$ denotes the standard natural numbers.

$k$ will denote a non-negative integer or $\infty^\#$.

If $f$ is a function then $\mathbf{Dom}(f)$ will denote its domain and the support of $f$, denoted by $\mathbf{supp}(f)$, is defined to be the closure of the set $\{x \in \mathbf{Dom}(f) : f(x) \neq 0\}$ in $\mathbf{Dom}(f)$.

For two functions $f, g : U \to *\mathbb{C}_c^\#$, the following notation defines external canonical pairing:

$$ \langle f, g \rangle = \mathrm{Ext} - \int_U f(x)g(x)d^\#x. \qquad (1.1) $$

A multi-index of size $n \in \mathbb{N}^\#$ is an element in $\mathbb{N}^{\#n}$, if the size of multi-indices is omitted then the size should be assumed to be $n$. The length of a multi-index $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{N}^{\#n}$ is defined as $\mathrm{Ext}-\sum_{i=1}^n \alpha_i$ and denoted by $|\alpha|$. Multi-indices are particularly useful when

dealing with functions of several variables, in particular we introduce the following canonical notations for a given multi-index $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{N}^{\#n}$,

$$ \begin{gathered} x^{\alpha} = x_1^{\alpha_1} \cdots x_n^{\alpha_n}, \\ \partial^{\#\alpha} = \frac{\partial^{\#|\alpha|}}{\partial x_1^{\alpha_1} \cdots \partial x_n^{\alpha_n}} \end{gathered} \qquad (1.2) $$

We also introduce a partial order of all multi-indices by $\beta \ge \alpha$ if and only if $\beta_i \ge \alpha_i$ for all $1 \le i \le n$. When $\beta \ge \alpha$ we define their multi-index binomial coefficient as:

$$ \binom{\beta}{\alpha} = \binom{\beta_1}{\alpha_1} \cdots \binom{\beta_n}{\alpha_n}. $$

1. Let $k \in \mathbb{N}^\# \cup \infty^\#$.

2. Let $C^{\#k}(U)$ denote the vector space of all $k$-times #-continuously #-differentiable $*\mathbb{R}_c^\#$-valued or $*\mathbb{C}_c^\#$-valued functions on $U$.
of $\zeta^{\#}$-measurability of A is symmetric in A and $A^c \triangleq X\backslash A$. Next, if $A, B \in \Sigma$ and $E \subseteq X$,

$$ \zeta^{\#}(E) = \zeta^{\#}(E \cap A) + \zeta^{\#}(E \cap A^c) = \zeta^{\#}(E \cap A \cap B) + \zeta^{\#}(E \cap A \cap B^c) + \zeta^{\#}(E \cap A^c \cap B^c) + \zeta^{\#}(E \cap A^c \cap B^c). $$

But $(A \cup B) = (A \cap B) \cup (A \cap B^c) \cup (A^c \cap B)$ so, by subadditivity,

$$ \zeta^{\#}(E \cap A \cap B) + \zeta^{\#}(E \cap A \cap B^c) + \zeta^{\#}(E \cap A^c \cap B) \geq \zeta^{\#}(E \cap (A \cup B)), $$

and hence $\zeta^{\#}(E) \geq \zeta^{\#}(E \cap (A \cup B)) + \zeta^{\#}(E \cap (A \cup B)^c)$.

It follows that $A \cup B \in \Sigma$, so $\Sigma$ is an algebra. Moreover, if $A, B \in \Sigma$ and

$$ A \cap B = \emptyset, \zeta^{\#}(A \cup B) = \zeta^{\#}((A \cup B) \cap A) + \zeta^{\#}((A \cup B) \cap A^c) = \zeta^{\#}(A) + \zeta^{\#}(B), $$

so $\zeta^{\#}$ is hyperfinitely additive on $\Sigma$.

To show that $\Sigma$ is a $\sigma^{\#}$-algebra, it suffices to show that $\Sigma$ is closed under $\ast$-countable disjoint unions. If $(A_j)_{j=1}^\infty$ is a sequence of disjoint sets in $\Sigma$, set

$$ B_n = \bigcup_{j=1}^{n} A_j \cap B = \bigcup_{j=1}^{\infty} A_j. \text{ Then, for any admissible } E \subseteq X, $$

$$ \zeta^{\#}(E \cap B_n) = \zeta^{\#}(E \cap B_n \cap A_n) + \zeta^{\#}(E \cap B_n \cap A_n^c) = \zeta^{\#}(E \cap A_n) + \zeta^{\#}(E \cap B_{n-1}), $$

so a hyperfinite induction shows that $\zeta^{\#}(E \cap B_n) = \text{Ext}-\sum_{j=1}^n \zeta^{\#}(E \cap A_j)$. Therefore

$$ \zeta^{\#}(E) = \zeta^{\#}(E \cap B_n) + \zeta^{\#}(E \cap B_n^c) \geq \text{Ext}-\sum_{j=1}^{n} \zeta^{\#}(E \cap A_j) + \zeta(E \cap B^c) $$

and, letting $n \to \infty$, we obtain

$$ \zeta^{\#}(E) \geq \text{Ext}-\sum_{j=1}^{\infty} \zeta^{\#}(E \cap A_j) + \zeta^{\#}(E \cap B^c) \geq \zeta^{\#}\left(\bigcup_{j=1}^{\infty} E \cap A_j\right) + \zeta^{\#}(E \cap B^c) = \zeta^{\#}(E \cap B) + \zeta^{\#}(E \cap B^c) \geq \zeta^{\#}(E). $$

Thus the inequalities in this last calculation become equalities. It follows $B \in \Sigma$.

Taking $E = B$ we have $\zeta^{\#}(B) = \text{Ext}-\sum_{j=1}^{\infty} \zeta^{\#}(A_j)$, so $\zeta^{\#}$ is $\sigma^{\#}$-additive on $\Sigma$. Finally, if

$\zeta^{\#}(A) = 0$ then we have for any admissible set $E \subseteq X$

$$ \zeta^{\#}(E) \leq \zeta^{\#}(E \cap A) + \zeta^{\#}(E \cap A^c) = \zeta^{\#}(E \cap A^c) \leq \zeta^{\#}(E), \text{ so } A \in \Sigma. $$

Therefore $\zeta^{\#}(E \cap A) = 0$ and $\zeta^{\#}|_{\Sigma}$ is a complete $\#$-measure.

Combination of Proposition 5.1.1 and Theorem 5.1.1 gives the following corollary which is also called generalized Caratheodory's theorem.

**Corollary 5.1.1** Let $G \subseteq P(X)$ be a set such that $\emptyset \in G, X \in G$, and let $\rho: G \to [0, \infty]$ satisfy $\rho(\emptyset) = 0$. Then the family $\Sigma$ of all $\rho^*\#$-measurable sets (where $\rho^*$ is given by (5.1.1)) is a $\sigma^\#$-algebra, and the restriction $\rho^*|_{\Sigma}$ of $\rho^*$ to $\Sigma$ is a complete $\#$-measure.

**Definition 5.1.4** Let $\bar{A}$ be an algebra of subsets of $X$, i.e. $\bar{A}$ contains $\emptyset$ and $X$, and $\bar{A}$ is closed under hyperfinite intersections and complements. A function

$$ \zeta: \bar{A} \to [0, \infty] \text{ is called a } \#\text{-premeasure if } \zeta(\emptyset) = 0 \text{ and } \zeta\left(\bigcup_{j=1}^{\infty} A_j\right) = \text{Ext}-\sum_{j=1}^{\infty} \zeta(A_j) \text{ for any disjoint sequence } (A_j)_{j \in \mathbb{N}} \text{ of elements of } \bar{A} \text{ such that } \bigcup_{j=1}^{\infty} A_j \in \bar{A}. $$

**Theorem 5.1.2** If $\zeta$ is a $\#$-premeasure on an algebra $\bar{A} \subseteq P(X)$ and

$$ \zeta^*: P(X) \to [0, \infty] \text{ is given by (5.1.1) then } \zeta^*|_{A} = \zeta \text{ and every } A \in \bar{A} \text{ is } \zeta^*\#\text{-measurable.} $$

## § 5.2. The Lebesgue and Lebesgue – Stieltjes $\#$-measure
For any #-compact subset $K \subseteq U$, let $C^{\#k}(K)$ and $C^{\#k}(K; U)$ both denote the vector space of all those functions $f \in C^{\#k}(U)$ such that **supp**($f$) $\subseteq K$.

Note that $C^{\#k}(K)$ depends on both $K$ and $U$ but we will only indicate $K$, where in particular, if $f \in C^{\#k}(K)$ then the domain of $f$ is $U$ rather than $K$. We will use the notation

$C^{\#k}(K; U)$ only when the notation $C^{\#k}(K)$ risks being ambiguous.

Every $C^{\#k}(K)$ contains the constant 0 map, even if $K = \emptyset$.

Let $C_c^{\#k}(U)$ denote the set of all $f \in C^{\#k}(U)$ such that $f \in C^{\#k}(K)$ for some #-compact subset $K$ of $U$.

Equivalently, $C_c^{\#k}(U)$ is the set of all $f \in C^{\#k}(U)$ such that $f$ has #-compact support.

$C_c^{\#k}(U)$ is equal to the union of all $C^{\#k}(K)$ as $K \subseteq U$ ranges over all #-compact subsets of $U$. If $f$ is a $*\mathbb{R}_c^\#$-valued function on $U$, then $f$ is an element of $C_c^{\#k}(U)$ if and only if $f$ is a $C^{\#k}$ bump function. Every $*\mathbb{R}_c^\#$-valued test function on $U$ is always also a $*C_c^\#$-valued test function on $U$.

For all $j, k \in \mathbb{N}$ and any #-compact subsets $K$ and $L$ of $U$, we have:

$$C^{\#k}(K) \subseteq C_c^{\#k}(U) \subseteq C^{\#k}(U);$$

$$C^{\#k}(K) \subseteq C^{\#k}(L) \text{ if } K \subseteq LC^{\#k}(K) \subseteq C^{\#j}(K) \text{ if } j \le k;$$

$$C_c^{\#k}(U) \subseteq C_c^{\#j}(U) \text{ if } j \le k;$$

$$C^{\#k}(U) \subseteq C^{\#j}(U) \text{ if } j \le k.$$

**Definition1.1.** Elements of $C_c^{\#\infty}(U)$ are called $*\mathbb{R}_c^\#$-valued test functions on $U$ and $C_c^{\#\infty}(U)$ is

called the space of $*\mathbb{R}_c^\#$-valued test functions on $U$. We will use both $D(U)$ and $C_c^{\#\infty}(U)$ to denote this space.

**Definition1.2.** Distributions on $U$ are #-continuous $*\mathbb{R}_c^\#$-valued linear functionals on $C_c^{\#\infty}(U)$ when this vector space is endowed with a particular topology called the canonical

LF-topology.

The following proposition states two necessary and sufficient conditions for the #-continuity of a linear functional on $C_c^{\#\infty}(U)$ that are often straightforward to verify.

**Proposition1.1.** A linear functional $T$ on $C_c^{\#\infty}(U)$ is #-continuous, and therefore a distribution, if and only if either of the following equivalent conditions are satisfied:

1. For every #-compact subset $K \subseteq U$ there exist constants $C > 0$ and $N \in \mathbb{N}$ dependent on $K$ such that for all $f \in C_c^{\#\infty}(U)$ with support contained in $K$

$$|T(f)| \le C \sup \{|\partial^{\#a} f(x)|: x \in U, |a| \le N\}.$$

2. For every #-compact subset $K \subseteq U$ and every sequence $\{f_i\}_{i=1}^{\infty}$ in $C_c^{\#\infty}(U)$ whose supports are contained in $K$, if $\{\partial^{\#a} f_i\}_{i=1}^{\infty}$ #-converges uniformly to zero on $U$ for every multi-index $\alpha$, then $\#\lim_{i \to \infty} T(f_i) = 0$.

# §2. The non-Archimedean external $*\mathbb{R}_c^\#$-Valued Schwartz distributions.

Defined below are the tempered distributions, which form a subspace of $D^{\#'}(*\mathbb{R}_c^{\#n})$, the space of distributions on $*\mathbb{R}_c^{\#n}$. This is a proper subspace: while every tempered distribution is a distribution and an element of $D^{\#'}(*\mathbb{R}_c^{\#n})$ the converse is not true.

Tempered distributions are useful if one studies the Fourier transform since all tempered distributions have a Fourier transform, which is not true for an arbitrary
on $\mathbb{R}^#_c$.

The most important application of generalized Caratheodory's theorem is the construction of the Lebesgue #-measure on $\mathbb{R}^#_c$. Take $G$ as the set of all intervals $[a,b]$, where $a,b \in \mathbb{R}^#_c \cup \{-\infty, +\infty\}$ and $[a,b] = \emptyset$ if $a > b$. Define the function $\rho: G \to \mathbb{R}^#_c \cup \{\infty\}$ by

$$ \forall a \forall b (a \le b) [\rho([a,b]) = b - a] \text{ and } \forall a \forall b (a > b) [\rho([a,b]) = 0]. \quad (5.2.1) $$

The function $\rho$ has the obvious extension (which we denote also by $\rho$) to the algebra $A$ generated by all intervals, and this extension is a #-premeasure on $A$. The $\sigma$#-algebra $\Sigma$ given by Corollary 5.1.1 is called the the Lebesgue $\sigma$#-algebra in $R$, and the restriction of $\rho^*$ to $\Sigma = \Sigma(\mathbb{R}^#_c)$ is called the Lebesgue #-measure on $\mathbb{R}^#_c$ and is denoted by $\mu^#$. By Theorem 5.1.2, $\mu^#$ is the unique extension of $\rho$. By the construction, $B^#(\mathbb{R}^#_c) \subseteq \Sigma(\mathbb{R}^#_c)$. Hence the Lebesgue #-measure is a Borel #-measure. It can be shown that $B^#(\mathbb{R}^#_c) \neq \Sigma(\mathbb{R}^#_c)$ and that the Lebesgue #-measure can be obtained also as the completion of any Borel #-measure $\omega^#$ such that $\omega^#([a,b]) = b - a(\forall a \le b)$.

The notion of the Lebesgue measure on $\mathbb{R}^#_c$ has the following generalization. Suppose that $\mu^#$ is a $\sigma$#-finite Borel measure on $\mathbb{R}^#_c$, and let $\forall x \in \mathbb{R}^#_c$

$$ F(x) = \mu^#((-\infty, x]) \quad (5.2.2) $$

Then $F$ is increasing and right #-continuous. Moreover, if $b > a$, $(-\infty, b] = (-\infty, a] \cup (a, b]$, so $\mu^#((a, b]) = F(b) - F(a)$.

Our procedure used above can be to turn this process around and construct a measure $\mu$ starting from an increasing, right-continuous function $F$. The special case $F(x) = x$ will yield the usual Lebesgue #-measure. As building blocks we can use the left-#-open, right-#-closed intervals in $\mathbb{R}^#_c$ i.e. sets of the form $(a, b]$ or $(a, \infty)$ or $\emptyset$, where $-\infty \le a < b < \infty$. We call such sets $h$-intervals. The family $A$ of all finite disjoint unions of $h$-intervals is an algebra, moreover, the $\sigma$#-algebra generated by $A$ is the #-Borel algebra $B^#(\mathbb{R}^#_c)$.

**Lemma 5.2.1.** Given an increasing and right #-continuous function $F: \mathbb{R}^#_c \to \mathbb{R}^#_c$, if $(a_j, b_j](j = 1, \dots, n), n \in \mathbb{N}$ are disjoint $h$-intervals, let

$$ \mu_0^# \left( \bigcup_{j=1}^n (a_j, b_j] \right) = \mathrm{Ext} - \sum_{j=1}^n [F(b_j) - F(a_j)], \quad (5.2.3) $$

and let $\mu_0^#(\emptyset) = 0$. Then $\mu_0^#$ is a #-premeasure.

**Lemma 5.2.2.** Assume that $\{(a_a, b_a)|a \in G\}$ is a hyperfinite or $*$-countable family of intervals in $\mathbb{R}^#_c$ such that $[0, 1] \subseteq \cup_{a \in G} (a_a, b_a)$ then $\mathrm{Ext} - \sum_{a \in G} |a_a - b_a| > 1$.

**Theorem 5.2.1** If $F: \mathbb{R}^#_c \to \mathbb{R}^#_c$ is any increasing, right #-continuous function, there is a unique Borel #-measure $\mu_F^#$ on $\mathbb{R}^#_c$ such that $\forall a \forall b (a, b \in \mathbb{R}^#_c)$

$$ \mu_F^#((a, b]) = F(b) - F(a). $$

If $G$ is another such function then $\mu_F^# = \mu_G^# \text{ iff } F - G$ is constant.

Conversely, if $\mu^#$ is a Borel #-measure on $\mathbb{R}^#_c$ that is hyperfinite on all #-bounded #-Borel sets, and we define $F(x) = \mu^#((0, x])$ if $x > 0$, $F(x) = 0$ if $x = 0$, $F(x) = -\mu^#((x, 0])$ if $x < 0$,

then $F$ is increasing and right #-continuous function, and $\mu^# = \mu_F^#$.

**Proof:** Each $F$ induces a #-premeasure on $B^#(\mathbb{R}^#_c)$ by Lemma 5.1.1. It is clear that $F$ and $G$ induce the same #-premeasure iff $F - G$ is constant, and that these
distribution in $\mathcal{D}'(*\mathbb{R}_c^{\#n})$.

## § 2.1. Schwartz space $S^{#}(*\mathbb{R}_c^{\#n})$.

**Definition 2.1.** A function $f: X \to *\mathbb{R}_c^\#$ defined on some set $X$ is called finitely bounded (or bounded) if the set of its values is finitely bounded, i.e., $f(X) \subset [a, b]$ where $a, b \in *\mathbb{R}_c^{\#}$, $c, \text{fin}$. In other words, there exists a finite hyperreal number $M \in *\mathbb{R}_c^{\#}$ such that

$$|f(X)| \le M. \qquad (2.1)$$

**Definition 2.2.** A function $f: X \to *\mathbb{R}_c^\#$ defined on some set $X$ is called hyper finitely bounded (or hyper bounded) if the set of its values is hyper finitely bounded, i.e., $f(X) \subset [a, b]$ where $a, b \in *\mathbb{R}_c^\# \setminus *\mathbb{R}_c^{\#}$, $c, \text{fin}$. In other words, there exists a hyperfinite hyperreal number $M \in *\mathbb{R}_c^\# \setminus *\mathbb{R}_c^{\#}$ such that $|f(X)| \le M$.

**Definition 2.3.** For $n \in \mathbb{N}^\#$, an #-integrable function $\phi: *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^\#$ is called #-rapidly decreasing if for all $\alpha \in \mathbb{N}^n$ the product function $x \mapsto x^\alpha \phi(x)$ is a finitely bounded or hyper finitely bounded function.

**Remark 2.1.** If $\phi$ is a #-rapidly decreasing function, then its integral exists

$$\mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} \phi(x) d^{\#n} x < \infty^\# \qquad (2.2)$$

In fact for all $\alpha \in \mathbb{N}^n$ the integral of $x \mapsto x^\alpha \phi(x)$ exists

$$\mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} x^\alpha \phi(x) d^{\#n} x < \infty^\# \qquad (2.3)$$

**Definition 2.4.** The Schwartz space, $S^{#}(*\mathbb{R}_c^{\#n})$, is the space of all #-smooth functions in $C^{\#\infty}(*\mathbb{R}_c^{\#n})$ that are rapidly decreasing at #-infinity along with all partial #-derivatives. Thus

$\phi: *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^\#$ is in the Schwartz space provided that any #-derivative of $\phi$, multiplied with any power of $|x|$, #-converges to 0 as $|x| \to \infty^\#$. These functions form a #-complete TVS with a suitably defined family of seminorms. More precisely, for any multi-indices $\alpha$ and $\beta$ define:

$$p_{\alpha, \beta}(\phi) = \sup_{x \in *\mathbb{R}_c^{\#n}} |x^\alpha \partial^{\#\beta} \phi(x)|. \qquad (2.1)$$

Then $\phi$ is in the Schwartz space $S^{#}(*\mathbb{R}_c^{\#n})$ if all the values satisfy: $p_{\alpha, \beta}(\phi) < \infty^\#$. Thus

$$S^{#}(*\mathbb{R}_c^{\#n}, *\mathbb{R}_c^\#) \triangleq \{\phi \in C^{\#n}(*\mathbb{R}_c^{\#n}, *\mathbb{R}_c^\#) | \forall \alpha, \beta \in \mathbb{N}^{\#n} (p_{\alpha, \beta}(\phi) < \infty^\#)\}.$$

Similarly

$$S^{#}(*\mathbb{R}_c^{\#n}, *\mathbb{C}_c^\#) \triangleq \{\phi \in C^{\#n}(*\mathbb{R}_c^{\#n}, *\mathbb{C}_c^\#) | \forall \alpha, \beta \in \mathbb{N}^{\#n} (p_{\alpha, \beta}(\phi) < \infty^\#)\}$$

The family of seminorms $p_{\alpha, \beta}(\cdot)$ defines a locally convex topology on the Schwartz space $S^{#}(*\mathbb{R}_c^{\#n})$.

