Then $g_1, g_2, \dots, g_n$ are the component functions of $G = (g_1, g_2, \dots, g_n)$.
We say that

$$ \#\lim_{U \to \#} U_0 G(U) = L = (L_1, \dots, L_n) \qquad (18.61) $$

if $\#\lim_{U \to \#} U_0 g_i(U) = L_i, 1 \le i \le n$ and that $G$ is $\#$-continuous at $U_0$ if $g_1, g_2, \dots, g_n$ are each $\#$-continuous at $U_0$.

**Theorem 18.25.** For a vector-valued function $G, \#\lim_{U \to \#} U_0 G(U) = L$ if and only if for each $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $\|G(U) - L\| < \varepsilon$ whenever $0 < \|U - U_0\| < \delta$ and $U \in D_G$. Similarly, $G$ is $\#$-continuous at $U_0$ if and only if for each $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $\|G(U) - G(U_0)\| < \varepsilon$ whenever $\|U - U_0\| < \delta$ and $U \in D_G$.

**Theorem 18.26.** Let $f$ be a $\mathbb{R}^n$-valued function defined on a subset of $\mathbb{R}^n$, and let the vector-valued function $G = (g_1, g_2, \dots, g_n)$ be defined on a domain $D_G$ in $\mathbb{R}^n$. Let the set $T = \{U | U \in D_G \text{ and } G(U) \in D_f\}$, be nonempty; and define the $\mathbb{R}^n$-valued composite function $h = f \circ G$ on $T$ by $h(U) = f(G(U)), U \in T$. Now suppose that $U_0 \in T$ and is a $\#$-limit point of $T$, $G$ is $\#$-continuous at $U_0$, and $f$ is $\#$-continuous at $X_0 = G(U_0)$. Then $h$ is $\#$-continuous at $U_0$.

**Theorem 18.27.** If $f$ is $\#$-continuous on a $\#$-compact set $S \subset \mathbb{R}^n$, then $f$ is bounded or hyperbounded on $S$.

**Theorem 18.28.** Let $f$ be $\#$-continuous on a compact set $S \subset \mathbb{R}^n$ and $\alpha = \inf_{X \in S} f(X)$, $\beta = \sup_{X \in S} f(X)$. Then $f(X_1) = \alpha$ and $f(X_2) = \beta$ for some $X_1$ and $X_2$ in $S$.

**Theorem 18.29.** (Intermediate Value Theorem) Let $f$ be $\#$-continuous on a region $S \subset \mathbb{R}^n$. Suppose that $A$ and $B$ are in $S$ and $f(A) < u < f(B)$. Then $f(C) = u$ for some $C \in S$.

**Definition 18.28.** $f$ is uniformly $\#$-continuous on a subset $S$ of its domain in $\mathbb{R}^n$ if for every $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $|f(X) - f(X_0)| < \varepsilon$ whenever $\|X - X_0\| < \delta$ and $X, X_0 \in S$.

**Theorem 18.30.** If $f$ is $\#$-continuous on a $\#$-compact set $S \subset \mathbb{R}^n$ then $f$ is uniformly $\#$-continuous on $S$.

## 18.8. Partial $\#$-Derivatives and the $\#$-Differential

**Definition 18.29.** Let $\Phi$ be a unit vector and $X$ a point in $\mathbb{R}^n$. The directional $\#$-derivative of $f(X)$ at $X$ in the direction of $\Phi$ is defined by

$$ \frac{\partial'^{\#} f(X)}{\partial'^{\#} \Phi} = \# - \lim_{t \to \# 0} \frac{f(X + t\Phi) - f(X)}{t} \qquad (18.62) $$

if the $\#$-limit exists. That is, $\partial'^{\#} f(X)/\partial'^{\#} \Phi$ is the ordinary derivative of the function $H(t) = f(X + t\Phi)$ at $t = 0$, if $H^{\#}(t)$ exists. The directional $\#$-derivatives that we are most interested in are those in the directions of the unit vectors $E_i, 1 \le i \le n$, where all components of $E_i$ are zero except for the $i$-th, which is 1.