For $n = 1$, the seminorms are norms on the Schwartz space $S^{#}(*\mathbb{R}_c^\#)$. One can also use the following family of seminorms to define the topology:

$$|f|_{m,k} = \sup_{|p| \le m} \left( \sup_{x \in *\mathbb{R}_c^{\#n}} \{(1 + |x|)^k |(\partial^{\#\alpha} f)(x)|\} \right), k, m \in \mathbb{N}^\#. \qquad (2.2)$$
#-premeasures are $\sigma^#$-finite (since $\ast\mathbb{R}_c^# = \bigcup_{j=-\infty}^{\infty}(j, j+1]$). The first two assertions

follow now from Lemma 5.2.2. As for the last one, the monotonicity of $\mu^#$ implies the monotonicity of $F$, and the #-continuity of $\mu^#$ from above and from below implies the right #-continuity of $F$ for $x \ge 0$ and $x < 0$. It is evident that $\mu^# = \mu_F^#$ on algebra $A$, and hence $\mu^# = \mu_F^#$ on $B^#(\ast\mathbb{R}_c^#)$ (accordingly to Lemma 5.2.4). Lebesgue – Stieltjes #-measures possess some important and useful regularity properties.

Let us fix a complete Lebesgue – Stieltjes #-measure $\mu^#$ on $\ast\mathbb{R}_c^#$ associated to an increasing, right #-continuous function $F$. We denote by $\Sigma_{\mu^#}$ the Lebesgue algebra correspondent to $\mu^#$. Thus, for any $E \in \Sigma_{\mu^#}$,

$$ \begin{aligned} \mu^#(E) &= \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{\infty} [F(b_j) - F(a_j)] \middle| E \subseteq \bigcup_{j=1}^{\infty} (a_j, b_j] \right\} = \\ &= \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{\infty} \mu_F^#((a_j, b_j]) \middle| E \subseteq \bigcup_{j=1}^{\infty} (a_j, b_j] \right\} \end{aligned} \quad (5.2.4) $$

if infinum in RHS of (5.2.4) exists. Since $B^#(\ast\mathbb{R}_c^#) \subseteq \Sigma_{\mu^#}$, we may replace in the second formula for $\mu^#(E)$ h-intervals by #-open intervals, namely

**Lemma 5.2.3** For any $E \in \Sigma_{\mu^#}$,

$$ \mu^#(E) = \inf \left\{ \mathrm{Ext} - \sum_{j=1}^{\infty} \mu_F^#((a_j, b_j)) \middle| E \subseteq \bigcup_{j=1}^{\infty} (a_j, b_j) \right\}. \quad (5.2.5) $$

**Theorem 5.2.2** If $E \in \Sigma_{\mu^#}$ then

$$ \begin{aligned} E \in \Sigma_{\mu^#} &= \inf\{\mu^#(U) : U \supseteq E \text{ and } U \text{ is #-open}\} = \\ &= \sup\{\mu^#(K) : K \subseteq E \text{ and } K \text{ is #-compact}\}. \end{aligned} \quad (5.2.6) $$

**Proof.** By Lemma 5.2.2, for any $\varepsilon \approx, \varepsilon > 0$, there exist intervals $(a_j, b_j)$ such that

$E \subseteq \bigcup_{j=1}^{\infty}(a_j, b_j)$ and $\mu^#(E) \le \mathrm{Ext} - \sum_{j=1}^{\infty}\mu^#((a_j, b_j)) + \varepsilon$. If $U = \bigcup_{j=1}^{\infty}(a_j, b_j)$ then $U$ is #-open, $E \subseteq U$, and $\mu^#(U) \le \mu^#(E) + \varepsilon$. On the other hand, $\mu^#(U) \ge \mu^#(E)$ whenever $E \subseteq U$ so the first equality is valid.

For the second one, suppose first that $E$ is bounded in $\ast\mathbb{R}_c^#$. If $E$ is #-closed then $E$ is #-compact and the equality is obvious. Otherwise, given $\varepsilon \approx, \varepsilon > 0$, we can choose an #-open $U$, $(\#\bar{E})\setminus E \subseteq U$, such that $\mu^#(U) \le \mu^#((\#\bar{E})\setminus E) + \varepsilon$.

Let $K = (\#\bar{E})\setminus U$. Then $K$ is #-compact, $K \subseteq E$, and

$$ \begin{aligned} \mu^#(K) &= \mu^#(E) - \mu^#(E \cap U) = \mu^#(E) - [\mu^#(U) - \mu^#(U\setminus E)] \ge \\ &\ge \mu^#(E) - \mu^#(U) + \mu^#((\#\bar{E})\setminus E) \ge \mu^#(E) - \varepsilon. \end{aligned} $$

If $E$ is unbounded in $\ast\mathbb{R}_c^#$, let $E_j = E \cap (j, j+1]$. By the preceding argument, for any $\varepsilon \approx, \varepsilon > 0$, there exist a #-compact $K_j \subseteq E_j$ with $\mu^#(K_j) \ge \mu^#(E_j) - \varepsilon 2^{-j}$. Let

$$ H_n = \bigcup_{j=-n}^{j=n} K_j. \text{ Then } H_n \text{ is #-compact, } H_n \subseteq E, \text{ and } \mu^#(H_n) \ge \mu^# \bigcup_{j=-n}^{j=n} (E_j) - \varepsilon. $$

Since $\mu^#(E) = \#\mathrm{lim}_{n \to \infty} \mu^#\left(\bigcup_{j=-n}^{j=n} E_j\right)$, the result follows.
Otherwise, one can define a norm on $S^{#}(*\mathbb{R}_c^{\#n})$ by

$$ \|\phi\|_k = \max_{|\alpha|+|\beta| \le k} \sup_{x \in *\mathbb{R}_c^{\#n}} |x^\alpha \partial^\beta \phi(x)|, k \ge 1. \qquad (2.3) $$

The Schwartz space $S^{#}(*\mathbb{R}_c^{\#n})$ is a Fréchet space (that is, a #-complete metrizable locally convex space). Because the Fourier transform changes $\partial^{\#\alpha}$ into multiplication by $x^\alpha$ and vice versa, this symmetry implies that the Fourier transform of a Schwartz function is also a Schwartz function.

**Definition 2.5.** A sequence $\{f_i\}_{i=1}^{\infty}$ #-converges to 0 in $S^{#}(*\mathbb{R}_c^{\#n})$ if and only if the functions $(1+|x|)^k(\partial^{\#p}f_i)(x)$ #-converge to 0 uniformly in the whole of $*\mathbb{R}_c^{\#n}$, which implies that such a sequence must converge to zero in $C^{\infty}(*\mathbb{R}_c^{\#n})$.

The subset of all #-analytic Schwartz functions is #-dense in $S^{#}(*\mathbb{R}_c^{\#n})$

The Schwartz space is nuclear and the tensor product of two maps induces a canonical surjective TVS-isomorphisms $S^{#}(*\mathbb{R}_c^{\#m}) \hat{\otimes} S^{#}(*\mathbb{R}_c^{\#n}) \to S^{#}(*\mathbb{R}_c^{\#m+n})$, where $\hat{\otimes}$ represents the #-completion of the injective tensor product

## § 2.2.Schwartz space $S_{fin}^{\#}(*\mathbb{R}_c^{\#n})$

**Definition 2.6.** For $n \in \mathbb{N}$, an $*\mathbb{R}_c^{\#n}$-valued and #-integrable function

$\phi : *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^{\#n}$ is called #-rapidly decreasing if for all $\alpha \in \mathbb{N}^n$ the product function $x \mapsto x^\alpha \phi(x)$ is a finitely bounded function.

**Remark 2.2.** If $\phi$ is a #-rapidly decreasing $*\mathbb{R}_c^{\#n}$-valued function, then its integral exists and finite, i.e.,

$$ \mathrm{Ext-} \int_{*\mathbb{R}_c^{\#n}} \phi(x) d^{\#n}x \in *\mathbb{R}_c^{\#n}. \qquad (2.2) $$

In fact for all $\alpha \in \mathbb{N}^n$ the integral of $x \mapsto x^\alpha \phi(x)$ exists and finite, i.e.,

$$ \mathrm{Ext-} \int_{*\mathbb{R}_c^{\#n}} x^\alpha \phi(x) d^{\#n}x \in *\mathbb{R}_c^{\#n}. \qquad (2.3) $$

It follows from () that for all $\alpha \in \mathbb{N}^n$ and for any $R \in *\mathbb{R}_c^{\#n} \setminus *\mathbb{R}_c^{\#n}$

$$ \mathrm{Ext-} \int_{*\mathbb{R}_c^{\#n} \setminus B(R)} x^\alpha \phi(x) d^{\#n}x \approx 0 \qquad (2.3) $$

where $B(R) \triangleq \{x \in *\mathbb{R}_c^{\#n} | |x| \le R\}$

**Definition 2.7.** The Schwartz space, $S_{fin}^{\#}(*\mathbb{R}_c^{\#n})$, is the space of all $*\mathbb{R}_c^{\#n}$-valued #-smooth functions that are rapidly decreasing at #-infinity along with all partial #-derivatives any finite order $1 \le m < \infty$.

Thus

$\phi : *\mathbb{R}_c^{\#n} \to \mathbb{R}_c^{\#n}$ is in the Schwartz space provided that any #-derivative of $\phi$, multiplied with any power of $|x|$, #-converges to 0 as $|x| \to \infty$. These functions form a #-complete TVS with a suitably defined family of seminorms. More precisely, for any multi-indices $\alpha$ and $\beta$ define:

$$ p_{\alpha, \beta}(\phi) = \sup_{x \in *\mathbb{R}_c^{\#n}} |x^\alpha \partial^{\#\beta} \phi(x)|. \qquad (2.1) $$

## § 2.3.Non-Archimedean tempered distributions $S^{\#}(*\mathbb{R}_c^{\#n})$.
**Theorem 5.2.3.** If $E \subseteq *\mathbb{R}_c^\#$, the following are equivalent:

(a) $E \in \Sigma_{\mu^\#}$;

(b) $E = \forall N_1$, where $V$ is a $G_{\delta^\#}$-set and $\mu^\#(N_1) = 0$;

(c) $E = H \cup N_2$, where $H$ is an $F_{\sigma^\#}$-set and $\mu^\#(N_2) = 0$.

**Theorem 5.2.4.** If $E \in \Sigma_{\mu^\#}$ and $\mu^\#(E) < *\infty$ then, for every $\varepsilon \approx, \varepsilon > 0$, there is a set $A$ that is a hyperfinite union of #-open intervals such that $\mu^\#(E\Delta A) < \varepsilon$.

**Lemma 5.2.4** Let $A \subseteq P(X)$ be an algebra, let $\mu_0^\#$ be a $\sigma^\#$-finite #-premeasure on $A$, and let $\Omega$ be the $\sigma^\#$-algebra generated by $A$. Then there exists a unique extension of $\mu_0^\#$ to a #-measure $\mu^\#$ on $\Omega$.

§ 5.3. Product #-measures.

**Definition 5.3.1.** Let $\{(X_\alpha, \mathcal{F}_\alpha, \mu_\alpha^\#)\}_{\alpha \in \Delta}$ be a nonempty family of #-measure spaces. We define the family $\Omega$ of blocks:

$$A(A_{a_1}, A_{a_2}, \dots, A_{a_n}) := \\ = A_{a_1} \times A_{a_2} \times \dots \times A_{a_n} \times \mathrm{Ext} - \prod_{\substack{a \in \Delta, a \neq a_k, 1 \le k \le n}} X_a, \qquad (5.3.1)$$

where $A_{a_k} \in \mathcal{F}_{a_k}$ and define a function

$$\mu_{\Omega}^{\#} : \Omega \to *\mathbb{R}_{c}^{\#} \cup \{ *\infty \} := \\ \mu^{\#}(A_{a_1}) \times \mu^{\#}(A_{a_2}) \times \dots \times \mu^{\#}(A_{a_n}) \times \left[ \mathrm{Ext} - \prod_{\substack{a \in \Delta, a \neq a_k, 1 \le k \le n}} \mu^{\#}(X_a) \right]. \qquad (5.3.2)$$

This function possesses an extension (by #-additivity) on the #-algebra $A$ generated by $\Omega$. It is easily to show that $\mu_{\Omega}^\#$ is a #-premeasure on $A$.

**Definition 5.3.2** The #-measure $\hat{\mu}^\#$ on the $\sigma^\#$-algebra $\Sigma$ generated by $A$ accordingly to **Theorem 2.1.3** is called the product #-measure of $\{\mu_\alpha^\#\}_{\alpha \in \Delta}$, and the triple

$(\prod_{\alpha \in \Delta} X_\alpha, \Sigma, \hat{\mu}^\#)$ is called the product of #-measure spaces $(X_\alpha, \Sigma_\alpha, \mu_\alpha^\#)$.

We denote the $\sigma^\#$-algebra $\Sigma$ by $\bigotimes_{\alpha \in \Delta} \Sigma_\alpha$, and the #-measure $\hat{\mu}^\#$ by $\bigotimes_{\alpha \in \Delta} \mu_\alpha^\#$.

**Definition 5.3.3.** If $E \subseteq X_1 \times X_2$ and $x_1 \in X_1, x_2 \in X_2$, we define

$$E_{x_1} = \{x \in X_2 : (x_1, x) \in E\} \text{ and } E^{x_2} = \{x \in X_1 : (x, x_2) \in E\}.$$

If $f: X_1 \times X_2 \to *\mathbb{R}_c^\#$ is a function, we define $f_{X_1}: X_2 \to *\mathbb{R}_c^\#$ and $f^{X_2}: X_1 \to *\mathbb{R}_c^\#$ by $f_{X_1}(x) = f(x_1, x)$ and $f^{X_2}(x) = f(x, x_2)$.

**Theorem 5.3.1.** (The generalized Fubini's theorem) Let $\mu_1^\#, \mu_2^\#$ be $\sigma^\#$-hyperfinite #-measures on $(X_1, \mathcal{F}_1)$ and $(X_2, \mathcal{F}_2)$,

$$(X_1 \times X_2, \mathcal{F}_1 \otimes \mathcal{F}_2, \mu_1^\# \otimes \mu_2^\#) = (X_1, \mathcal{F}_1, \mu_1^\#) \times (X_2, \mathcal{F}_2, \mu_2^\#), \qquad (5.3.3)$$

and let $f \in L_1^\#(X_1 \times X_2, \mathcal{F}_1 \otimes \mathcal{F}_2, \mu_1^\# \otimes \mu_2^\#)$. Then $f_{X_1} \in L_1^\#(X_2, \mathcal{F}_2, \mu_2^\#)$, $\mu_1^\#$-#-a.e., and $f^{X_2} \in L_1^\#(X_1, \mathcal{F}_1, \mu_1^\#)$, $\mu_2^\#$-#-a.e., and

$$\mathrm{Ext} - \int_{X_1 \times X_2} f d^\#(\mu_1^\# \otimes \mu_2^\#) = \mathrm{Ext} - \int_{X_2} \left[ \mathrm{Ext} - \int_{X_1} f^{X_2} d^\# \mu_1^\# \right] d^\# \mu_2^\# = \\ = \mathrm{Ext} - \int_{X_1} \left[ \mathrm{Ext} - \int_{X_2} f_{X_1} d^\# \mu_2^\# \right] \qquad (5.3.4)$$

**Lemma 5.3.1.** Let $(X_1, \Sigma_1, \mu_1^\#)$ and $(X_2, \Sigma_2, \mu_2^\#)$ be #-measure spaces, $E \in \Sigma_1 \otimes \Sigma_2$,
A non-Archimedean tempered distribution is a distribution $u \in D'(*\mathbb{R}_c^{\#n})$ that does not "grow too fast" – at most polynomial (or tempered) growth – at $\#$-infinity in all directions; in particular it is only defined on $*\mathbb{R}_c^{\#n}$, not on any $\#$-open subset.
Formally, a tempered distribution is a $\#$-continuous linear functional on the Schwartz space $S^#(*\mathbb{R}_c^{\#n})$ of smooth functions with $\#$-rapidly decreasing $\#$-derivatives. The space of tempered distributions (with its natural topology) is denoted $S^{\#'}(*\mathbb{R}_c^{\#n})$.
Every $\#$-compactly supported distribution is a tempered distribution, yielding an inclusion $\mathcal{E}^{\#'}(*\mathbb{R}_c^{\#n}) \hookrightarrow S^{\#'}(*\mathbb{R}_c^{\#n})$.

### § 3. The Fourier transform on $S^#(*\mathbb{R}_c^{\#n})$, $S_{\text{fin}}^#(*\mathbb{R}_c^{\#n})$

We begin by defining the Fourier transform, and the inverse transform, on $S^#(*\mathbb{R}_c^{\#n})$,

$n \in \mathbb{N}^\#$, the Schwartz space of $C^\infty$ functions of rapid decrease.

**Definition 3.1.** Suppose $f \in S^#(*\mathbb{R}_c^{\#n})$. The Fourier transform of $f(x)$ is the function $\hat{f}(\lambda)$

given by

$$ \hat{f}(\lambda) = \frac{1}{(2\pi\#)^{n/2}} \left( \text{Ext-} \int_{*\mathbb{R}_c^{\#n}} f(x)[\text{Ext-exp}(-ix \cdot \lambda)]d^{\#n}x \right), \quad (3.1) $$

where $x \cdot \lambda = \text{Ext-}\sum_{i=1}^n x_i \lambda_i$. The inverse Fourier transform of $f$, denoted by $\check{f}$, is the function

$$ \check{f}(\lambda) = \frac{1}{(2\pi\#)^{n/2}} \left( \text{Ext-} \int_{*\mathbb{R}_c^{\#n}} f(x)[\text{Ext-exp}(ix \cdot \lambda)]d^{\#n}x \right). \quad (3.2) $$

We will usually write $\hat{f} = \mathcal{F}[f]$ and $\check{f} = \mathcal{F}^{-1}[f]$.

Since every function in Schwartz space is in $\mathcal{L}_1^#(*\mathbb{R}_c^{\#n})$, the above integrals (1.1) and (1.2) make sense.

We will use the standard multi-index notation. A multi-index $\alpha = (\alpha_1, \dots, \alpha_n)$, $n \in \mathbb{N}^\#$ is an $n$-tuple of nonnegative integers. The collection of all multi-indices will be denoted by $I_+^\alpha$. The symbols $|\alpha|$, $x^\alpha$, $D^{\#\alpha}$, and $x^2$ are defined as follows:

$$ |\alpha| = \text{Ext-} \sum_{i=1}^{n} \alpha_i $$

$$ \begin{gathered} x^\alpha = \text{Ext-} \prod_{i=1}^{n} x_i^{\alpha_i} \quad \text{or} \quad \text{Ext-}(x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}) \quad \text{or} \quad \text{simbolically} \quad x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n} \\ D^{\#\alpha} f(x) = \text{Ext-} \prod_{i=1}^{n} \frac{\partial^{\#\alpha_i} f(x)}{\partial x^{\alpha_i}} \quad \text{or} \quad \text{simbolically} \quad D^{\#\alpha} f(x) = \frac{\partial^{\#|\alpha|} f(x)}{\partial x^{\alpha_1} \partial x^{\alpha_2} \cdots \partial x^{\alpha_n}} \end{gathered} \quad (3.3) $$

$$ x^2 = \text{Ext-} \sum_{i=1}^{n} x_i^2. $$

**Lemma 1.1.** The maps $f \mapsto \hat{f}$ and $f \mapsto \check{f}$ are $\#$-continuous linear transformations of $S^#(*\mathbb{R}_c^{\#n})$ into $S^#(*\mathbb{R}_c^{\#n})$. Furthermore, if $\alpha$ and $\beta$ are multi-indices, then

$$ ((i\lambda)^\alpha D^{\#\beta} \hat{f})(\lambda) = \overline{D^{\#\alpha} ((-ix)^\beta f(x))}(\lambda). \quad (3.4) $$
and let $f$ be a $\Sigma_1 \otimes \Sigma_2$-measurable function on $X_1 \times X_2$, then:

(a) $E_{x_1} \in \Sigma_2$ for all $x_1 \in X_1$ and $E_{x_2} \in \Sigma_1$ for all $x_2 \in X_2$;

(b) $f_{x_1}$ is $\Sigma_2$-measurable and $f_{x_2}$ is $\Sigma_1$-measurable for all $x_1 \in X_1$ and $x_2 \in X_2$.

**Proof.** Denote by $A$ the collection of all $A \subseteq X_1 \times X_2$ such that $A_{x_1} \in \Sigma_2$ and $A^{x_2} \in \Sigma_1(\forall x_1 \in X_1, x_2 \in X_2)$.

The family $A$ contains all rectangles. Thus, since

$$ \left[ \bigcup_{n=1}^{\infty} A_n \right]_{x_1} = \bigcup_{n=1}^{\infty} [A_n]_{x_1}, [B_n]^{x_2} = [B_n]^{x_2} \qquad (5.3.5) $$

and

$$ [X_1 \times X_2 \setminus A]_{x_1} = X_2 \setminus A_{x_1}, [X_1 \times X_2 \setminus A]^{x_2} = X_1 \setminus A^{x_2}, \qquad (5.3.6) $$

$A$ is a $\sigma^{\#}$-algebra. So $\Sigma_1 \otimes \Sigma_2 \subseteq A$, and (a) is proved. Now the part (b) follows from (a) due to $f_{x_1}^{-1}(A) = [f^{-1}(A)]_{x_1}$ and $[f^{x_2}]^{-1}(A) = [f^{-1}(A)]^{x_2}(\forall A \subseteq \mathbb{R}_c^{\#})$.

**Definition 5.3.4** A family $M \subseteq P(X)$ is called a monotone class if $M$ is closed under $*$-countable increasing unions and $*$-countable decreasing intersections.

**Lemma 5.3.2.** If $A \subseteq P(X)$ is an algebra then the monotone class generated by $A$ coincides with the $\sigma^{\#}$-algebra generated by $A$.

**Lemma 5.3.3.** Let $(X_1, \Sigma_1, \mu_1^{\#})$ and $(X_2, \Sigma_2, \mu_2^{\#})$ be $*$-measure spaces, $E \in \Sigma_1 \otimes \Sigma_2$. Then the functions $x_1 \to \mu_2^{\#}(E_{x_1})$ and $x_2 \to \mu_1^{\#}(E_{x_2})$ are $*$-measurable on $(X_1, \Sigma_1)$ and $(X_2, \Sigma_2)$, and

$$ \mu_1^{\#} \otimes \mu_2^{\#}(E) = \mathrm{Ext} - \int_{X_2} \mu_1^{\#}(E^{x_2}) d^{\#} \mu_2^{\#} = \mathrm{Ext} - \int_{X_1} \mu_2^{\#}(E_{x_1}) d^{\#} \mu_1^{\#}. \qquad (5.3.7) $$

**Proof.** First we consider the case when $\mu_1^{\#}$ and $\mu_2^{\#}$ are finite. Let $A$ be the family of all $E \in \Sigma_1 \otimes \Sigma_2$ for which (5.3.7) is true. If $E = A \times B$, then $\mu_1^{\#}(E^{x_2}) = \mu_1^{\#}(A)\chi_B(x_2)$ and $\mu_2^{\#}(E_{x_1}) = \mu_2^{\#}(B)\chi_A(x_1)$, so $E \in A$. By additivity, it follows that hyperfinite disjoint unions of rectangles are in $A$ so, by Lemma 5.3.2, bit will suffice to show that $A$ is a monotone class. If $(E_n)_{n=1}^{\infty}$ is an increasing hyper infinite sequence in $A$ and $E = \bigcup_{n=1}^{\infty} E_n$, then the function $f_n(x_2) = \mu_1^{\#}((E_n)^{x_2})$

are $*$-measurable and increase pointwise to $f(y) = \mu_1^{\#}(E^{x_2})$. Hence $f$ is $*$-measurable and, by the monotone convergence theorem,

$$ \mathrm{Ext} - \int_{X_2} \mu_1^{\#}(E^{x_2}) d\mu_2^{\#} = \mathrm{\#-lim}_{n \to *\infty} \mathrm{Ext} X_1 \int_{X_2} \mu_1^{\#}((E_n)^{x_2}) d\mu_2^{\#} = \mathrm{\#-lim}_{n \to *\infty} \mu_1^{\#} \times \mu_2^{\#}(E_n) = \mu_1^{\#} \times \mu_2^{\#}(E). \qquad (5.3.8) $$

Likewise $\mu_1^{\#} \times \mu_2^{\#}(E) = \mathrm{Ext} - \int_{X_1} \mu_2^{\#}(E_x) d\mu_1^{\#}$, so $E \in A$. Similarly, if $(E_n)_{n=1}^{\infty}$ is a decreasing hyper infinite sequence in $A$ and $E = \bigcap_{n=1}^{\infty} E_n$, the function $x_2 \to \mu_1^{\#}((E_1)^{x_2})$ is in $L_1^{\#}(\mu_2^{\#})$ because $\mu_1^{\#}((E_1)x_2) \le \mu_1^{\#}(X_1) < *\infty$ and $\mu_2^{\#}(X_2) < *\infty$, so the dominated convergence theorem can be applied to show that $E \in A$. Thus, $A$ is a monotone class, and the proof is complete for the case of finite $*$-measure spaces. Finally, if $\mu_1^{\#}$ and $\mu_2^{\#}$ are $\sigma^{\#}$-finite, we can write $X_1 \times X_2$ as the union of an increasing hyper infinite sequence $(X_1^j \times X_2^j)_{j=1}^{\infty}$ of rectangles of finite or hyperfinite
**Proof** The map $f \mapsto \hat{f}$ is clearly linear. Since

$$((i\lambda)^a D^{\#\beta} \hat{f})(\lambda) = \frac{1}{(2\pi^\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} (\lambda^\alpha)(-ix)^\beta f(x) [\mathrm{Ext} - \exp(-ix\cdot\lambda)] f(x) d^{\#n}x \right) = \frac{1}{(2\pi^\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} \frac{1}{(-i)^\alpha} (D_x^{\#\alpha} [\mathrm{Ext} - \exp(-ix\cdot\lambda)]) (-ix)^\beta f(x) d^{\#n}x \right) = \frac{(-i)^\alpha}{(2\pi^\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}x} \frac{1}{(2\pi^\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} [\mathrm{Ext} - \exp(-ix\cdot\lambda)] D_x^{\#\alpha} ((-ix)^\beta f(x)) d^{\#n}x \right) \right). \quad (3.5)$$

We conclude that

$$\|\hat{f}\|_{\alpha, \beta} = \sup_{\lambda \in *\mathbb{R}_c^{\#n}} |\lambda^\alpha (D^{\#\beta} \hat{f})(\lambda)| \le \frac{1}{(2\pi^\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} |D_x^{\#\alpha} (x^\beta f(x))| d^{\#n}x \right) < \infty^\# \quad (3.6)$$

so $f \mapsto \hat{f}$ takes $S^\#(*\mathbb{R}_c^{\#n})$ into $S^\#(*\mathbb{R}_c^{\#n})$, and we have also proven (1.4). Furthermore, if $k$ is large enough, $\int (1+x^2)^{-k} d^{\#n}x < \infty^\#$ so that

$$\|\hat{f}\|_{\alpha, \beta} \le \frac{1}{(2\pi^\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} \frac{(1+x^2)^{-k}}{(1+x^2)^{-k}} |D_x^{\#\alpha} ((-ix)^\beta f(x))| d^{\#n}x \right) \le \frac{1}{(2\pi^\#)^{n/2}} \left( \mathrm{Ext} - \int_{*\mathbb{R}_c^{\#n}} (1+x^2)^{-k} d^{\#n}x \right) \sup_{x \in *\mathbb{R}_c^{\#n}} \left\{ (1+x^2)^{+k} |D_x^{\#\alpha} ((-ix)^\beta f(x))| \right\}. \quad (3.7)$$

Using generalized Leibnitz's rule we easily conclude that there exist multi-indices $\alpha_j, \beta_j$ and constants $c_j$ so that

$$\|\hat{f}\|_{\alpha, \beta} \le \sum_{j=1}^{M} c_j \|f\|_{\alpha_j, \beta_j}. \quad (1.8)$$

Thus the map $f \mapsto \hat{f}$ is bounded and therefore #-continuous. The proof for $f \mapsto \check{f}$ is the same.

**Theorem 1.1.** (Generalized Fourier inversion theorem) The Fourier transform (3.1) is a linear bicontinuous bijection from $S^\#(*\mathbb{R}_c^{\#n})$ onto $S^\#(*\mathbb{R}_c^{\#n})$. Its inverse map is the inverse Fourier transform, i.e., $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ and $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$.

**Proof.** We will prove that $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$. The proof that $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$ is similar. $\mathcal{F}(\mathcal{F}^{-1}[f]) = f$ implies that $\mathcal{F}[f]$ is surjective and $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ implies that $\mathcal{F}[f]$ is injective. Since $\mathcal{F}[f]$ and $\mathcal{F}^{-1}[f]$ are #-continuous maps of $S^\#(*\mathbb{R}_c^{\#n})$ onto $S^\#(*\mathbb{R}_c^{\#n})$, it is sufficient to prove that $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$ for $f$ contained in the dense set $C_0^\infty(*\mathbb{R}_c^{\#n})$. Let $C_\epsilon, \epsilon \approx 0$ be the cube of volume $(2/\epsilon)^n$ centered at the origin in $*\mathbb{R}_c^{\#n}$. Choose $\epsilon \approx 0$ infinite small enough so that the support of $f$ is contained in $C_\epsilon$. Let $K_\epsilon = \{\mathbf{k} \in *\mathbb{R}_c^{\#n} | \text{ each } k_i/\epsilon\pi^\# \in \mathbf{k} \text{ is an integer}\}$
#-measure. If $E \in \Sigma_1 \otimes \Sigma_2$, the preceding argument applies to $E \cap (X_1^j \times X_2^j)$ for each $j$ gives us

$$ \mu_1^{\#} \times \mu_2^{\#}(E \cap (X_1^j \times X_2^j)) = \mathrm{Ext}- \int_{X_2} \mu_1^{\#}(E^{x_2} \cap X_1^j) \mu_2^{\#} = \mathrm{Ext}- \int_{X_1} \mu_2^{\#}(E_{x_1} \cap X_2^j) \mu_1^{\#}. \quad (5.3.9) $$

The application of the monotone convergence theorem then yields the desired result.

**Lemma 5.3.3.** (Generalized Tonelli's theorem) Let $(X_1, \Sigma_1, \mu_1^{\#})$ and $(X_2, \Sigma_2, \mu_2^{\#})$ be #-measure spaces, and $f: X_1 \times X_2 \to \mathbb{R}_{c+}^\#$ be a $\Sigma_1 \otimes \Sigma_2$-#-measurable function. Then the functions

$$ f_{\mu_2^{\#}}(x_1) = \mathrm{Ext}- \int_{X_2} f_{x_1} d^{\#} \mu_2^{\#} \quad \text{and} \quad f_{\mu_1^{\#}}(x_2) = \mathrm{Ext}- \int_{X_1} f^{x_2} d^{\#} \mu_1^{\#} \quad (5.3.10) $$

are $\Sigma_1$-#-measurable and $\Sigma_2$-#-measurable, respectively, and

$$ \begin{aligned} \mathrm{Ext}- \int_{X_1 \times X_2} f d^{\#} \mu_1^{\#} \otimes \mu_2^{\#} &= \mathrm{Ext}- \int_{X_2} \left[ \mathrm{Ext}- \int_{X_1} f^{x_2} d^{\#} \mu_1^{\#} \right] d^{\#} \mu_2^{\#} = \\ &= \mathrm{Ext}- \int_{X_1} \left[ \mathrm{Ext}- \int_{X_2} f_{x_1} d^{\#} \mu_2^{\#} \right] d^{\#} \mu_1^{\#}. \end{aligned} \quad (5.3.11) $$

**Proof:** In the case when $f$ is a characteristic function, the statement of this lemma follows from Lemma 5.3.3. Therefore, by linearity, it holds also for nonnegative simple functions. If a nonnegative #-measurable function $f$ is arbitrary, there exists a sequence of nonnegative simple functions which increase pointwise to $f$, say $(f_n)_{n=1}^\infty$. By the monotone convergence theorem,

$$ \begin{aligned} \mathrm{Ext}- \int_{X_1} f_{\mu_2^{\#}} d^{\#} \mu_1^{\#} &= \#\lim_{n \to *\infty} \left[ \mathrm{Ext}- \int_{X_1} f_{\mu_2^{\#}}^n d^{\#} \mu_1^{\#} \right] = \\ &= \#\lim_{n \to *\infty} \left[ \mathrm{Ext}- \int_{X_1 \times X_2} f_n d^{\#} \mu_1^{\#} \otimes \mu_2^{\#} \right] \end{aligned} \quad (5.3.12) $$

and

$$ \begin{aligned} \mathrm{Ext}- \int_{X_2} f_{\mu_1^{\#}} d^{\#} \mu_2^{\#} &= \#\lim_{n \to *\infty} \left[ \mathrm{Ext}- \int_{X_2} f_{\mu_1^{\#}}^n d^{\#} \mu_2^{\#} \right] = \\ &= \#\lim_{n \to *\infty} \left[ \mathrm{Ext}- \int_{X_1 \times X_2} f_n d^{\#} \mu_1^{\#} \otimes \mu_2^{\#} \right], \end{aligned} \quad (5.3.13) $$

where

$$ f_{\mu_2^{\#}}^n(x_1) = \mathrm{Ext}- \int_{X_1} [f_n]_{x_1} d^{\#} \mu_2^{\#}, f_{\mu_1^{\#}}^n(x_2) = \mathrm{Ext}- \int_{X_2} [f_n]^{x_2} d^{\#} \mu_1^{\#}. \quad (5.3.14) $$

This proves (5.3.11) and the lemma.
$$f(x) = \mathrm{Ext} - \sum_{\mathbf{k} \in \mathbb{K}_{\varepsilon}} \left( \left( \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)], f \right) \left( \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \right) \right) \quad (3.9)$$

is just the hyper infinite Fourier series of $f$ which #-converges uniformly in $C_\varepsilon$ to $f$ since $f$ is #-continuously #-differentiable. Thus

$$f(x) = \mathrm{Ext} - \sum_{\mathbf{k} \in \mathbb{K}_{\varepsilon}} \frac{\hat{f}(k)[\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)]}{(2\pi\#)^{n/2}} (\varepsilon\pi_{\#})^n. \quad (3.10)$$

Since $\mathbb{R}^{\#n}$ is the disjoint union of the cubes of volume $(\varepsilon\pi_{\#})^n$ centered about the points in $\mathbb{K}_{\varepsilon}$, the right-hand side of (1.10) is just a hyper finite Riemann sum for the integral of the function $\hat{f}(k)[\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)]$. By the lemma 3.1, $\hat{f}(k)[\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \in S^{\#}(\mathbb{R}^{\#n})$, so the hyperfinite Riemann sums (1.10) #-converge to the integral. Thus $\mathcal{F}^{-1}(\mathcal{F}[f]) = f$.

**Corollary 3.1.** Suppose $f \in S^{\#}(\mathbb{R}^{\#n})$. Then

$$\mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |f(x)|^2 d^{\#n}x = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |f(k)|^2 d^{\#n}k. \quad (3.11)$$

**Proof.** This is really a corollary of the proof rather than the statement of Theorem 1.1. If $f$ has #-compact support, then for $\varepsilon \approx 0$ small enough,

$$f(x) = \mathrm{Ext} - \sum_{\mathbf{k} \in \mathbb{K}_{\varepsilon}} \left( \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)], f \right) \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \quad (3.12)$$

Since $\left\{ \left( \frac{1}{2} \varepsilon \right)^{n/2} [\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)] \right\}_{\mathbf{k} \in \mathbb{K}_{\varepsilon}}$ is an orthonormal basis for $\mathcal{L}_2^{\#}(C_\varepsilon)$,

$$\mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |f(x)|^2 d^{\#n}x = \mathrm{Ext} - \int_{C_\varepsilon} |f(x)|^2 d^{\#n}x = \sum_{\mathbf{k} \in \mathbb{K}_\varepsilon} \left| \left(\frac{1}{2}\varepsilon\right)^{n/2} ([\mathrm{Ext} - \exp(i\mathbf{k} \cdot x)], f(x)) \right|^2 = \\ \sum_{\mathbf{k} \in \mathbb{K}_\varepsilon} |\hat{f}(k)|^2 (\varepsilon\pi_\#)^n \xrightarrow{\varepsilon \to 0} \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} |f(k)|^2 d^{\#n}k. \quad (3.13)$$

This proves the corollary for $f \in C_0^{\#}(\mathbb{R}_c^{\#n})$. Since $f \mapsto \hat{f}$ and $\|\cdot\|_2$ are #-continuous on $S^{\#}(\mathbb{R}_c^{\#n})$ and $C_0^{\#}(\mathbb{R}_c^{\#n})$ is #-dense, the result holds for all of $S^{\#}(\mathbb{R}_c^{\#n})$.

**Definition 3.2.** Let $T \in S^{\#}(\mathbb{R}_c^{\#n})$ the Fourier transform of $T$, denoted by $\hat{T}$ or $\mathcal{F}[T]$, is the tempered distribution defined by $\hat{T}(\varphi) = T(\hat{\varphi})$.

Suppose that $h, \varphi \in S^{\#}(\mathbb{R}_c^{\#n})$, then by the polarization identity and the corollary to Theorem 1.1 we have $(h, \varphi) = (\hat{h}, \hat{\varphi})$. Substituting $\overline{\mathcal{F}[g]} = \mathcal{F}^{-1}[g]$ for $h$, we obtain

$$T_{\hat{g}}(\varphi) = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} \hat{g}(x)\varphi(x)d^{\#n}x = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} g(x)\hat{\varphi}(x)d^{\#n}x = T_g(\hat{\varphi}) = \hat{T}_g(\varphi).$$

where $T_{\hat{g}}$ and $T_g$ are the distributions corresponding to the functions $\hat{g}$ and $g$ respectively. This shows that the Fourier transform on $S^{\#}(\mathbb{R}_c^{\#n})$ extends the transform we previously defined on $S^{\#}(\mathbb{R}_c^{\#n})$.

**Theorem 3.2.** The Fourier transform is a one-to-one linear bijection from $S^{\#}(\mathbb{R}_c^{\#n})$ to $S^{\#}(\mathbb{R}_c^{\#n})$ which is the unique weakly #-continuous extension of the Fourier transform on $S^{\#}(\mathbb{R}_c^{\#n})$.

**Proof.** If hyper infinite sequence $\{\varphi_n\}_{n \in \mathbb{N}^\#}$ #-convergence to $\varphi \in S^\#$, then by Theorem 1.1, hyper infinite sequence $\{\hat{\varphi}_n\}_{n \in \mathbb{N}^\#}$ #-convergence to $\hat{\varphi} \in S^\#$, so
**Proof** of Theorem 5.3.1. Since an $^*\mathbb{R}_c^\#$-valued function $f$ is Lebesgue #-integrable iff its positive $f^+$ and negative $f^-$ parts are #-integrable, it is sufficient to prove the theorem only for nonnegative function $f \in L_1^\#(X_1 \times X_2, \Sigma, \mu_1^\# \otimes \mu_2^\#)$.
But this was exactly done in Lemma 5.3.3.

## § 5.4. Lebesgue #-measure and integral in $^*\mathbb{R}_c^{\#n}$.

In this section, we study $^*\mathbb{R}_c^{\#n}$, $n \in ^*\mathbb{N}$ and functions from $^*\mathbb{R}_c^{\#n}$ to $^*\mathbb{R}_c^\#$ from the point of view of the Lebesgue #-measure and Lebesgue integration. All results presented below possess obvious $^*\mathbb{C}_c^{\#n}$-valued analogs. Then we define and study generalized Cantor sets which are interesting from the point of view of the set topology and the #-measure theory. Cantor sets are #-closed #-Borel nowhere #-dense subsets of the interval $[0, 1]$ or, more generally, of a Hausdorff #-space.

**Definition 5.4.1.** The Lebesgue #-measure $\mu^{\#n}$ on $^*\mathbb{R}_c^{\#n}$ is the #-completion of the product of the Lebesgue #-measure on $^*\mathbb{R}_c^{\#n}$ according to Definition 5.3.1. The domain $\Sigma^n$ of $\mu^\#$ (of course, $B^\#(*^\mathbb{R}_c^{\#n}) \subseteq \Sigma^n$) is the class of Lebesgue #-measurable sets in $^*\mathbb{R}_c^{\#n}$. We write $d^\#x^n$ for $d^\#\mu^{\#n}$ and

$$ \mathrm{Ext}-\int f(x)d^\#x^n = \mathrm{Ext}-\int fd^\#\mu^{\#n}. $$

We extend some of the results of previous section to the $n$-dimensional case with $n \in ^*\mathbb{N}$. If $E = \mathrm{Ext}-\prod_{j=1}^n E_j$ is a block in $^*\mathbb{R}_c^{\#n}$, we call sets $E_j \subseteq ^*\mathbb{R}_c^{\#n}$ the sides of the block $E$.