**Definition 18.30.** Since $X$ and $X + tE_i$ differ only in the $i$-th coordinate, $\partial'^{\#} f(X)/\partial'^{\#} E_i$ is called the partial $\#$-derivative of $f$ with respect to $x_i$ at $X$. It is also denoted by $\partial'^{\#} f(X)/\partial'^{\#} x_i$ or $f_{x_i}^{\#}(X)$, thus,

$$ \frac{\partial'^{\#} f(X)}{\partial'^{\#} x_i} = f_{x_i}^{\#}(X) = \# - \lim_{t \to \# 0} \frac{f(\{x_i + t\}_{i \in n}) - f(\{x_i\}_{i \in n})}{t} \qquad (18.63) $$
it is an r th-order partial derivative of $f(\mathbf{X})$.

**Theorem 18.32.** Suppose that $f, f_x^{\#}, f_y^{\#}$, and $f_{xy}^{\#}$ exist on a \#-neighborhood $\Omega$ of $(x_0, y_0)$, and $f_{xy}^{\#}$ is \#-continuous at $(x_0, y_0)$. Then $f_{yx}^{\#}$ $(x_0, y_0)$ exists, and

$$f_{yx}^{\#}(x_0, y_0) = f_{xy}^{\#}(x_0, y_0). \qquad (18.70)$$

**Theorem 18.33.** Suppose that $f$ and all its partial \#-derivatives of order $r$ are \#-continuous on an \#-open subset $S$ of $\mathbb{R}_c^{\#n}$. Then

$$f_{x_1, x_2, \dots, x_r}^{\#}(\mathbf{X}) = f_{x_1, x_2, \dots, x_r}^{\#}(\mathbf{X}), \mathbf{X} \in \mathcal{S}, \qquad (18.71)$$

if each of the variables $x_1, x_2, \dots, x_n$ appears the same number of times in $\{x_{i_1}, x_{i_2}, \dots, x_{i_r}\}$ and $\{x_{j_1}, x_{j_2}, \dots, x_{j_r}\}$. If this number is $r_k$, we denote the common value of the two sides of (18.71) by

$$\frac{\partial^{r^{\#}} f(\mathbf{X})}{\partial^{\prime \#} x_{i_r} \partial^{\prime \#} x_{i_{r-1}} \dots \partial^{\prime \#} x_{i_1}}. \qquad (18.72)$$

**Definition 18.31.** A function $f(\mathbf{X})$ is \#-differentiable at $\mathbf{X}_0 = (x_{10}, x_{20}, \dots, x_{n0})$ if there are constants $m_1, m_2, \dots, m_n$ such that

$$\#\lim_{\|\mathbf{X}-\mathbf{X}_0\| \to 0} \frac{f(\mathbf{X}) - f(\mathbf{X}_0) - (\mathrm{Ext} - \sum_{i=1}^n m_i (x_i - x_{i0}))}{\|\mathbf{X} - \mathbf{X}_0\|} = 0. \qquad (18.73)$$

**Theorem 18.34.** If $f$ is differentiable at $\mathbf{X}_0 = (x_{10}, x_{20}, \dots, x_{n0})$, then $f_{x_0}^{\#}(\mathbf{X}_0)$, $1 \le i \le n$, exist and the constants $m_i$, $1 \le i \le n$, in Eq. (18.73) are given by

$$m_i = f_{x_{i0}}^{\#}(\mathbf{X}_0). \qquad (18.74)$$

**Theorem 18.35.** If $f$ is \#-differentiable at $\mathbf{X}_0$, then $f$ is \#-continuous at $\mathbf{X}_0$.

**Definition 18.32.** A linear function $L: \mathbb{R}_c^{\#n} \to \mathbb{R}_c^{\#}$ is a $\mathbb{R}_c^{\#}$-valued function of the form

$$L(\mathbf{X}) = \mathrm{Ext} - \sum_{i=1}^{n} m_i x_i, \qquad (18.75)$$

where $m_i$, $1 \le i \le n$ are constants. From Definition 14.31, $f$ is \#-differentiable at $\mathbf{X}_0$ if and only if there is a linear function $L$ such that $f(\mathbf{X}) - f(\mathbf{X}_0)$ can be approximated so well near $\mathbf{X}_0$ by

$$f(\mathbf{X}) - f(\mathbf{X}_0) = L(\mathbf{X} - \mathbf{X}_0) + E(\mathbf{X}) \|\mathbf{X} - \mathbf{X}_0\|, \qquad (18.76)$$

where

$$\#\lim_{\|\mathbf{X}-\mathbf{X}_0\| \to 0} E(\mathbf{X}) = 0. \qquad (18.77)$$

**Remark 18.3.** Theorem 18.34 implies that if $f$ is \#-differentiable at $\mathbf{X}_0$, then there is exactly one linear function $L$ that satisfies (18.76) and (18.77).