**Theorem 5.4.1.** Let $E \in \Sigma^n$. Then

(a) $\mu^{\#n}(E) = \inf\{\mu^{\#n}(U) : E \subseteq U, U \text{ #-open}\} = \sup\{\mu^{\#n}(K) : K \subseteq E, K \text{ #-compact}\}$;

(b) $E = A_1 \cup N_1 = A_2 \cap W_2$, where $A_1$ is an $F_{\sigma}^{\#n}$ set, $A_2$ is a $G_{\delta}^{\#n}$ set, and
$\mu^{\#n}(N_1) = \mu^{\#n}(N_2) = 0$;

(c) If $\mu^{\#n}(E) < *\infty$ then, for any $\varepsilon \approx 0, \varepsilon > 0$, there is a hyperfinite family $\{R_j\}_{j=1}^N$ of disjoint blocks, whose sides are intervals such that $\mu^{\#n}(E\Delta \cup_{j=1}^N R_j) < \varepsilon$.

**Proof:** By the definition of product #-measures, if $E \in \Sigma^n$ and $\varepsilon \approx 0, \varepsilon > 0$, there is a $*-\text{countable family } \{T_j\}_{j=1}^{*\infty}$ of blocks such that $E \subseteq \cup_{j=1}^{*\infty} T_j$ and

$$ \mathrm{Ext}-\sum_{j=1}^{*\infty} \mu^{\#n}(T_j) \leq \mu^{\#n}(E) + \varepsilon. $$

For each $j$, by applying Theorem 5.2.3 to the sides of $R_j$, we can find blocks $U_j \supseteq F_j$ whose sides are #-open sets such that $\mu^{\#n}(U_j) < \mu(T_j) + \varepsilon 2^{-j}$.
If $U = \cup_{j=1}^{*\infty} U_j$ then $U$ is #-open and

$$ \mu^{\#n}(U) \leq \mathrm{Ext}-\sum_{j=1}^{*\infty} \mu^{\#n}(U_j) \leq \mu^{\#n}(E) + 2\varepsilon. $$

This proves the first equation in part (a). The second equation and part (b) follow as in the proofs of Theorems 2.1.6 and 2.1.7.

Next, if $\mu^{\#n}(E) < *\infty$ then $\mu^{\#n}(U_j) < *\infty$ for all $j$. Since the sides of $U_j$ are $*-\text{countable unions of #-open intervals}$, by taking suitable hyperfinite subunions, we obtain blocks $V_j \subseteq U_j$ whose sides are hyperfinite unions of intervals such that $\mu^{\#n}(V_j) \geq \mu^{\#n}(U_j) - \varepsilon 2^{-j}$. If $N \in ^*\mathbb{N}$ is sufficiently hyperfinite large, we have

$$ \mu^{\#n} \left( E \setminus \bigcup_{j=1}^{N} V_j \right) \leq \mu^{\#n} \left( \bigcup_{j=1}^{N} U_j \setminus V_j \right) + \mu^{\#n} \left( \bigcup_{j=N+1}^{*\infty} U_j \right) < 2\varepsilon $$

and
$T(\hat{\phi}_n) \to_{\#} T(\hat{\phi})$ for each $T \in S^{\#}$. Thus $\#\lim_{n \to \infty} T(\hat{\phi}_n) = T(\hat{\phi})$, which shows that $T$ is a $\#$-continuous linear functional on $S^{\#}$. Furthermore, if $T_n \xrightarrow{w} T$, then $\hat{T}_n \xrightarrow{w} \hat{T}$ because $T(\hat{\phi}_n) \to_{\#} T(\hat{\phi})$ implies $\hat{T}(\phi_n) \to_{\#} \hat{T}(\phi)$. Thus $T \mapsto \hat{T}$ is weakly $\#$-continuous.

**Definition 3.3.** Suppose that $f,g \in S^{\#}(\mathbb{R}_c^{\#n})$. Then the convolution of $f$ and $g$, denoted by $f * g$, is the function

$$ (f * g)(y) = \mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} f(y-x)g(x)d^{\#n}x. \qquad (3.14) $$

Convolutions frequently occur when one uses the Fourier transform because the Fourier transform takes products into convolutions.

**Theorem 3.3.** (a) For each $f \in S^{\#}(\mathbb{R}_c^{\#n})$, $g \mapsto f * g$ is a $\#$-continuous map of $S^{\#}(\mathbb{R}_c^{\#n})$ into $S^{\#}(\mathbb{R}_c^{\#n})$.

(b) $\hat{fg} = (2\pi_{\#})^{-n/2}\hat{f} * \hat{g}$ and $\hat{f} * \hat{g} = (2\pi_{\#})^{n/2}\hat{f}\hat{g}$.

(c) For $f,g,h \in S^{\#}(\mathbb{R}_c^{\#n})$, $f * g = g * f$ and $f * (g * h) = (f * g) * h$.

**Definition 3.4.** Suppose that $f \in S^{\#}(\mathbb{R}_c^{\#n})$, $T \in S^{\#}(\mathbb{R}_c^{\#n})$ and let $\tilde{f}(x)$ denote the function, $f(-x)$. Then, the convolution of $T$ and $f$ denoted $T * f$ is the distribution in, $S^{\#}(\mathbb{R}_c^{\#n})$ given by $(T * f)(\varphi) = T(\tilde{f} * \varphi)$ for all $\varphi \in S^{\#}(\mathbb{R}_c^{\#n})$.

The fact that $g \to \tilde{f} * g$ is a $\#$-continuous transformation guarantees that $T * f \in S^{\#}(\mathbb{R}_c^{\#n})$.

**Abbreviation 3.1.** Let $f_y$ denote the function $f_y(x) = f(x - y)$ and $\tilde{f}_y$ the function $f(y - x)$. When $f$ is given by a longe expression $(\cdots)$, we will sometimes write $(\cdots)^{\sim}$ rather than $(\cdot \cdot \cdot)$.

**Theorem 3.4.** For each $f \in S^{\#}(\mathbb{R}_c^{\#n})$ the map $T \to T * f$ is a weakly $\#$-continuous map of $S^{\#}(\mathbb{R}_c^{\#n})$ into $S^{\#}(\mathbb{R}_c^{\#n})$ which extends the convolution on $S^{\#}(\mathbb{R}_c^{\#n})$.

Furthermore,
(a) $T * f$ is a polynomially bounded $C^{\infty}$ function. In fact, $(T * f)(y) = T(\tilde{f}_y)$ and $D^{\#\beta}(T * f) = (D^{\#\beta}T) * f = T * D^{\#\beta}f$;

(b) $(T * f) * g = T * (f * g)$;

(c) $\hat{T} * \hat{f} = (2\pi_{\#})^{n/2}\hat{f}\hat{T}$.

**Theorem 3.5.** Let $T \in S^{\#}(\mathbb{R}_c^{\#n})$ and $f \in S^{\#}(\mathbb{R}_c^{\#n})$. Then $\hat{f}T \in O_M^n$ and $\hat{f}T(k) = (2\pi_{\#})^{n/2}T(f[\mathrm{Ext}\text{-exp}(-ik \cdot x)])$. In particular, if $T$ has $\#$-compact support and $\psi \in S^{\#}(\mathbb{R}_c^{\#n})$ is identically one on a $\#$-neighborhood of the support of $T$, then

$$ \hat{T}(k) = (2\pi_{\#})^{n/2} T(\psi[\mathrm{Ext}\text{-exp}(-ik \cdot x)]). \qquad (3.15) $$

**Proof** By Theorem 3.4.c and the Fourier inversion formula we have

$$ \hat{f}T = (2\pi_{\#})^{n/2} \hat{f} * \hat{T}. \text{ Thus } \hat{f}T \in O_M^n \text{ and } \hat{f}T(k) = (2\pi_{\#})^{n/2} \hat{T}(\hat{f}_k) = (2\pi_{\#})^{n/2} T(f[\mathrm{Ext}\text{-exp}(-ik \cdot x)]) $$

**Remark 3.1.** We remark that one can also define the convolution of a distribution $T \in D^{\#}(\mathbb{R}_c^{\#n})$ with an $f \in D^{\#}(\mathbb{R}_c^{\#n})$ by $(T * f)(y) = T(\tilde{f}_y)$.

**Definition 3.5.** Let $j(x)$ be a positive $C^{\infty}$ function whose support lies in the sphere of radius one about the origin in $\mathbb{R}_c^{\#n}$ and which satisfies $\mathrm{Ext} - \int_{\mathbb{R}_c^{\#n}} j(x)d^{\#n}x = 1$. The function $j_{\varepsilon}(x) = \varepsilon^{-n}j(x/\varepsilon)$, $\varepsilon \approx 0$ is called an approximate identity.

**Proposition 3.1.** Suppose $T \in S^{\#}(\mathbb{R}_c^{\#n})$ and let $j_{\varepsilon}(x)$ be an approximate identity. Then
$$ \mu^{\#n} \left( \bigcup_{j=1}^{N} V_j \setminus E \right) \leq \mu^{\#n} \left( \bigcup_{j=1}^{\infty} U_j \setminus E \right) < \varepsilon, $$

so $\mu^{\#n}(E\Delta \cup_{j=1}^N V_j) < 3\varepsilon$. Since $\cup_{j=1}^N V_j$ can be expressed as a hyperfinite disjoint union of rectangles whose sides are intervals, we have proved (c).

## § 5.5. Lebesgue #-integrable functions on $\mathbb{R}_c^{\#n}$

Let $\mu^{\#n}$ be the Lebesgue #-measure in $\mathbb{R}_c^{\#n}$. The set $M(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ of all $*\mathbb{R}_c^{\#n}$-valued $\mu^{\#n}$-measurable functions on $*\mathbb{R}_c^{\#n}$ is a vector space (addition and scalar multiplication are pointwise). By $L_1^*(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ we denote its subspace of all Lebesgue #-integrable functions (with finite in $*\mathbb{R}_c^{\#n}$-integral). Now write $f \approx g$ for $f$ and $g$ in $M(*\mathbb{R}_c^{\#n}, \mu^{\#n})$, whenever $f$ and $g$ differ only on a $\mu^{\#n}$-null set (a set of $\mu^{\#n}$-measure zero). It is easily seen that $\approx$ is an equivalencerelation. Let $L_0 = L_0(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ be the set of equivalence classes of functions in $M(*\mathbb{R}_c^{\#n}, \mu^{\#n})$. We denote the equivalence classes of $f, g, \dots$ by $[f], [g], \dots$. The set $L_0$ becomes a vector space over field $*\mathbb{R}_c^\#$ by defining $[f] + [g] = [f + g]$ and $\alpha[f] = [\alpha f]$ for a real $\alpha \in *\mathbb{R}_c^\#$. Observe that these definitions do not depend on the choice of $f$ and $g$ in their equivalence classes. The same is true for the partial order in $L_0$, if we define $[f] \le [g]$ to mean $f(x) \le g(x)$ for all $x \in *\mathbb{R}_c^{\#n}$ except a null set. In practice, the elements of $L_0 = L_0(*\mathbb{R}_c^{\#n}, \mu^{\#n})$ are usually denoted by $f, g, \dots$ and treated as if they were functions instead of equivalence classes of functions.

**Definition 5.5.1.**

**Theorem 5.5.1.** If $f \in L_1^*(\mu^{\#n})$ and $\varepsilon \approx 0, \varepsilon > 0$, there is a simple function $\varphi = \text{Ext}-\sum_{j=1}^N \alpha_j \chi_{R_j}$, where each $R_j$ is a product of intervals such that $\text{Ext}-\int |f-\varphi| d^\# \mu^{\#n} < \varepsilon$, and there is a #-continuous function $g$ vanishing outside of a bounded in $*\mathbb{R}_c^{\#n}$ set such that $\text{Ext}-\int |f-g| d^\# \mu^{\#n} < \varepsilon < \varepsilon$.

**Proof.** By the definition of Lebesgue #-integrable functions, we can approximate $f$ by simple functions in $L_1^#$-norm. Then use Theorem 5.4.1 to approximate a simple function by a function $\varphi$ of the desired form. Finally, use the generalized Urysohn Lemma to approximate such $\varphi$ by a #-continuous function.

**Theorem 5.5.2.** The Lebesgue #-measure on $*\mathbb{R}_c^{\#n}$ is translation-invariant. Namely, let $a \in *\mathbb{R}_c^{\#n}$. Define the shift $\tau_a : *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^{\#n}$ by $\tau_a(x) = x + a$.

(a) If $E \in \mathcal{L}^{\#n}$ then $\tau_a(E) \in \mathcal{L}^{\#n}$ and $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$;

(b) If $f : *\mathbb{R}_c^{\#n} \to *\mathbb{R}_c^\#$ is Lebesgue #-measurable then so is $f \circ \tau_a$. Moreover, if either $f \ge 0$ or $f \in L_1^*(\mu^{\#n})$ then

$$ \text{Ext}-\int (f \circ \tau_a) d^\# \mu^{\#n} = \text{Ext}-\int f d^\# \mu^{\#n}. \qquad (5.5.1) $$

**Proof.** Since $\tau_a$ and its inverse $\tau_{-a}$ are #-continuous, they preserve the class of #-Borel sets. The formula $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$ follows easily from the trivial one dimensional variant of this result if $E$ is a block. For a general #-Borel set $E$, the formula $\mu^{\#n}(\tau_a(E)) = \mu^{\#n}(E)$ follows from the previous step, since $\mu^{\#n}$ is determined by its action on blocks. Assertion (a) now follows immediately.

If $f$ is Lebesgue #-measurable and $B$ is a #-Borel set in $*\mathbb{R}_c^\#$, we have $f^{-1}(B) = E \cup N$, where $E$ is #-Borel and $\mu^\#(N) = 0$. But $\tau_a^{-1}(E)$ is Borel and $\mu^\#(\tau_a^{-1}(N)) = 0$, so $(f \circ \tau_a)^{-1}(B) \in \Sigma^n$ and $f$ is Lebesgue #-measurable. The equality (5.5.1) reduces to the equality $\mu^{\#n}(\tau_{-a}(E)) = \mu^{\#n}(E)$ when $f = \chi_E$. It is true for simple functions by linearity,
$T * j_\varepsilon(x) \to_\# T$ weakly as $\varepsilon \to_\# 0$.

**Proof.** If $\varphi \in S^#(*\mathbb{R}_c^{\#n})$, then $(T * j_\varepsilon)(\varphi) = T(\tilde{j}_\varepsilon * \varphi)$, so it is sufficient to show that $\tilde{j}_\varepsilon * \varphi \to_\# \varphi$ in $S^#(*\mathbb{R}_c^{\#n})$. To do this it is sufficient to show that $(2\pi_\#)^{n/2} \hat{j}_\varepsilon \hat{\varphi} \to_\# \hat{\varphi}$ in $S^#(*\mathbb{R}_c^{\#n})$. Since $\hat{j}_\varepsilon(\lambda) = j(\varepsilon\lambda)$ and $j(0) = (2\pi_\#)^{n/2}$, it follows that $(2\pi_\#)^{n/2} \hat{j}_\varepsilon(x)$
\#-converges to 1 uniformly on \#-compact sets and is uniformly bounded. Similarly, $D^{\#\alpha} \hat{j}_\varepsilon$ \#-converges uniformly to zero. We conclude that $(2\pi_\#)^{n/2} \hat{j}_\varepsilon \hat{\varphi} \to_\# \hat{\varphi}$.

**Theorem 3.6 (The generalized Plancherel theorem)** The Fourier transform extends uniquely to a unitary map of $\mathcal{L}_2^#(*\mathbb{R}_c^{\#n})$ onto $\mathcal{L}_2^#(*\mathbb{R}_c^{\#n})$. The inverse transform extends uniquely to its adjoint.

**Proof** The corollary to Theorem 3.1 states that if $f \in S^#(*\mathbb{R}_c^{\#n})$, then $\|f\|_2 = \|\hat{f}\|_2$. Since $\mathcal{F}[S^#] = S^#$ is a surjective isometry on $\mathcal{L}_2^#(*\mathbb{R}_c^{\#n})$.

**Theorem 3.7 (The generalized Riemann-Lebesgue lemma)** The Fourier transform extends uniquely to a bounded map from $\mathcal{L}_1^#(*\mathbb{R}_c^{\#n})$ into $C^\infty#(*\mathbb{R}_c^{\#n})$, the \#-continuous functions vanishing at $\infty$.

**Proof** For $f \in S^#(*\mathbb{R}_c^{\#n})$, we know that $\hat{f} \in S^#(*\mathbb{R}_c^{\#n})$ and thus $\hat{f} \in C^\infty#(*\mathbb{R}_c^{\#n})$. The estimate is trivial. The Fourier transform is thus a bounded linear map from a \#-dense set of $\mathcal{L}_1^#(*\mathbb{R}_c^{\#n})$ into $C^\infty#(*\mathbb{R}_c^{\#n})$. By the generalized B.L.T. theorem, extends uniquely to a bounded linear transformation of $C^\infty#(*\mathbb{R}_c^{\#n})$ into $C^\infty#(*\mathbb{R}_c^{\#n})$.

**Remark 3.2.** We remark that the Fourier transform takes $\mathcal{L}_1^#(*\mathbb{R}_c^{\#n})$ into, but not onto $C^\infty#(*\mathbb{R}_c^{\#n})$.

A simple argument with test functions shows that the extended transform on $\mathcal{L}_1^#(*\mathbb{R}_c^{\#n})$ and $\mathcal{L}_2^#(*\mathbb{R}_c^{\#n})$ is the restriction of the transform on $S^#(*\mathbb{R}_c^{\#n})$, but it is useful to have an explicit integral representation. For $f \in \mathcal{L}_1^#(*\mathbb{R}_c^{\#n})$, this is easy since we can find $f_m \in S^#(*\mathbb{R}_c^{\#n})$ so that $\#\text{lim}_{m \to \infty} \|f - f_m\|_1 = 0$. Then, for each $\lambda$,

$$ 
\begin{aligned}
f(\lambda) = \#\text{lim}_{m \to \infty} \hat{(f_m(\lambda))} = & \#\text{lim}_{m \to \infty} \left\{ \frac{1}{(2\pi_\#)^{n/2}} \left( \text{Ext} - \int_{*\mathbb{R}_c^{\#n}} [\text{Ext}\text{-exp}(-ik \cdot x)]f_m(x)d^\#x \right) \right\} = \\
& \frac{1}{(2\pi_\#)^{n/2}} \left( \text{Ext} - \int_{*\mathbb{R}_c^{\#n}} [\text{Ext}\text{-exp}(-ik \cdot x)]f(x)d^\#x \right).
\end{aligned} 
\quad (3.16) 
$$

So, the Fourier transform of a function in $\mathcal{L}_1^#(*\mathbb{R}_c^{\#n})$ is given by the usual formula. Next, suppose $f \in \mathcal{L}_2^#(*\mathbb{R}_c^{\#n})$ and let

$$ \chi_R(x) = \begin{cases} 1 & \text{if } |x| \le R \\ 0 & \text{if } |x| > R \end{cases} \quad (3.17) $$

Then $\chi_R f \in \mathcal{L}_1^#(*\mathbb{R}_c^{\#n})$ and $\#\text{lim}_{R \to \infty} \chi_R f = f$ in $\mathcal{L}_2^\#$, so by the generalized Plancherel theorem $\#\text{lim}_{R \to \infty} \widehat{\chi_R f} = \hat{f}$ in $\mathcal{L}_2^\#$. Thus

$$ f(\lambda) = \#\text{lim}_{R \to \infty} \frac{1}{(2\pi_\#)^{n/2}} \left( \text{Ext} - \int_{|x| \le R} [\text{Ext}\text{-exp}(-ik \cdot x)]f(x)d^\#x \right) \quad (3.18) $$

where by $\#\text{lim}_{R \to \infty}$ we mean the \#-limit in the $\mathcal{L}_2^\#$-norm. Sometimes we will dispense
and hence for nonnegative #-measurable functions by the definition of #-integral.
Taking positive and negative parts of real and imaginary parts, we obtain the result
for $f \in L_1^\#(\mu^{\#n})$.

**Theorem 5.5.3.** (Generalized Lusin's theorem) If $f$ is a Lebesgue #-measurable function on $\mathbb{R}^{\#n}$ and $\varepsilon \approx 0, \varepsilon > 0$ then there exist a #-measurable set $A \subseteq \mathbb{R}^{\#n}$ such that $\mu^{\#n}(\mathbb{R}^{\#n} \setminus A) \le \varepsilon$ and the restriction of $f$ onto $A$ is #-continuous.

## §6. $\tilde{\mathbb{R}}_c^\#$-Valued measurable functions

**Definition 6.1** ($\sigma^\#$-algebra). Let $X$ be any set. We denote by $2^X = P(X) = \{A : A \subset X\}$ the set of all subsets of $X$. A family $\mathcal{F} \subset 2^X$ is called a $\sigma^\#$-algebra (on $X$) if:

(i) $\emptyset \in \mathcal{F}$;

(ii) $\mathcal{F}$ is closed under complements, i.e. $A \in \mathcal{F}$ implies $X \setminus A \in \mathcal{F}$;

(iii) $\mathcal{F}$ is closed under hypercountable unions, i.e. if $(A_n)_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence in $\mathcal{F}$ then $\bigcup_{n \in \mathbb{N}^\#} A_n \in \mathcal{F}$.

**Proposition 6.1.** If $\mathcal{F}$ is a $\sigma^\#$-algebra on $X$ then:

1. $\mathcal{F}$ is closed under hypercountable intersections, i.e. if $(A_n)_{n \in \mathbb{N}^\#}$ is a hyper infinite sequence in $\mathcal{F}$ then $\bigcap_{n \in \mathbb{N}^\#} A_n \in \mathcal{F}$.

2. $X \in \mathcal{F}$.

3. $\mathcal{F}$ is closed under hyperfinite unions and hyperfinite intersections.

4. $\mathcal{F}$ is closed under set differences.

5. $\mathcal{F}$ is closed under symmetric differences.

**Proposition 6.2.** Suppose $\mathcal{F} \subset 2^X$ is a family of subsets satisfying the following:

1. $\emptyset \in \mathcal{F}$;

2. $\mathcal{F}$ is closed under complements;

3. $\mathcal{F}$ is closed under hyperinfinite intersections.

Then $\mathcal{F}$ is a $\sigma^\#$-algebra.

**Proposition 6.3.** If $(\mathcal{F}_\alpha)_{\alpha \in I}$ is a collection of $\sigma^\#$-algebras on $X$, then $\bigcap_\alpha \mathcal{F}_\alpha$ is also a $\sigma^\#$-algebra on $X$.

**Proposition 6.4.** ($\sigma^\#$-algebra generated by subsets). Let $K$ be a collection of subsets of $X$. There exists a $\sigma^\#$-algebra, denoted $\sigma^\#(K)$ such that $K \subset \sigma^\#(K)$ and for every other $\sigma^\#$-algebra $\mathcal{F}$ such that $K \subset \mathcal{F}$ we have that $\sigma^\#(K) \subset \mathcal{F}$
We call $\sigma^\#(K)$ the $\sigma^\#$-algebra generated by $K$.

**Proof.** Define $\sigma^\#(K) \triangleq \bigcap \{\mathcal{F} | \mathcal{F}$ is a $\sigma^\#$-algebra on $X, K \subset \mathcal{F}\}$. This is a $\sigma^\#$-algebra with the required properties.

**Proposition 6.5.** If $K \subset \mathcal{L}$ then $\sigma^\#(K) \subset \sigma^\#(\mathcal{L})$. Also, if $K \subset \mathcal{F}$ and $\mathcal{F}$ is a $\sigma^\#$-algebra, then $\sigma^\#(K) \subset \mathcal{F}$.

**Definition 6.2.** (Borel $\sigma^\#$-algebra). Given a topological space $X$, the Borel $\sigma^\#$-algebra is the $\sigma^\#$-algebra generated by the open sets. It is denoted $B^\#(X)$.
Specifically in the case $X = \mathbb{R}^{\#d}$, $d \in \mathbb{N}^\#$ we have that

$\tilde{B}_d^\# \triangleq B^\#(\mathbb{R}_c^{\#d}) = \sigma^\#(U|U \text{ is an } \#\text{-open set})$.

A Borel-#-measurable set, i.e. a set in $B^\#(X)$, is called a #-Borel set.

**Measurable functions.** Let $f$ be a $\mathbb{R}_c^\#$-valued function defined on a set $X$. We suppose that some $\sigma^\#$-algebra $\Omega \subseteq P(X)$ is fixed.

**Definition 6.3.** We say that $f$ is #-measurable, if $f^{-1}([a, b]) \in \Omega$ for any hyperreals
with $|x| \le R$ and just write

$$f(\lambda) = \#\lim_{R \to \infty} \frac{1}{(2\pi\lambda)^{n/2}} \left( \mathrm{Ext} - \int [\mathrm{Ext} - \exp(-ik \cdot x)]f(x)d^\#x \right) \quad (3.19)$$

for functions $f \in \mathcal{L}_2^{\#}(\mathbb{R}_c^{\#n})$.

We have proven above that $\mathcal{F} : \mathcal{L}_2^{\#}(\mathbb{R}_c^{\#n}) \to \mathcal{L}_2^{\#}(\mathbb{R}_c^{\#n})$ and $\mathcal{F} : \mathcal{L}_1^{\#}(\mathbb{R}_c^{\#n}) \to \mathcal{L}_{\infty}^{\#}(\mathbb{R}_c^{\#n})$ and in both cases is a bounded operator.

**Theorem 3.8 (Generalized Hausdorff-Young inequality)** Suppose $1 \le q \le 2$, and $p^{-1} + q^{-1} = 1$. Then the Fourier transform is a bounded map of $\mathcal{L}_p^{\#}(\mathbb{R}_c^{\#n})$ to $\mathcal{L}_q^{\#}(\mathbb{R}_c^{\#n})$ and its norm is less than or equal to $(2\pi\lambda)^{n(1/2-1/q)}$.

# Chapter V.Non-Archiedean Hilbert Spaces over field $\widehat{\mathbb{C}}_c^\#$.

## § 1. Non-Archiedean Hilbert Spaces over field $\widehat{\mathbb{C}}_c^\#$
Basics.

**Definition 1.1. (i)** Let $H$ be external hyper infinite dimensional vector space over field $\widehat{\mathbb{C}}_c^\# = \widehat{\mathbb{R}}_c + i\widehat{\mathbb{R}}_c$. An inner #-product(or non-Archiedean inner product) on $H$

is a $\widehat{\mathbb{C}}_c^\#$-valued function, $\langle \cdot, \cdot \rangle_\# : H \times H \to \mathbb{C}_c$, such that

(1) $\langle ax + by, z \rangle_\# = \alpha \langle x, z \rangle + \beta \langle y, z \rangle_\#$, i.e. $x \to \langle x, z \rangle_\#$ is linear.

(2) $\overline{\langle x, y \rangle_\#} = \langle y, x \rangle_\#$.

(3) $\|x\|_\#^2 = \langle x, x \rangle_\# \ge 0$ with equality $\|x\|_\#^2 = 0$ iff $x = 0$.

Notice that combining properties (1) and (2) that $x \to \langle z, x \rangle$ is anti-linear for fixed $z \in H$, i.e. $\langle z, ax + by \rangle_\# = \bar{a}\langle z, x \rangle_\# + \bar{b}\langle z, y \rangle_\#$.

(ii) Let $\{a_n\}_{n=0}^k$ be finite sequence in $H$, $\{a_n\}_{n=0}^k \subset H$.

We define external hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^k} \subset H$ by

$$\begin{align}
\{A_n; k\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^k} = \nonumber \\
&= (a_0, a_1, \dots, a_m, \dots, a_{k-1}, \hat{a}_k). \tag{0.1} \\
\hat{a} &= (a, a, \dots), a \in H. \nonumber
\end{align}$$

(iii) Let $\{a_n\}_{n=0}^\infty$ be countable sequence in $H : \{a_n\}_{n=0}^\infty \subset H$.

We define hyper infinite sequence $\{A_n\}_{n=0}^{\infty} = \overbrace{\{a_n\}_{n=0}^\infty} \subset H$ by

$$\begin{align}
\{A'_n; \infty\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^{\infty}} = \nonumber \\
&= (\hat{a}_0, a_1, \dots, a_k, \dots, \{a_n\}_{n=0}^{\infty}, \overbrace{\{a_n\}_{n=0}^{\infty}}). \tag{0.2}
\end{align}$$

(iv) Let $\{a_n\}_{n=0}^N$, $N \in \mathbb{N}$ be external hyperfinite sequence in $H : \{a_n\}_{n=0}^N \subset H$.

We define hyper infinite sequence $\overbrace{\{a_n\}_{n=0}^N} \subset H$ by

$$\begin{align}
\{A_n; N\}_{n=0}^{\infty} &= \overbrace{\{a_n\}_{n=0}^N} = \nonumber \\
&= (a_0, a_1, \dots, a_m, \dots, a_{N-1}a_N, \hat{a}_N). \tag{0.3}
\end{align}$$

(v) Let $\{a_n\}_{n=0}^k$, $k \in \mathbb{N}$ be finite sequence in $H$, $\{a_n\}_{n=0}^N \subset H$.
$a, b \in *\tilde{\mathbb{R}}_c^\#$ such that $a < b$.

The following three propositions are obvious.

**Proposition 6.7.** Let $f: X \to *\tilde{\mathbb{R}}_c^\#$ be a function. Then the following conditions are equivalent:

(a) $f$ is $\#$-measurable;

(b) $f^{-1}([0, b)) \in \Omega$ for any hyperreal $b \in *\tilde{\mathbb{R}}_c^\#$;

(c) $f^{-1}((b, \infty)) \in \Omega$ for any hyperreal $b \in *\tilde{\mathbb{R}}_c^\#$;

(d) $f^{-1}(B) \in \Omega$ for any $B \in B(R)$.

**Proposition 6.8.** Let $f$ and $g$ be $\#$-measurable functions, then

(a) $\alpha \times f + \beta \times g$ is $\#$-measurable for any $\alpha, \beta \in *\tilde{\mathbb{R}}_c^\#$;

(b) functions $\max\{f, g\}$ and $f \times g$ are $\#$-measurable.

In particular, functions $f^+ := \max\{f, 0\}, f^- := (-f)^+$, and $|f| := f^+ + f^-$ are $\#$-measurable.

## §7. *$\tilde{\mathbb{R}}_c^\#$-Valued $\#$-Measures and measure $\#$-space

**Definition 7.1.** A pair $(X, \mathcal{F})$ where $\mathcal{F}$ is a $\sigma^\#$-algebra on $X$ is called a $\#$-measurable space. Elements of $\mathcal{F}$ are called $\#$-measurable sets.

Given a $\#$-measurable space $(X, \mathcal{F})$, a function $\mu^\# : \mathcal{F} \to [0, \infty^\#] \subset *\tilde{\mathbb{R}}_c^\#$ is called a $\#$-measure on $(X, \mathcal{F})$ if

1. $\mu^\#(\emptyset) = 0$;

2. (Hyper infinite additivity) For all hyper infinite sequences $(A_n)_{n \in \mathbb{N}^\#} \subset \mathcal{F}$ of pairwise disjoint sets in $\mathcal{F}$, we have that $\mu^\#\left(\bigcup_{n \in \mathbb{N}^\#} A_n\right) = \text{Ext}-\sum_{n \in \mathbb{N}^\#} \mu^\#(A_n)$.

$(X, \mathcal{F}, \mu^\#)$ is called a $\#$-measure space.

**Definition 7.2.** A measure space $(X, \mathcal{F}, \mu^\#)$ is called: (a) hyperfinite if $\mu^\#(X) < \infty^\#$.

(b) It is called $\sigma^\#$-hyperfinite if $X = \bigcup_{n \in \mathbb{N}^\#} A_n$ where $A_n \in \mathcal{F}$ and $\mu^\#(A_n) < \infty^\#$ for all $n \in \mathbb{N}^\#$.

**Definition 7.3.** Let $\Sigma$ be a $\sigma^\#$-algebra of subsets of a set $X$, and let $E = (E, \|\cdot\|_\#)$ be a non-Archimedean Banach space. A function $\mu^\# : \Sigma \to E \cup \{\infty\}$ is called a vector-valued $\#$-measure (or $E$-valued measure) if

(a) $\mu^\#(\emptyset) = 0$;

(b) $\mu^\#\left(\bigcup_{n \in \mathbb{N}^\#} A_n\right) = \text{Ext}-\sum_{n \in \mathbb{N}^\#} \mu^\#(A_n)$ for any pairwise disjoint sequence $A_n, n \in \mathbb{N}^\#$, $A_n \subseteq \Sigma$;

(c) for any $S \in \Sigma$, $\mu^\#(S) = \infty$, there exists $B \in \Sigma$ such that $B \subseteq S$ and $0 < \|\mu^\#(B)\|_\# < *\infty$.

**Definition 7.4.** (a) A function $\mu^\# : \mathcal{F} \to *\tilde{\mathbb{C}}_c^\# \cup \{\infty\}$ is called a complex $\#$-measure if

1. $\mu^\#(\emptyset) = 0$,

2. $\mu^\#\left(\bigcup_{n \in \mathbb{N}^\#} A_n\right) = \text{Ext}-\sum_{n \in \mathbb{N}^\#} \mu^\#(A_n)$ for any sequence $A_n, n \in \mathbb{N}^\#$ of pairwise disjoint sets from $\mathcal{F}$, and, for any $A \in \mathcal{F}$, $\mu^\#(A) = *\infty$, there exists $B \in \mathcal{F}$ such that $B \subseteq A$ and $0 < |\mu^\#(B)|_\# < *\infty$.
We define external finite sum $Ext-\sum_{n=0}^{n=k} a_n$ by

$$ Ext-\sum_{n=0}^{n=k} a_n = \overbrace{\{c_n\}_{n=0}^k}^{k} = (c_0, c_1, \dots, c_k, \dots, \hat{c}_k) \in [[\hat{c}_k]] \quad (0.4) $$

$$ \text{where } c_0 = a_0, c_j = Ext-\sum_{n=0}^{n=j} a_n, 0 \le j \le k. $$

(vi) Let $\{a_n\}_{n=0}^\infty$ be countable sequence in $H : \{a_n\}_{n=0}^\infty \subset H$. We define external countable sum $Ext-\sum_{n=0}^{n=\infty} a_n$ by

$$ \begin{aligned} Ext-\sum_{n=0}^{n=\infty} a_n &= \overbrace{\{c_n\}_{n=0}^\infty}^{k} = \\ &= (c_0, c_1, \dots, c_k, \dots, \{c_n\}_{n=0}^\infty, \overbrace{\{c_n\}_{n=0}^\infty}^{k}) \in [[\overbrace{\{c_n\}_{n=0}^\infty}^{k}]] \end{aligned} \quad (0.5) $$

$$ \text{where } c_0 = a_0, c_k = Ext-\sum_{n=0}^{n=k} a_n, k \in \mathbb{N}. $$

(vii) Let $\{a_n\}_{n=0}^{n=N}$, $N \in *NN$ be external hyperfinite sequence in $H : \{a_n\}_{n=0}^N \subset H$.

We define external hyperfinite sum $Ext-\sum_{n=0}^{n=N} a_n$ by

$$ Ext-\sum_{n=0}^{n=N} a_n = \overbrace{\{c_n\}_{n=0}^{N}}^{N} = (c_0, c_1, \dots, c_k, \dots, c_N, \hat{c}_N) \in [[\hat{c}_N]] \quad (0.6) $$

$$ \text{where } c_0 = a_0, c_k = Ext-\sum_{n=0}^{n=k} a_n, 0 \le k \le N, c_N = Ext-\sum_{n=0}^{n=N} a_n. $$

(viii) Let $\{a_n\}_{n=0}^{n=N}$, $N \in *NN$ be external hyperfinite sequence in $H : \{a_n\}_{n=0}^N \subset H$ such that $a_n \equiv 0$ for all $n \in *NN$. We assume that

$$ Ext-\sum_{n=0}^{n=N} a_n = Ext-\sum_{n=0}^{n=\infty} a_n. \quad (0.7) $$

**Remark 1.1.** (i) Let $\{x_i\}_{i=1}^N \subset H$ and $\{y_i\}_{i=1}^N \subset H, N \in *N$ by external hyperfinite sequences; let $\{\alpha_i\}_{i=1}^N \subset \overline{\mathbb{C}}_c^\#$ and $\{\beta_i\}_{i=1}^N \subset \overline{\mathbb{C}}_c^\#$. Then the equality holds

$$ Ext-\sum_{i=1}^{N} \langle \alpha_i x_i + \beta_i y_i, z \rangle_{\#} = Ext-\sum_{i=1}^{N} \alpha_i \langle x_i, z \rangle_{\#} + Ext-\sum_{i=1}^{N} \beta_i \langle y_i, z \rangle_{\#} \quad (0.8) $$

(ii) Let $\{x_{ij}\} \subset H, N, K \in *N, K \le j \le i \le N$, by external hyperfinite sequences; let $\{\alpha_{ij}\}_{i=1}^N \subset \overline{\mathbb{C}}_c^\#$. Then the equality holds

$$ \left\langle Ext-\sum_{K \le j \le i \le N} \alpha_{ij} x_{ij}, z \right\rangle_{\#} = Ext-\sum_{i=K}^{N} \left( Ext-\sum_{i=K}^{N} \alpha_{ij} \langle x_{ij}, z \rangle_{\#} \right). \quad (0.9) $$

(iii) Let $\{x_i\}_{i=1}^\infty \subset H$ by external hyperinfinite sequence in $H$. We call $\{x_i\}_{i=1}^\infty$ a Cauchy hyperinfinite sequence if for any $\varepsilon \approx 0, \varepsilon > 0$ there is $N \in *NN$ such that for any $m, n > N, \|x_n - x_m\|_\# < \varepsilon$.

(iv) We now stand ready to give construction of $H^\#$. The members of $H^\#$ will be constructed as equivalence classes of Cauchy hyperinfinite sequences in $H$. Let $C(H)$ denote the set of all Cauchy hyperinfinite sequences in $H$. We must define an equivalence relation on $C(H)$. For $s \in C(H)$, denote by $[s]$ the set of all elements in $C(H)$

that are related to $s$. Then for any $s, t \in C(H)$, either $[s] = [t]$ or $[s]$ and $[t]$ are disjoint. Let $\{x_i\}_{i=1}^\infty$ and $\{y_i\}_{i=1}^\infty$ be in $C(H)$. Say they are equivalent (i.e. related $\{x_i\}_{i=1}^\infty \sim \{y_i\}_{i=1}^\infty$) if $\|x_n - y_n\|_\# \to_\# 0_{\overline{\mathbb{R}}_c}$; i.e. if the hyperinfinite sequence $\|x_n - y_n\|_\#$-tends to $0_{\overline{\mathbb{R}}_c}$ as $n \to \infty$.

(vi) Definition (iv) yields an equivalence relation $(\cdot \sim \cdot)$ on $C(H)$.
(b) A function $\mu^# : \mathcal{F} \to *\tilde{\mathbb{R}}_c^# \cup \{*\infty\}$ is called a signed #-measure if

$$ \mu^#(\emptyset) = 0 $$

$$ \mu^#\left(\bigcup_{n \in \mathbb{N}^#} A_n\right) = \text{Ext-}\sum_{n \in \mathbb{N}^#} \mu^#(A_n) \text{ for any sequence } A_n, n \in \mathbb{N}^# \text{ of pairwise disjoint sets from } \mathcal{F}, \text{ and, for any } A \in \mathcal{F}, \mu^#(A) = *\infty, \text{ there exists } B \in \mathcal{F} \text{ such that } B \subseteq A \text{ and } 0 < |\mu^#(B)| < *\infty. $$

**Definition 7.5.** If a certain property involving the points of #-measure space is true, except a subset having #-measure zero, then we say that this property is true #-almost everywhere (abbreviated as #-a.e.).

**Proposition 7.5.** Let $\mu^#$ be a #-measure on a $\sigma^#$-algebra $\mathcal{F}, A_n \in \mathcal{F}$, and $A_n \to A$. Then $A \in \mathcal{F}$ and $\mu^#(A) = \text{-lim}_{n \to *\infty} \mu^#(A_n)$. In particular, if $(B_n)_{n=1}^{*\infty}$ is a decreasing hyper infinite sequence of elements of $\mathcal{F}$ such that $\bigcap_{n=1}^{*\infty} B_n = \emptyset$, then $\mu^#(B_n) \to_# 0$.

**Definition 7.6.** If $\mathcal{F}$ is a $\sigma^#$-algebra of subsets of $X$ and $\mu^#$ is a #-measure on $\mathcal{F}$, then the triple $(X, \mathcal{F}, \mu)$ is called a #-measure space. The sets belonging to $\mathcal{F}$ are called #-measurable sets because the #-measure is defined for them.

# 8. #-Convergence of functions and the generalized Egoroff theorem.

**Definition 8.1.** Let $f_n, n \in \mathbb{N}^#$ be a hyper infinite sequence of $*\tilde{\mathbb{R}}_c^#$-valued functions defined on $X$. We say that:

1. $f_n \to_# f$ pointwise, if $f_n(x) \to_# f(x)$ for all $x \in X$;

2. $f_n \to_# f$ almost #-everywhere (#-a.e.), if $f_n(x) \to_# f(x)$ for all $x \in X$ except a set of #-measure 0;

3. $f_n \to_# f$ uniformly, if for any $\varepsilon > 0$, $\varepsilon \approx 0$ there is $n(\varepsilon)$ such that $\sup\{|f_n(x) - f(x)|: x \in X\} \le \varepsilon$ for all $n \ge n(\varepsilon)$.

**Theorem 8.1.** (generalized Egoroff's theorem) Suppose that $\mu^#(X) < *\infty$, $\{f_n\}_{n=1}^{*\infty}$ and $f$ are #-measurable functions on $X$ such that $f_n \to_# f$ #-a.e. Then, for every $\varepsilon \approx 0$, there exists $E \subseteq X$ such that $\mu^#(E) < \varepsilon$ and $f_n \to_# f$ uniformly on $E^c = X \setminus E$.

**Proof:** Without loss of generality, we may assume that $f_n \to_# f$ everywhere on $X$ and (by replacing $f_n$ with $f_n - f$) that $f \equiv 0$. For $k, n \in *\mathbb{N}$, let

$$ E_n(k) := \bigcup_{m=n}^{*\infty} \{x : |f_m(x)| \ge k-1\}. \text{ Then, for a fixed } k, E_n(k) \text{ decreases as } n \text{ increases, and } \bigcap_{n=1}^{*\infty} E_n(k) = \emptyset. \text{ Since } \mu^#(X) < *\infty, \text{ we conclude that } \mu^#(E_n(k)) \to_# 0 \text{ as } n \to *\infty. $$

Given $\varepsilon \approx 0$, choose $n_k$ such that $\mu^#(E_n(k)) < \varepsilon \times 2^{-k}$, and set $E = \bigcup_{n=1}^{*\infty} E_n(k)$. Then $\mu^#(E) < \varepsilon$, and we have $|f_n(x)| < k^{-1}(\forall n > n_k, x \notin E)$.

Thus $f_n \to_# 0$ uniformly on $X \setminus E$.

## Generalized exhaustion argument.

Let $(X, \Sigma, \mu^#)$ be a $\sigma^#$-finite #-measure space. Given a hyper infinite sequence $(U_n)_{n=1}^{*\infty} \subseteq \Sigma$, a set $A \in \Sigma$ is called $(U_n)_n$-bounded if there exists $n \in *\mathbb{N}$ such that $A \subseteq U_n$ $\mu^#$-almost everywhere.
**Proof.** We need to show that this relation is reflexive, symmetric, and transitive.

* Reflexive: $x_n - x_n = 0_{\mathbb{R}_c^*}$, $n \in \mathbb{N}$ and the sequence all of whose terms are $0_{\mathbb{R}_c^*}$ clearly #-converges to $0_{\mathbb{R}_c^*}$. So $\{x_i\}_{i=1}^* \in \mathbb{R}_c^*$ is related to $\{x_i\}_{i=1}^*$.

* Symmetric: Suppose $\{x_i\}_{i=1}^* \in \mathbb{R}_c^*$ is related to $\{y_i\}_{i=1}^*$, so $\|x_n - y_n\|_# \to_# 0_{\mathbb{R}_c^*}$. But $y_n - x_n = -(x_n - y_n)$, and since only the absolute value $\|x_n - y_n\|_# = \|y_n - x_n\|_#$ comes into play in Definition (iv), it follows that $\|y_n - x_n\|_# \to_# 0_{\mathbb{R}_c^*}$ as well. Hence, $\{y_i\}_{i=1}^*$ is related to $\{x_i\}_{i=1}^*$.

* Transitive: Suppose $\{x_i\}_{i=1}^* \in \mathbb{R}_c^*$ is related to $\{y_i\}_{i=1}^*$, and $\{y_i\}_{i=1}^* \in \mathbb{R}_c^*$ is related to $\{z_i\}_{i=1}^*$. This means that $\|x_n - y_n\|_# \to_# 0_{\mathbb{R}_c^*}$ and $\|y_n - z_n\|_# \to_# 0_{\mathbb{R}_c^*}$.

To be fully precise, let us fix $\varepsilon \approx 0, \varepsilon > 0$; then there exists an $N \in \mathbb{N}$ such that for all $n > N$, $\|x_n - y_n\|_# < \varepsilon/2$; also, there exists an $M$ such that for all $n > M$, $\|y_n - z_n\|_# < \varepsilon/2$. Well, then, as long as $n$ is bigger than both $N$ and $M$, we have that $\|x_n - z_n\|_# = \|(x_n - y_n) + (y_n + z_n)\|_# \le \|x_n - y_n\|_# + \|y_n - z_n\|_# < \varepsilon/2 + \varepsilon/2 = \varepsilon$.

So, choosing $L$ equal to the max($N, M$), we see that given $\varepsilon \approx 0, \varepsilon > 0$ we can always choose $L$ so that for $n > L$, $\|x_n - z_n\|_# < \varepsilon$. This means that $\|x_n - z_n\|_# \to_# 0_{\mathbb{R}_c^*}$ i.e. $\{x_i\}_{i=1}^*$ is related to $\{z_i\}_{i=1}^*$.

So, we really have an equivalence relation $(\cdot \sim \cdot)$, and therefore the set $C(H)$ is partitioned into disjoint subsets (equivalence classes). We will denote that partition $C(H)/\sim$ by $H^#$

$$ C(H)/\sim \triangleq H^# \qquad (0.10) $$

(vii) assume that $[s] = [\{x_i\}_{i=1}^*] \in H^# \text{ and } [t] = [\{y_j\}_{j=1}^*] \in H^#$, we define inner #-product $\langle [s], [t] \rangle_#$ on $H^#$ by

$$ \langle [\{x_i\}_{i=1}^*], [\{y_j\}_{j=1}^*] \rangle_# = \mathrm{Ext} - \sum_{i,j=1}^{i,j=*\infty} \langle x_i, y_j \rangle_# \qquad (0.11) $$

In particular if for all $i \neq j$ $\langle x_i, y_j \rangle_# = 0_{\mathbb{R}_c^*}$

$$ \langle [\{x_i\}_{i=1}^*], [\{y_j\}_{j=1}^*] \rangle_# = \mathrm{Ext} - \sum_{i=1}^{i=*\infty} \langle x_i, y_i \rangle_# \qquad (0.12) $$

**Remark 1.2.** The following formula useful:

$$ \|x + y\|_#^2 = \langle x + y, x + y \rangle_# = \|x\|_#^2 + \|y\|_#^2 + \langle x, y \rangle_# + \langle y, x \rangle_# = \|x\|_#^2 + \|y\|_#^2 + 2 \mathrm{Re}\langle x, y \rangle_# \quad (1.1) $$

**Theorem 1.1.** (Generalized Schwarz Inequality). Let $(H, \langle \cdot, \cdot \rangle_#)$ be an inner #-product space and $x, y \in H$. Assume that:

(1) at least one of hyperreals $\|x\|_#$, $\|y\|_#$ is invertible in $\mathbb{R}_c^*$ then

$$ |\langle x, y \rangle_#| \le \|x\|_# \times \|y\|_# \qquad (1.2) $$

and equality holds iff $x$ and $y$ are linearly dependent.

(2) both of hyperreals $\|x\|_#$, $\|y\|_#$ is not invertible in $\mathbb{R}_c^*$ then

$$ |\langle x, y \rangle_#| \times 1_{\mathbb{R}_c^*} \le \|x\|_# \times \|y\|_# \qquad (1.2') $$

**Proof.** (1) If $y = 0$, the result holds trivially. So assume that $y \neq 0$ and $\|y\|_#$ is invertible in $\mathbb{R}_c^*$. First off notice that if $x = \alpha y$ for some $\alpha \in \mathbb{C}_c^*$, then $\langle x, y \rangle = \alpha \|y\|_#^2$ and hence
**Theorem 8.2.** (Generalized Exhaustion theorem) Let $(Y_n)_{n=1}^{*\infty} \subseteq \Sigma$ be a hyper infinite sequence satisfying $Y_n \uparrow X$ and $\mu^#(Y_n) < *\infty$ for all $n \in *\mathbb{N}$. Let $P$ be some property of $(Y_n)_n$-bounded

#-measurable sets, such that $A \in P$ iff $B \in P$ for all $B$, $\mu^#(A\Delta B) = 0$. Suppose that any $(Y_n)_n$-bounded set $A$, $\mu^#(A) > 0$, has a subset $B \in \Sigma$, $\mu^#(B) > 0$ with the property $P$. Moreover, assume that either

(a) $A_1 \cup A_2 \in P$ for every $A_1, A_2 \in P$, or

(b) $\bigcup_{n \in *\mathbb{N}} B_n \in P$ for every at most hyper infinite family $(B_n)_n$ of pairwise disjoint sets possessing the property $P$.

Then there exists hyper infinite sequence $(X_n)_{n=1}^{*\infty} \subseteq \Sigma$ such that $X_n \uparrow X$, and
$P \ni X_n \subseteq Y_n$

for all $n \in *\mathbb{N}$. Moreover, there exists a pairwise disjoint sequence $(A_n)_{n=1}^{*\infty} \subseteq \Sigma$
such that $\bigcup_{n \in *\mathbb{N}} A_n = X$ and $A_n \in P$ for all $n \in *\mathbb{N}$.

**Proof:** Let $A$ be a $(Y_n)_n$-bounded set with $\mu^#(A) > 0$. Denote

$$P_A := \{B \in P : B \subseteq A\} \wedge m(A) := \sup\{\mu^#(B) : B \in P_A\}.$$

I(a) Suppose $P$ satisfies (a). Then there exists a sequence $(F_n)_{n=1}^{*\infty} \subseteq P_A$ such that $m(A) = \text{-lim}_{n \to *\infty} \mu^#(F_n)$, We may assume, that $F_n \uparrow$. By Proposition 2.5 the set $F = \bigcup_{n=1}^{*\infty} F_n$ satisfies $\mu^#(F) = m(A)$. We show that $\mu^#(A) = m(A)$. If not then $\mu^#(AF) > 0$. The set $A\F$ has a subset of positive #-measure $F_0 \in P$. Then $F_n \cup F_0 \in P_A$ and $\mu^#(F_n \cup F_0) > m(A)$ for a sufficiently large $n \in *\mathbb{N}$, which contradicts to the definition of $m(A)$. Therefore, $\mu^#(A) = m(A)$.

Now we apply this for $A = Y_n$. Thus, there exists hyper infinite sequence $(X'_n)_n \subseteq \Sigma$ such that $X'_n \subseteq Y_n$, $X_n, n \in P$, and $\mu^#(Y_n \setminus X'_n) < n^{-1}$ for all $n \in *\mathbb{N}$. By (a), we may assume that $X'_n \uparrow$. The set $X'_0 = \bigcup_{n=1}^{*\infty} X'_n$ satisfies $Y_n \setminus X'_0 \subseteq Y_n \setminus X'_n$, so $\mu^#(Y_n \setminus X'_0) < n^{-1}$ for all $n \in *\mathbb{N}$. Then $\mu^#(Y_n \setminus X'_0) = 0$, and $\mu^#((\bigcup_{n=1}^{*\infty} Y_n) \setminus X'_0) = 0$, or $\mu^#(XXX'_0) = 0$.

Let $X_n = (X'_n \cup (XXX'_0) \cap Y_n)$, then the hyper infinite sequence $(X_n)_n$ has the required properties. The desired pairwise disjoint sequence $(A_n)_{n=1}^{*\infty}$ is given recurrently by $A_1 = X_1$ and $A_{k+1} = X_{k+1} \setminus A_k$.

I(b) Suppose $P$ satisfies (b). Let $F_A$ be the family of all pairwise disjoint families of elements of $P_A$ of nonzero measure. Then $F_A$ is ordered by inclusion and, obviously, satisfies the conditions of the Zorn lemma. Therefore, we have a maximal element in $F_A$, say $\Delta$. Then $\Delta$ is at most hyper infinite family, say $\Delta = \{D_n\}_n$. By (b), its union $D = \bigcup_n D_n$ is an element of $P_A$ as well. If $D$ is a proper subset of $A$, then $\mu^#(A\Delta) > 0$. The set $A\Delta$ has a subset $F \in P$ of the positive measure. Then $\Delta_1 := \Delta \cup \{F\}$ is an element of $F_A$ which is strictly greater then $\Delta$. The obtained contradiction, shows that $A \in P$ for every $(Y_n)_n$-bounded set $A$. So, we may take $X_n = Y_n$ for each $n \in *\mathbb{N}$.

Now we apply this for $A = Z_m = Y_m \cup \bigcup_{k=1}^{m-1} Y_k$ be a pairwise disjoint union, where $D_n^m \in P$ for all $n, m \in *\mathbb{N}$. The family $\{D_n^m\}_{n,m}$ is an at most hyper infinite disjoint decomposition of $X$, say $\{D_n^m\}_{n,m} = (A_n)_{n=1}^{*\infty}$. The sequence $(A_n)_{n=1}^{*\infty}$ satisfies the required properties.

**Theorem 8.3.** (The generalized Borel-Cantelli lemma) Let $(X, \Sigma, \mu^#)$ be a #-measure space. Assume that $\{A_n\}_n \subseteq \Sigma$ and $\text{Ext}-\sum_{n=1}^{*\infty} \mu(A_n) < *\infty$ then $\lim_{n \to *\infty} \mu^#(A_n) = 0$.

## §9. Vector-valued #-measures
$$|\langle x, y \rangle_{\#}| = |\alpha| \|y\|_{\#}^2 = \|x\|_{\#} \|y\|_{\#}$$

Note that in this case $\alpha = \langle x, y \rangle \|y\|^2$. Now suppose that $x \in H$ is arbitrary, let $z = x - \|y\|_{\#}^{-2} \langle x, y \rangle_{\#} y$. So $z$ is the orthogonal projection of $x$ onto $y$. Then

$$ 
\begin{aligned}
0 \le \|z\|_{\#}^2 &= \left\| x - \frac{\langle x, y \rangle_{\#}}{\|y\|_{\#}^2} y \right\|_{\#}^2 \\
&= \|x\|_{\#}^2 + \frac{|\langle x, y \rangle_{\#}|^2}{\|y\|_{\#}^4} \|y\|_{\#}^2 - 2 \operatorname{Re} \left\langle x, \frac{\langle x, y \rangle_{\#}}{\|y\|_{\#}^2} y \right\rangle \\
&= \|x\|_{\#}^2 - \frac{|\langle x, y \rangle_{\#}|^2}{\|y\|_{\#}^2}.
\end{aligned} 
\quad (1.3) $$

from (1.3) it follows that $0 \le \|y\|_{\#}^2 \|x\|_{\#}^2 - |\langle x, y \rangle_{\#}|^2$ with equality iff $z = 0$ or equivalently iff $x = \|y\|_{\#}^{-2} \langle x, y \rangle_{\#} y$.

(2) Let $z = x - (\|y\|_{\#}^{-1*})^2 \langle x, y \rangle_{\#} y$. So $z$ is the orthogonal projection of $x$ onto $y$. Then

$$ 
\begin{aligned}
0 \le \|z\|_{\#}^2 &= \left\| x - \langle x, y \rangle_{\#} (\|y\|_{\#}^{-1*})^2 y \right\|_{\#}^2 \\
&= \|x\|_{\#}^2 + |\langle x, y \rangle_{\#}|^2 (\|y\|_{\#}^{-1*})^4 \|y\|_{\#}^2 - 2 \operatorname{Re} \langle x, \langle x, y \rangle_{\#} (\|y\|_{\#}^{-1*})^2 y \rangle \\
&= \|x\|_{\#}^2 - |\langle x, y \rangle_{\#}|^2 (\|y\|_{\#}^{-1*})^2.
\end{aligned} 
\quad (1.3') $$

From (1.3') it follows that $0 \le \|y\|_{\#}^2 \times \|x\|_{\#}^2 - |\langle x, y \rangle_{\#}|^2 \times \tilde{1}_{\#}^{\cdot \cdot \cdot}$.

**Corollary 1.1.** Let $(H^{\#}, \langle \cdot, \cdot \rangle)$ be an inner #-product space and $\|x\|_{\#} := \sqrt{\langle x, x \rangle_{\#}}$. Then $\|\cdot\|_{\#}$ is a $\ast\mathbb{R}_c$-valued #-norm on $H^{\#}$. Moreover $\langle \cdot, \cdot \rangle_{\#}$ is #-continuous on $H^{\#} \times H^{\#}$, where $H$ is viewed as the #-normed space $(H^{\#}, \|\cdot\|_{\#})$.

**Proof.** The only non-trivial thing to verify that $\|\cdot\|_{\#}$ is a #-norm is the triangle inequality:

$$ \|x+y\|^2 = \|x\|^2 + \|y\|^2 + 2 \operatorname{Re} \langle x, y \rangle_{\#} \le \|x\|^2 + \|y\|^2 + 2 \|x\|_{\#} \|y\|_{\#} = (\|x\|_{\#} + \|y\|_{\#})^2 $$
where we have made use of Schwarz's inequality. Taking the square root of this inequality shows $\|x+y\| \le \|x\| + \|y\|$. For the #-continuity assertion:

$$ 
\begin{aligned}
|\langle x, y \rangle_{\#} - \langle x', y' \rangle_{\#}| &= |\langle x - x', y \rangle_{\#} + \langle x', y - y' \rangle_{\#}| \le \|y\|_{\#} \|x - x'\|_{\#} + \|x'\|_{\#} \|y - y'\|_{\#} \\
&\le \|y\|_{\#} \|x - x'\|_{\#} + (\|x\|_{\#} + \|x - x'\|_{\#}) \|y - y'\|_{\#} = \|y\|_{\#} \|x - x'\|_{\#} + \|x\|_{\#} \|y - y'\|_{\#} \\
&\quad + \|x - x'\|_{\#} \|y - y'\|_{\#} \text{ from which it follows that } \langle \cdot, \cdot \rangle \text{ is #-continuous.}
\end{aligned} 
$$

**Definition 1.2.** Let $(H, \langle \cdot, \cdot \rangle_{\#})$ be an inner #-product space, we say $x, y \in H$ are orthogonal and write $x \perp y$ iff $\langle x, y \rangle_{\#} = 0$. More generally if $A \subset H$ is a set, $x \in H$ is orthogonal to $A$ and write $x \perp A$ iff $\langle x, y \rangle = 0$ for all $y \in A$. Let $A_{\perp} = \{x \in H : x \perp A\}$ be the set of vectors orthogonal to $A$. We also say that a set $S \subset H$ is orthogonal if $x \perp y$ for all $x, y \in S$ such that $x \neq y$. If $S$ further satisfies, $\|x\|_{\#} = 1$ for all $x \in S$, then $S$ is said to be orthonormal.

**Proposition 1.1.** Let $(H, \langle \cdot, \cdot \rangle_{\#})$ be an inner product space then

(1) (Parallelogram Law)

$$ \|x+y\|_{\#}^2 + \|x-y\|_{\#}^2 = 2\|x\|_{\#}^2 + 2\|y\|_{\#}^2 \quad (1.4) $$

for all $x, y \in H$.

(2) (Pythagorean Theorem) If $S \subset H$ is a hyperfinite orthonormal set, then

$$ \left\| \operatorname{Ext} - \sum_{x \in S} x \right\|_{\#}^2 = \operatorname{Ext} - \sum_{x \in S} \|x\|_{\#}^2 \quad (1.5) $$

(3) If $A \subset H^{\#}$ is a set, then $A_{\perp}$ is a #-closed linear subspace of $H^{\#}$.
In this section, we extend the notion of a measure. Then we study the basic operations with signed measures and present the Jordan decomposition theorem.

## 9.1. Vector-valued, signed and complex #-measures.

Let $\Sigma^#$ be a $\sigma^#$-algebra of subsets of a set $X$, and let $E^# = (E^# , \|\cdot\|_{\#})$ be a non-Archimedean Banach space over field $\tilde{\mathbb{R}}_c^#$.

**Definition 9.1** A function $\mu^# : \Sigma^# \to E^# \cup \{^*\infty\}$ is called a vector-valued #-measure (or $E^#$-valued measure) if

(a) $\mu^#(\emptyset) = 0$;

(b) $\mu^#(\cup_{k=1}^\infty A_k) = \text{Ext}-\sum_{k=1}^\infty \mu^#(A_k)$ for any pairwise disjoint sequence $(A_k)_{k \in \Sigma^#}$;

(c) for any $A \in \Sigma^#$, $\mu^#(A) = ^*\infty$, there exists $B \in \Sigma^#$ such that $B \subseteq A$ and $0 < \|\mu^#(B)\|_{\#} < ^*\infty$.

**Example 9.1** Take $\Sigma^# = P(^*\mathbb{N})$, and $c_0^#$ is the non-Archimedean Banach space of all #-convergent $\mathbb{C}_c^#$-valued hyper infinite sequences with a fixed element $(\alpha_n)_n \in c_0^#$. Define now for any $A \subseteq \mathbb{N}\psi(A) := (\beta_n)_n$,

where $\beta_n = \alpha_n$ if $n \in A$ and $\beta_n = 0$ if $n \notin A$. Then $\psi$ is a $c_0^#$-valued #-measure on $P(^*\mathbb{N})$.

**Example 9.2** Let $X$ be a set and let $\Omega$ be a $\sigma^#$-algebra in $P(X)$. Then for any family $\{\mu_k\}_{k=1}^m$ of finite #-measures on $\Omega$ and for any family $\{w_k\}_{k=1}^m$ of vectors of $\mathbb{R}_c^{\#n}$, the $\mathbb{R}_c^{\#n}$-valued #-measure $\Psi$ on $\Omega$ is defined by the formula

$$ \Psi(E) = \text{Ext}-\sum_{k=1}^{m} \mu_k(E) \times w_k, (E \in \Omega). $$

**Example 9.3** Let $X$ be a set and let $\Omega$ be a $\sigma^#$-algebra in $P(X)$. Then for any family $\{\mu_k\}_{k=1}^m$ of finite #-measures on $\Omega$, for any family $\{A_k\}_{k=1}^m$ of pairwise disjoint sets in $\Omega$, and for any family $\{w_k\}_{k=1}^m$ of $\tilde{\mathbb{R}}_c^{\#n}$, $n \in ^*\mathbb{N}$, the $\mathbb{R}_c^{\#n}$-valued #-measure $\Phi$ on $\Omega$ is defined by the formula $\Phi(E) = \text{Ext}-\sum_{k=1}^{m} \mu_k(E \cap A_k) \times w_k$, ($E \in \Omega$).

# §10. The $^*\tilde{\mathbb{R}}_c^#$-valued Lebesgue #-Integral

In the following consideration, we fix a $\sigma^#$-finite #-measure space $(X, \mathcal{F}, \mu^#)$.

**Definition 10.1.** Let $A_i \in \mathcal{F}, i = 1, \dots, n \in ^*\mathbb{N}$, be such that $\mu^#(A_i) < ^*\infty$ for all $i$, and $A_i \cap A_j = \emptyset$ for all $i \neq j$. The external function

$$ f(x) = \text{Ext}-\sum_{i=1}^{n} \lambda_i \chi_{A_i}(x), \qquad (10.1) $$

$\lambda_i \in ^*\tilde{\mathbb{R}}_c^#$, is called a simple external function. The Lebesgue external integral (Lebesgue #-integral) of a simple external function $f(x)$ is defined as

$$ \text{Ext}-\int_X f(x)d^\#\mu^# = \text{Ext}-\sum_{i=1}^{n} \lambda_i \mu^#(A_i). \qquad (10.2) $$

The Lebesgue external integral of a simple function is well defined.

**Notation 10.1.** Let $A_i \in \mathcal{F}, i = 1, \dots, n \in ^*\mathbb{N}$, be such that $\mu^#(A_i) < ^*\infty$ for all $i$, and $A_i \cap A_j = \emptyset$ for all $i \neq j$. Let $f_1(x), f_2(x)$ be a simple external function such that

$$ (i) \ 0 \le f_1(x) \le f_2(x) \text{ and } (ii) f_1(x) = \text{Ext}-\sum_{i=1}^{n} \lambda_{1,i} \chi_{A_i}(x), f_2(x) = \text{Ext}-\sum_{i=1}^{n} \lambda_{2,i} \chi_{A_i}(x). $$

$$ \text{Ext}-\sum_{i=1}^{n} \lambda_{1,i} \le \text{Ext}-\sum_{i=1}^{n} \lambda_{2,i}, \qquad (10.3) $$

then we will write $f_1(x) \le f_2(x)$.

**Definition 10.2.** Suppose that $\mu^#$ is hyperfinite. Let $f: X \to ^*\tilde{\mathbb{R}}_c^#$ be an arbitrary
**Proof.** We will assume that $H^\#$ is a complex Hilbert space with $\widehat{\mathbb{C}}^\#$-valued inner product, the real case being easier. Statements (1) and (2) are proved by the following elementary computations:

$$ \|x+y\|_{\#}^{2} = \|x\|_{\#}^{2} + \|y\|_{\#}^{2} + 2\operatorname{Re}\langle x, y \rangle_{\#} + \|x\|_{\#}^{2} + \|y\|_{\#}^{2} - 2\operatorname{Re}\langle x, y \rangle_{\#} = 2\|x\|_{\#}^{2} + 2\|y\|_{\#}^{2} \quad (1.6) $$

and

$$ \begin{aligned} \left\| \operatorname{Ext} - \sum_{x \in S} x \right\|_{\#}^{2} &= \left\langle \operatorname{Ext} - \sum_{x \in S} x, \operatorname{Ext} - \sum_{y \in S} y \right\rangle_{\#} = \operatorname{Ext} - \sum_{x,y \in S} \langle x, y \rangle_{\#} = \\ &= \operatorname{Ext} - \sum_{x \in S} \langle x, x \rangle_{\#} = \operatorname{Ext} - \sum_{x \in S} \|x\|_{\#}^{2}. \end{aligned} \quad (1.7) $$

Item 3. is a consequence of the #-continuity of $\langle \cdot, \cdot \rangle_{\#}$ and the fact that $A^{\perp} = \cap_{x \in A} \operatorname{Ker}(\langle \cdot, x \rangle)$ where $\operatorname{Ker}(\langle \cdot, x \rangle) = \{y \in H | \langle y, x \rangle_{\#} = 0\}$ is a #-closed subspace of $H$.

**Definition 1.3.** A non-Archiedean Hilbert space $H^\#$ is an inner #-product space $(H, \langle \cdot, \cdot \rangle_\#)$ such that the induced Hilbertian #-norm is #-complete.

**Example 1.3.** Let $(X, M, \mu^\#)$ be a #-measure space then $H^\# = L_2^\#(X, M, \mu^\#)$ with inner #-product $\langle f, g \rangle_\# = \operatorname{Ext} - \int_X f \bar{g} d^\# \mu^\#$ is a non-Archiedean Hilbert space. Note that every non-Archiedean Hilbert space $H^\#$ is “equivalent” to a Hilbert space of this form.

**Definition 1.4.** A subset $C$ of a non-Archiedean vector space $X$ is said to be convex if for all $x, y \in C$ the line segment $[x, y] = \{tx + (1-t)y : 0 \le t \le 1\}$ joining $x$ to $y$ is contained in $C$ as well. (Notice that any vector subspace of $X$ is convex.)

**Definition 1.5.** $M \subset H^\#$ is essentially #-closed if $\forall x(x \in H^\#) \exists [\inf_{z \in M} \|x-z\|_\#]$.

**Theorem 1.2.** Suppose that $H^\#$ is a non-Archiedean Hilbert space and $M \subset H^\#$ be a essentially #-closed convex subset of $H^\#$. Then for any $x \in H^\#$ there exists a unique $y \in M$ such that $\|x-y\|_\# = \operatorname{dist}(x, M) = \inf_{z \in M} \|x-z\|_\#$. Moreover, if $M$ is a vector subspace of $H^\#$, then the point $y$ may also be characterized as the unique point in $M$ such that $(x-y) \perp M$.

**Proof.** (1) Uniqueness: By replacing $M$ by $M-x = \{m-x|m \in M\}$ we may assume $x=0$. Let $\delta = \operatorname{dist}(0, M) = \inf_{m \in M} \|m\|_\#$ and $y, z \in M$. By the parallelogram law and the convexity of $M$ one obtains

$$ 2\|y\|_{\#}^{2} + 2\|z\|_{\#}^{2} = \|y+z\|_{\#}^{2} + \|y-z\|_{\#}^{2} = 4\left\|\frac{y+z}{2}\right\|_{\#}^{2} + \|y-z\|_{\#}^{2} \geq 4\delta^{2} + \|y-z\|_{\#}^{2}. \quad (1.8) $$

Hence if $\|y\|_\# = \|z\|_\# = \delta$, then $2\delta^2 + 2\delta^2 \geq 4\delta^2 + \|y-z\|_\#^2$, so that $\|y-z\|_\#^2 = 0$. Therefore, if a minimizer for $\operatorname{dist}(0, \cdot)|_M$ exists, it is unique.

(2) Existence: Let $y_n \in M$ be chosen such that $\|y_n\|_\# = \delta_n \to \delta \equiv \operatorname{dist}(0, M)$. Taking $y = y_m$ and $z = y_n$ in Eq.(1.8) shows $2\delta_m^2 + 2\delta_n^2 \geq 4\delta^2 + \|y_n - y_m\|_\#^2$. Passing to the #-limit $m, n \to \infty$ in this equation implies, $2\delta^2 + 2\delta^2 \geq 4\delta^2 + \#-\lim_{m,n \to \infty} \|y_n - y_m\|_\#^2$. Therefore $\{y_n\}_{n=1}^\infty$ is hyper infinite Cauchy sequence and hence #-convergent.Because $M$ is #-closed, $y = \#-\lim_{n \to \infty} \sup_{n \to \infty} y_n \in M$ and because $\|\cdot\|_\#$ is #-continuous, $\|y\|_\# = \#-\lim_{n \to \infty} \|y_n\|_\# = \delta = \operatorname{dist}(0, M)$. So $y$ is the desired point in $M$ which is closest to 0.

Now for the second assertion we further assume that $M$ is a #-closed subspace of $H$ and $x \in H^\#$. Let $y \in M$ be the closest point in $M$ to $x$. Then for $w \in M$, the
nonnegative bounded in $\mathbb{R}_c^\#$ #-measurable external function and let $(f_n)_{n \in \mathbb{N}}$, be a hyper infinite sequence of simple external functions which #-converges uniformly to $f$. Then the Lebesgue #-integral of $f$ is

$$ \mathrm{Ext-} \int_X f(x) d^\# \mu^\# = \#-\lim_{n \to \infty} \left( \mathrm{Ext-} \int_X f_n(x) d^\# \mu^\# \right). \quad (10.4) $$

**Remark 10.1.** It can be easily shown that the #-limit in Definition 11.2 exists and does not depend on the choice of a hyper infinite sequence $(f_n)_{n \in \mathbb{N}}$, and moreover, the hyper infinite sequence $(f_n)_{n \in \mathbb{N}}$ can be chosen such that $0 \le f_n \le f$ for all $n \in \mathbb{N}$.

**Notation 10.2.** Let $f_1 : X \to \tilde{\mathbb{R}}_c^\#$ and $f_2 : X \to \tilde{\mathbb{R}}_c^\#$ be an arbitrary nonnegative bounded in $\tilde{\mathbb{R}}_c^\#$ #-measurable external functions and let $(f_{1,n})_{n \in \mathbb{N}}$ and $(f_{2,n})_{n \in \mathbb{N}}$ be a hyper infinite sequences of simple external functions which #-converges uniformly to $f_1$ and to $f_2$ correspondingly. We assume that for all $n \in \mathbb{N}$ the inequality (10.3) is satisfied, then we will write $f_1(x) \le_s f_2(x)$.

**Definition 10.3.** Let $f : X \to \tilde{\mathbb{R}}_c^\#$ be a #-measurable function. Then the Lebesgue #-integral of $f$ is defined by

$$ \mathrm{Ext-} \int_X f(x) d^\# \mu^\# = \mathrm{Ext-} \int_X f^+(x) d^\# \mu^\# - \mathrm{Ext-} \int_X f^-(x) d^\# \mu^\#. \quad (10.5) $$

If both of these terms are finite or hyperfinite then the function $f$ is called #-integrable. In this case we write $f \in L_1^\#(X, \mathcal{F}, \mu^\#)$.

**Notation 10.3.** We will use the following notation. For any $A \in \mathcal{F}$:

$$ \mathrm{Ext-} \int_A f(x) d^\# \mu^\# = \mathrm{Ext-} \int_X f(x) \chi_A(x) d^\# \mu^\#. \quad (10.6) $$

**Lemma 10.1.** (1) Let $f : X \to \tilde{\mathbb{R}}_c^\#$ be an arbitrary nonnegative #-measurable function then

$$ \mathrm{Ext-} \int_X f(x) d^\# \mu^\# = \sup \left\{ \mathrm{Ext-} \int_X \varphi(x) d^\# \mu^\# \mid \varphi \text{ is a simple function such that } 0 \le \varphi(x) \le_s f(x) \right\}. \quad (11.7) $$

(2) If $f,g : X \to \tilde{\mathbb{R}}_c^\#$ are #-measurable, $g$ is #-integrable, and $|f(x)| \le_s g(x)$, then $f$ is #-integrable and

$$ \left| \mathrm{Ext-} \int_X f(x) d^\# \mu^\# \right| \le \mathrm{Ext-} \int_X g(x) d^\# \mu^\#. \quad (11.8) $$

(3) $\mathrm{Ext-} \int_X |f(x)| d^\# \mu^\# = 0$ if and only if $f(x) = 0$ #-a.e.

(4) If $f_1, f_2, \dots, f_n : X \to \tilde{\mathbb{R}}_c^\#$, $n \in \mathbb{N}$ are integrable then, for $\lambda_1, \lambda_2, \dots, \lambda_n \in \tilde{\mathbb{R}}_c^\#$, the linear combination $\mathrm{Ext-} \sum_{i=1}^n \lambda_i f_i$ is #-integrable and

$$ \mathrm{Ext-} \int_X \left( \mathrm{Ext-} \sum_{i=1}^n \lambda_i f_i \right) d^\# \mu^\# = \mathrm{Ext-} \sum_{i=1}^n \left( \mathrm{Ext-} \int_X \lambda_i f_i d^\# \mu^\# \right). \quad (11.9) $$

(5) Let $f \in L_1^\#(X, \mathcal{F}, \mu^\#)$, then the formula

$$ \nu^\#(A) = \mathrm{Ext-} \int_A f(x) d^\# \mu^\# = \mathrm{Ext-} \int_X f(x) \chi_A(x) d^\# \mu^\# \quad (11.10) $$

defines a signed #-measure on the $\sigma^\#$-algebra $\mathcal{F}$.

**Remark 11.2.** Assume that $f,g : X \to \tilde{\mathbb{R}}_c^\#$ are #-integrable functions and such that
function $g(t) = \|x - (y + tw)\|_{\#}^2 = \|x - y\|_{\#}^2 - 2t \operatorname{Re}\langle x - y, w \rangle_{\#} + t^2 \|w\|_{\#}^2$ has a minimum at $t = 0$. Therefore $0 = g'(0) = -2 \operatorname{Re}\langle x - y, w \rangle_{\#}$. Since $w \in M$ is arbitrary, this implies that $(x - y) \perp M$. Finally suppose $y \in M$ is any point such that $(x - y) \perp M$. Then for $z \in M$, by Pythagorean's theorem,

$\|x - z\|_{\#}^2 = \|x - y + y - z\|_{\#}^2 = \|x - y\|_{\#}^2 + \|y - z\|_{\#}^2 \ge \|x - y\|_{\#}^2$ which shows $[\operatorname{dist}(0, M)]^2 \ge \|x - y\|_{\#}^2$. That is to say $y$ is the point in $M$ closest to $x$.

**Definition 1.6.** $A : H^{\#} \to H^{\#}$ is a bounded in $*\mathbb{R}_c^{\#}$ operator if and only if there exists some $M \in *\mathbb{R}_c^{\#}$, $M > 0$ such that for all $x \in H^{\#}$, $\|Ax\|_{\#} \le M\|x\|_{\#}$. The smallest such $M$ if exists is called the operator #-norm of $A$ and denoted by $\|A\|_{\#OP}$ or $\|A\|_{\#}$. Thus

$$ \|A\|_{\#} = \sup_{\|x\|_{\#}=1} (\|Ax\|_{\#}) < *\infty \quad (1.9) $$

if supremum in RHS of (1.9) exists and $\sup_{\|x\|_{\#}=1}(\|Ax\|_{\#}) < *\infty$. Conversely if (1.9) holds

**Proposition 1.2.** A linear operator $A : H_1^{\#} \to H_2^{\#}$ between #-normed spaces is bounded in $*\mathbb{R}_c^{\#}$ if and only if it is #-continuous.

**Proof.** Suppose that $A$ is bounded in $*\mathbb{R}_c^{\#}$ Then, for all vectors $x, h \in H_1^{\#}$ with $h \approx 0$ non zero we have $\|A(x + h) - A(x)\|_{\#} = \|A(h)\|_{\#} \le M\|h\|_{\#}, M \in *\mathbb{R}_c^{\#}, M > 0$. Letting $h$ go to zero shows that $A$ is #-continuous at $x$. Moreover, since the constant $M$ does not depend on $x$, this shows that in fact $A$ is uniformly #-continuous, and even Lipschitz #-continuous.

Conversely, it follows from the #-continuity at the zero vector that there exists a $\varepsilon \approx 0$, $\varepsilon > 0$ such that $\|A(h)\|_{\#} = \|A(h) - A(0)\|_{\#} \le 1$ for all vectors $h \in H_1^{\#}$ with $\|h\|_{\#} \le \varepsilon$. Thus, for all non-zero $x \in H_1^{\#}$, one has

$$ \|Ax\|_{\#} = \left\| \frac{\|x\|_{\#}}{\varepsilon} A \left( \varepsilon \frac{\|x\|_{\#}}{x} \right) \right\|_{\#} = \frac{\|x\|_{\#}}{\varepsilon} \left\| A \left( \varepsilon \frac{\|x\|_{\#}}{x} \right) \right\|_{\#} \le \frac{\|x\|_{\#}}{\varepsilon}. $$

This proves that $A$ is bounded in $*\mathbb{R}_c^{\#}$.

**Definition 1.7.** Suppose that $A : H^{\#} \to H^{\#}$ is a bounded in $*\mathbb{R}_c^{\#}$ operator. The #-adjoint of $A$, denote $A^*$, is the unique operator $A^* : H^{\#} \to H^{\#}$ such that $\langle Ax, y \rangle_{\#} = \langle x, A^*y \rangle_{\#}$. (The proof that $A^*$ exists and is unique will be given in Proposition below.) A bounded in $*\mathbb{R}_c^{\#}$ operator $A : H^{\#} \to H^{\#}$ is self #-adjoint or Hermitian if $A = A^*$.

**Definition 1.8.** Let $H^{\#}$ be a non-Archimedean Hilbert space and $M \subset H$ be a #-closed subspace. The orthogonal projection of $H^{\#}$ onto $M$ is the function $P_M : H^{\#} \to H^{\#}$ such that for $x \in H^{\#}$, $P_M(x)$ is the unique element in $M$ such that $(x - P_M(x)) \perp M$.

**Proposition 1.3.** Let $H^{\#}$ be a non-Archimedean Hilbert space and $M \subset H^{\#}$ be a #-closed subspace. The orthogonal projection $P_M$ satisfies:

(1) $P_M$ is linear (and hence we will write $P_M x$ rather than $P_M(x)$.

(2) $P_M^2 = PM$ ($P_M$ is a projection).

(3) $P_M^* = P_M$, ($P_M$ is self-#-adjoint).

(4) $\operatorname{Ran}(P_M) = M$ and $\operatorname{ker}(P_M) = M^{\perp}$.

**Proof.** (1) Let $x_1, x_2 \in H^{\#}$ and $\alpha \in *\mathbb{R}_c^{\#}$, then $P_M x_1 + \alpha P_M x_2 \in M$ and $P_M x_1 + \alpha P_M x_2 - (x_1 + \alpha x_2) = [P_M x_1 - x_1 + \alpha (P_M x_2 - x_2)] \in M^{\perp}$ showing $P_M x_1 + \alpha P_M x_2 = P_M(x_1 + \alpha x_2)$, i.e. $P_M$ is linear.

(2) Obviously $\operatorname{Ran}(P_M) = M$ and $P_M x = x$ for all $x \in M$. Therefore $P_M^2 = P_M$.

(3) Let $x, y \in H^{\#}$, then since $(x - P_M x) \in M^{\perp}$ and $(y - P_M y) \in M^{\perp}$, $\langle P_M x, y \rangle_{\#} = \langle P_M x, P_M y + y - P_M y \rangle_{\#} = \langle P_M x, P_M y \rangle_{\#} = \langle P_M x + (x - P_M), P_M y \rangle_{\#} = \langle x, P_M y \rangle_{\#}$.

(4) It is clear that $\operatorname{Ran}(P_M) \subset M$. Moreover, if $x \in M$, then $P_M x = x$ implies
$0 \le f \le s$ g #-a.e., then

$$ \mathrm{Ext}-\int_X f(x)d^\#\mu^\# \le \mathrm{Ext}-\int_X g(x)d^\#\mu^\#. $$

#-Convergence theorem

**Definition 11.4.** We say that a hyper infinite sequence {$f_n$}$_{n=1}^{\infty}$ of #-integrable functions $L_1^#$ #-converges to $f$ (or #-converges in $L_1^+(X, \mathcal{F}, \mu^#)$) if

$$ \mathrm{Ext}-\int_X |f_n - f| d^\# \mu^\# \to_\# 0 \quad \text{as} \quad n \to^* \infty. \qquad (11.11) $$

**Theorem 11.1** (The monotone #-convergence theorem) If {$f_n$}$_{n=1}^{\infty}$ is a hyper infinite sequence in $L_1^+(X, \mathcal{F}, \mu^#)$ such that $f_j \le s$ $f_{j+1}$ for all $j$ and $f(x) = \sup_{n \in \mathbb{N}} f_n(x)$ then

$$ \mathrm{Ext}-\int_X f(x)d^\#\mu^\# = \#\mathrm{-lim}_{n \to^* \infty} \mathrm{Ext}-\int_X f_n(x)d^\#\mu^\#. \qquad (11.12) $$

**Proof:** The #-limit of the increasing sequence

($\ast$-finite or $\ast$-infinite) exists. Moreover by (3.2),

$$ \mathrm{Ext}-\int_X f_n(x)d^\#\mu^\# \le \mathrm{Ext}-\int_X f(x)d^\#\mu^\# $$

for all $n \in \mathbb{N}$, so

$$ \#\mathrm{-lim}_{n \to^* \infty} \left( \mathrm{Ext}-\int_X f_n(x)d^\#\mu^\# \right) \le \mathrm{Ext}-\int_X f(x)d^\#\mu^\#. $$

To establish the reverse inequality, fix $\alpha \in (0,1)$, let $\varphi$ be a simple function with $0 \le \varphi \le f$, and let $E_n = \{x : f_n(x) \ge \alpha\varphi(x)\}$. Then $(E_n)^{\infty}_{n=1}$ is an increasing hyper infinite sequence of #-measurable sets whose union is $X$, and we have

$$ \mathrm{Ext}-\int_X f_n(x)d^\#\mu^\# \ge \mathrm{Ext}-\int_{E_n} f_n(x)d^\#\mu^\# \ge \alpha \left( \mathrm{Ext}-\int_{E_n} \varphi(x)d^\#\mu^\# \right) \quad (11.13) $$

By (11.10) and by Proposition 2.5,

$$ \#\mathrm{-lim}_{n \to^* \infty} \left( \mathrm{Ext}-\int_{E_n} \varphi(x)d^\#\mu^\# \right) = \mathrm{Ext}-\int_X \varphi(x)d^\#\mu^\#, \qquad (11.14) $$

and hence

$$ \#\mathrm{-lim}_{n \to^* \infty} \left( \mathrm{Ext}-\int_{E_n} f_n(x)d^\#\mu^\# \right) \ge \alpha \left( \mathrm{Ext}-\int_X \varphi(x)d^\#\mu^\# \right). \qquad (10.15) $$

Since this is true for all $\alpha, 0 < \alpha < 1$, it remains true for $\alpha = 1$ :

$$ \#\mathrm{-lim}_{n \to^* \infty} \left( \mathrm{Ext}-\int_{E_n} f_n(x)d^\#\mu^\# \right) \ge \mathrm{Ext}-\int_X \varphi(x)d^\#\mu^\#. \qquad (10.16) $$

§ 11. #-Convergence in $\mathbb{R}^\#$-Valued #-measure.

**Definition 11.1.** We say that a hyper infinite sequence $(f_n)^{\infty}_{n=1}$ of #-measurable functions on $(X, M, \mu^#)$ is Cauchy in $\mathbb{R}^\#$-Valued #-measure if, for every $\varepsilon \approx 0, \varepsilon > 0$,

$$ \mu^\#(\{x : |f_n(x) - f_m(x)| \ge \varepsilon\}) \to_\# 0 \quad \text{as} \quad m, n \to^* \infty, \qquad (11.1) $$

and that $(f_n)^{\infty}_{n=1}$ #-converges in #-measure to $f$ if, for every $\varepsilon \approx 0, \varepsilon > 0$,

$$ \mu^\#(\{x : |f_n(x) - f(x)| \ge \varepsilon\}) \to_\# 0 \quad \text{as} \quad n \to^* \infty. \qquad (11.2) $$

**Proposition 11.1.** If $f_n \to_\# f$ in $L^1$ then $f_n \to_\# f$ in #-measure.

**Proof.** Let $E_{n,\varepsilon} = \{x : |f_n(x) - f(x)| \ge \varepsilon\}$. Then
that $\mathrm{Ran}(P_M) = M$. Now $x \in \mathrm{ker}(P_M)$ iff $P_M x = 0$ iff $x = x - 0 \in M^{\perp}$.

**Corollary 1.2.** Suppose that $M \subset H^\#$ is a proper closed subspace of a non-Archiedean Hilbert space $H^\#$, then $H^\# = M \oplus M^{\perp}$.

**Proof.** Given $x \in H^\#$, let $y = P_M x$ so that $x - y \in M^{\perp}$. Then $x = y + (x - y) \in M \oplus M^{\perp}$. If $x \in M \cap M^{\perp}$, then $x \perp x$, i.e. $\|x\|_\#^2 = \langle x, z \rangle_\# = 0$. So $M \cap M^{\perp} = \{0\}$.

**Proposition 1.4.** (Generalized Riesz Theorem). Let $H^{\#*}$ be the dual space of $H^\#$. The map

$$z \in H^{\#} \xrightarrow{j} \langle \cdot, z \rangle_{\#} \in H^{\#*} \qquad (1.9')$$

is a conjugate linear #-isometric isomorphism.

**Proof.** The map $j$ is conjugate linear by the axioms of the non-Archiedean inner products. Moreover, for $x, z \in H^\#$, $\langle x, z \rangle_\# \le \|x\|_\# \|z\|_\#$ for all $x \in H^\#$ with equality when $x = z$. This implies that $\|jz\|_{H^{\#*}} = \langle \cdot, z \rangle_{H^{\#*}} = \|z\|_\#$. Therefore $j$ is #-isometric and this shows that $j$ is injective. To finish the proof we must show that $j$ is surjective. So let $f \in H^{\#*}$ which we assume with out loss of generality is non-zero. Then $M = \mathrm{ker}(f)$ is a #-closed proper subspace of $H^\#$. Since, by Corollary 1.1, $H^\# = M \oplus M^{\perp}$, $f: H^\#/M \approx M^{\perp} \to *\mathbb{C}_c^\#$ is a linear isomorphism. This shows that $\dim(M^{\perp}) = 1$ and hence $H^\# = M \oplus *\mathbb{C}_c^\# x_0$ where $x_0 \in M^{\perp}\setminus\{0\}$. Alternatively, choose $x_0 \in M^{\perp}\setminus\{0\}$ such that $f(x_0) = 1$. For $x \in M^{\perp}$ we have $f(x - \lambda x_0) = 0$ provided that $\lambda = f(x)$. Therefore $x - \lambda x_0 \in M \cap M^{\perp} = \{0\}$, i.e. $x = \lambda x_0$. This again shows that $M^{\perp}$ is spanned by $x_0$. Choose $z = \lambda x_0 \in M^{\perp}$ such that $f(x_0) = \langle x_0, z \rangle$. (So $\lambda = \overline{f(x_0)}/\|x_0\|^2$.) Then for $x = m + \lambda x_0$ with $m \in M$ and $\lambda \in *\mathbb{C}_c^\#$, $f(x) = \lambda f(x_0) = \lambda \langle x_0, z \rangle_\# = \langle \lambda x_0, z \rangle_\# = \langle m + \lambda x_0, z \rangle_\# = \langle x, z \rangle_\#$ which shows that $f = jz$.

**Proposition 1.5.** (Adjoints). Let $H^\#$ and $K^\#$ be a non-Archiedean Hilbert spaces and $A: H^\# \to K^\#$ be a bounded in $*\mathbb{R}_c^\#$ operator. Then there exists a unique bounded operator $A^*: K^\# \to H^\#$ such that

$$\langle Ax, y \rangle_{K^\#} = \langle x, A^*y \rangle_{H^\#} \qquad (1.10)$$

for all $x \in H^\#$ and $y \in K^\#$. Moreover $(A + \lambda B)^* = A^* + \bar{\lambda}B^*$, $A^{**} \cong (A^*)^* = A$, $\|A^*\|_\# = \|A\|_\#$ and $\|A^*A\|_\# = \|A\|_\#^2$ for all $A, B \in L(H^\#, K^\#)$ and $\lambda \in *\mathbb{C}_c^\#$.

**Proof.** For each $y \in K^\#$, then map $x \to \langle Ax, y \rangle_{K^\#}$ is in $H^{\#*}$ and therefore there exists by Proposition 12.15 a unique vector $z \in H^\#$ such that $\langle Ax, y \rangle_{K^\#} = \langle x, z \rangle_{H^\#}$ for all $x \in H^\#$. This shows there is a unique map $A^*: K^\# \to H^\#$ such that $\langle Ax, y \rangle_{K^\#} = \langle x, A^*(y) \rangle_{H^\#}$ for all $x \in H^\#$ and $y \in K^\#$. To finish the proof, we need only show $A^*$ is linear and bounded in $*\mathbb{R}_c^\#$ operator. To see $A^*$ is linear, let $y_1, y_2 \in K^\#$ and $\lambda \in *\mathbb{C}_c^\#$, then for any $x \in H^\#$, $\langle Ax, y_1 + \lambda y_2 \rangle_{K^\#} = \langle Ax, y_1 \rangle_{K^\#} + \bar{\lambda} \langle Ax, y_2 \rangle_{K^\#} = \langle x, A^*(y_1) \rangle_{K^\#} + \bar{\lambda} \langle x, A^*(y_2) \rangle_{K^\#} = \langle x, A^*(y_1) + \lambda A^*(y_2) \rangle_{K^\#}$ and by the uniqueness of $A^*(y_1 + \lambda y_2)$ we find $A^*(y_1 + \lambda y_2) = A^*(y_1) + \lambda A^*(y_2)$.

This shows $A^*$ is linear and so we will now write $A^*y$ instead of $A^*(y)$. Since $\langle A^*y, x \rangle_{H^\#} = \langle x, A^*y \rangle_{H^\#} = \langle Ax, y \rangle_{K^\#} = \langle y, Ax \rangle_{K^\#}$ it follows that $A^{**} = A$. The assertion that $(A + \lambda B)^* = A^* + \bar{\lambda}B^*$ is left to the reader.

The following arguments prove the assertions about #-norms of $A$ and $A^*$:

$$\begin{align*}
\|A^*\|_\# &= \sup_{k \in K^\#, \|k\|_\# = 1} \|A^*k\|_\# = \sup_{k \in K^\#, \|k\|_\# = 1} \sup_{h \in H^\#, \|h\|_\# = 1} |\langle A^*k, h \rangle_\#| \\
&= \sup_{h \in H^\#, \|h\|_\# = 1} \sup_{k \in K^\#, \|k\|_\# = 1} |\langle k, Ah \rangle_\#| = \sup_{h \in H^\#, \|h\|_\# = 1} \|Ah\|_\# = \|A\|_\#, \\
\|A^*A\|_\# &\le \|A^*\|_\# \|A\|_\# = \|A\|_\#^2 \text{ and}
\end{align*}$$
$$ \mathrm{Ext}-\int_X |f_n - f| d\mu^{\#} \ge \mathrm{Ext}-\int_{E_{n,\varepsilon}} |f_n - f| d\mu^{\#} \ge \varepsilon \mu^{\#}(E_{n,\varepsilon}), $$

$$ \text{so } \mu(E_{n,\varepsilon}) \le \varepsilon^{-1} \mathrm{Ext}-\int_X |f_n - f| d\mu^{\#} \to_{\#} 0. $$

**Theorem 11.1.** Suppose that $(f_n)_{n=1}^{\infty}$ is Cauchy in #-measure. Then there is a #-measurable function $f$ such that $f_n \to_{\#} f$ in #-measure, and there is a hyper infinite subsequence $(f_{n_j})_{j \in \mathbb{N}}$ that #-converges to $f$ #-a.e. Moreover, if $f_n \to_{\#} g$ in #-measure then $g = f$ #-a.e.

**Proof.** We can choose a hyper infinite subsequence $(g_j)_j = (f_{n_j})_j$ of $(f_n)_{n=1}^{\infty}$ such that if $E_j = \{x : |g_j(x) - g_{j+1}(x)| \ge 2^{-j}\}$ then $\mu^{\#}(E_j) \le 2^{-j}$. If $F_k = \bigcup_{j=k}^{\infty} E_j$ then $\mu^{\#}(F_k) \le \mathrm{Ext}-\sum_{j=k}^{\infty} 2^{-j} = 2^{1-k}$, and if $x \notin F_k$ we have for $i \ge j \ge k$

$$ |g_j(x) - g_i(x)| \le \mathrm{Ext}-\sum_{l=j}^{i-1} |g_{l+1}(x) - g_l(x)| \le \mathrm{Ext}-\sum_{l=j}^{i-1} 2^{1-j}. \quad (11.3) $$

Thus $(g_j)_j$ is pointwise Cauchy on $F_k^c$. Let $F = \bigcap_{k=1}^{\infty} F_k = \limsup_j E_j$. Then $\mu^{\#}(F) = 0$, and if we set $f(x) = \lim_{j \to \infty} g_j(x)$ for $x \notin F$, and $f(x) = 0$ for $x \in F$, then $f$ is #-measurable and $g_j \to_{\#} f$ a.e. By (11.3), we have that $|g_j(x) - f(x)| \le 2^{-j}$ for $x \notin F_k$ and $j \ge k$. Since $\mu^{\#}(F_k) \to_{\#} 0$ as $k \to \infty$, it follows that $g_j \to_{\#} f$ in #-measure, because

$$ \{x : |f_n(x) - f(x)| \ge \varepsilon\} \subseteq \{x : |f_n(x) - g_j(x)| \ge (1/2)\varepsilon\} \cup \{x : |g_j(x) - f(x)| \ge (1/2)\varepsilon\}, \quad (11.4) $$

and the sets on the right both have infinite small #-measure when $n$ and $j$ are infinite large. Likewise, if $f_n \to_{\#} g$ in #-measure

$$ \{x : |f(x) - g(x)| \ge \varepsilon\} \subseteq \{x : |f(x) - f_n(x)| \ge (1/2)\varepsilon\} \cup \{x : |f_n(x) - g(x)| \ge (1/2)\varepsilon\} \quad (4.5) $$

for all $n \in \mathbb{N}$, hence $\mu^{\#}(\{x : |f(x) - g(x)| \ge \varepsilon\}) = 0$ for all $\varepsilon > 0$, and $f = g$ #-a.e.

**Theorem 3.2** Let $f_n \to_{\#} f$ in $L_1^\#$ then there is a hyper infinite subsequence $(f_{n_k})_k$ such that $f_{n_k} \to_{\#} f$ #-a.e.

**Proof.** Let $E_{n, \varepsilon} = \{x : |f_n(x) - f(x)| \ge \varepsilon\}$. Then

$$ \mathrm{Ext}-\int_X |f_n - f| d\mu^{\#} \ge \mathrm{Ext}-\int_{E_{n,\varepsilon}} |f_n - f| d\mu^{\#} \ge \varepsilon \mu^{\#}(E_{n,\varepsilon}), $$

so $\mu^{\#}(E_{n,\varepsilon}) \to_{\#} 0$. Then, by Theorem 3.1, there is a hyper infinite subsequence $(f_{n_k})_k$ such that $f_{n_k} \to_{\#} f$ #-a.e.

# 5. The Extension of #-Measure

## § 5.1. Outer #-measures.

**Definition 5.1.1.** Let $X$ be a nonempty set. An outer #-measure (or #-submeasure) on $X$ is a function $\zeta^{\#} : \tilde{P}(X) \to [0, \infty]$, $\tilde{P}(X) \subset P(X)$ that satisfies:

(a) $\zeta^{\#}(\emptyset) = 0$;

(b) $\zeta^{\#}(A) \le \zeta(B)$ if $A \subseteq B$;

(c) $\zeta^{\#}(\bigcup_{j=1}^{\infty} A_j) \le \mathrm{Ext}-\sum_{j=1}^{\infty} \zeta^{\#}(A_j)$ for all hyper infinite sequences $(A_j)_{j=1}^{\infty}$ in $\tilde{P}(X)$.
$$
\begin{align*}
\|A\|_{\#}^2 &= \sup_{h \in H^{\#}, \|h\|_{\#} = 1} |\langle Ah, Ah \rangle_{\#}| = \sup_{h \in H^{\#}, \|h\|_{\#} = 1} |\langle h, A^*Ah \rangle_{\#}| \le \sup_{h \in H^{\#}, \|h\|_{\#} = 1} \|A^*Ah\| = \\
&= \|A^*A\|_{\#}.
\end{align*}
$$

**Corollary 1.3.** Let $H^{\#}, K^{\#}, M^{\#}$ be non-Archiedean Hilbert space, $A, B \in L(H^{\#}, K^{\#})$, $C \in L(K^{\#}, M^{\#})$ and $\lambda \in *\tilde{C}_{c}^{\#}$. Then $(A + \lambda B)^{*} = A^{*} + \bar{\lambda}B^{*}$ and $(CA)^{*} =$
$= A^{*}C^{*} \in L(M^{\#}, H^{\#}).$

**Corollary 1.4.** Let $H^{\#} = *\tilde{C}_{c}^{\#}$ and $K^{\#} = *\tilde{C}_{c}^{\#}$ equipped with the canonical inner

products, i.e. $\langle z, w \rangle_{H^{\#}} = \text{Ext} - \sum_{1 \le i \le n} z_i \cdot \bar{w}_i$ for $z, w \in H^{\#}$. Let $A$ be an $m \times n$ external hyperfinite matrix thought of as a linear operator from $H^{\#}$ to $K^{\#}$. Then the hyperfinite matrix associated to $A^{*} : K^{\#} \to H^{\#}$ is the conjugate transpose of $A$.

**Corollary 1.5.** Let $K : L_2^\#(\nu^\#) \to L_2^\#(\mu^\#)$ be the operator defined in Corollary 1.3.
Then $K^* : L_2^\#(X, \mu^\#) \to L_2^\#(X, \nu^\#)$ is the operator given by

$$
K^*g(y) = \text{Ext} - \int_X \overline{k(x,y)} g(x) d^\# \mu^\#(x).
$$

**Definition 1.9.** $\{u_\alpha\}_{\alpha \in A} \subset H^\#$ is an orthonormal set if $u_\alpha \perp u_\beta$ for all $\alpha \neq \beta$ and $\|u_\alpha\|_\# = 1$.

**Proposition 1.6 (Generalized Bessel's Inequality).** Let $\{u_\alpha\}_{\alpha \in A}$ be an orthonormal set, then

$$
\mathrm{Ext} - \sum_{\alpha \in A} |\langle x, u_\alpha \rangle_\#|^2 \le \|x\|_\#^2 \quad (1.11)
$$

for all $x \in H^#$. In particular the set $\{\alpha \in A : \langle x, u_\alpha \rangle_\# \neq 0\}$ is at most $*$-countable, i.e.
$\mathrm{card}(A) = \mathrm{card}(\mathbb{N})$ for all $x \in H^#$.

**Proof.** Let $\Gamma \subset A$ be any hyperfinite set. Then

$$
0 \le \left\| x - \mathrm{Ext} - \sum_{\alpha \in \Gamma} \langle x, u_\alpha \rangle_\# u_\alpha \right\|_{\#}^2 = \|x\|_{\#}^2 - 2 \mathrm{Re} \left( \mathrm{Ext} - \sum_{\alpha \in \Gamma} \langle x, u_\alpha \rangle_\# \langle u_\alpha, x \rangle_\# \right) + \\
+ \mathrm{Ext} - \sum_{\alpha \in \Gamma} |\langle x, u_\alpha \rangle_\#|^2 = \|x\|_{\#}^2 - \mathrm{Ext} - \sum_{\alpha \in \Gamma} |\langle x, u_\alpha \rangle_\#|^2 \quad (1.12)
$$

and (1.12) gives that

$$
\mathrm{Ext} - \sum_{\alpha \in \Gamma} |\langle x, u_\alpha \rangle_\#|^2 \le \|x\|_\#^2. \quad (1.13)
$$

Taking the supremum of the inequality (1.13) of $\Gamma \subset \subset A$ then proves (1.11).

**Proposition 1.7.** Suppose $A \subset H^\#$ is an orthogonal set. Then $s = \text{Ext} - \sum_{v \in A} v$ exists in $H^\#$ iff $\text{Ext} - \sum_{v \in A} \|v\|_\#^2 < *\infty$. In particular $A$ must be at most a $*$-countable set.

Moreover, $\text{Ext} - \sum_{v \in A} \|v\|_\#^2 < *\infty$, then

$$
(1) \quad \|s\|_{\#}^{2} = \text{Ext} - \sum_{v \in A} \|v\|_{\#}^{2} \quad \text{and}
$$

$$
(2) \quad \langle s, x \rangle_{\#} = \mathrm{Ext} - \sum_{v \in A} \langle v, x \rangle_{\#} \quad \text{for all } x \in H^{\#}.
$$

Similarly if $\{v_n\}_{n=1}^{*\infty}$ is an orthogonal set, then $s = \text{Ext} - \sum_{n=1}^{*\infty} v_n$ exists in $H^\#$ iff
$\text{Ext} - \sum_{n=1}^{*\infty} v_n < *\infty$. In particular if $\text{Ext} - \sum_{n=1}^{*\infty} v_n$ exists, then it is independent of
rearrangements of $\{v_n\}_{n=1}^{*\infty}$.