This function is called the \#-differential of $f$ at $\mathbf{X}_0$. We will denote it by $d_{\mathbf{X}_0}^{\#} f$ and its value by $(d_{\mathbf{X}_0}^{\#} f)(\mathbf{X})$; thus,

$$(d_{\mathbf{X}_0}^{\#} f)(\mathbf{X}) = \mathrm{Ext} - \sum_{i=1}^{n} f_{x_{i0}}^{\#}(\mathbf{X}_0) x_i. \qquad (18.78)$$

For convenience in writing $d_{\mathbf{X}_0}^{\#} f$, and to conform with standard notation, we introduce the function $d^{\prime \#} x_i : \mathbb{R}_c^{\#n} \to \mathbb{R}_c^{\#}$ defined by $dx_i(\mathbf{X}) = x_i$. That is, $d^{\prime \#} x_i$ is the function whose value at a point in $\mathbb{R}_c^{\#n}$ is the i-th coordinate of the point. It is the \#-differential of the function $g_i(\mathbf{X}) = x_i$. From Eq. (18.78)

$$d_{\mathbf{X}_0}^{\prime \#} f = \mathrm{Ext} - \sum_{i=1}^{n} f_{x_{i0}}^{\prime \#}(\mathbf{X}_0) d^{\prime \#} x_i. \qquad (18.78)$$

## 19. \#-Analytic functions $f: \mathbb{C}_c^{\#} \to \mathbb{C}_c^{\#}$.
If $\mathbf{X} = (x, y)$, then we denote the partial #-derivatives accordingly; thus,

$$ \frac{\partial^{\prime\#} f(x, y)}{\partial^{\prime\#} x} = f_{x}^{\prime\#}(x, y) = \#- \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h} \quad (18.64) $$

and

$$ \frac{\partial^{\prime\#} f(x, y)}{\partial^{\prime\#} y} = f_{y}^{\prime\#}(x, y) = \#- \lim_{h \to 0} \frac{f(x, y+h) - f(x, y)}{h}. \quad (18.65) $$

**Theorem 18.31.** If $f_{x_i}^{\prime\#}(\mathbf{X})$ and $g_{x_i}^{\prime\#}(\mathbf{X})$ exist, then

$$ \frac{\partial^{\prime\#}(f+g)(\mathbf{X})}{\partial^{\prime\#}x_i} = f_{x_i}^{\prime\#}(\mathbf{X}) + g_{x_i}^{\prime\#}(\mathbf{X}), \quad \frac{\partial^{\prime\#}(f \times g)(\mathbf{X})}{\partial^{\prime\#}x_i} = f_{x_i}^{\prime\#}(\mathbf{X})g(\mathbf{X}) + g_{x_i}^{\prime\#}(\mathbf{X})f(\mathbf{X}), \quad (18.66) $$

and, if $g(\mathbf{X}) \neq 0$,

$$ \frac{\partial^{\prime\#}(f/g)(\mathbf{X})}{\partial^{\prime\#}x_i} = \frac{g(\mathbf{X})f_{x_i}^{\prime\#}(\mathbf{X}) - f(\mathbf{X})g_{x_i}^{\prime\#}(\mathbf{X})}{[g(\mathbf{X})]^2}. \quad (18.67) $$

If $f_{x_i}^{\prime\#}(\mathbf{X})$ exists at every point of a set $D \subset \mathbb{R}_c^{\#n}$, then it defines a function $f_{x_i}^{\prime\#}(\mathbf{X})$ on $D$. If this function has a partial #-derivative with respect to $x_j$ on a subset of $D$, we denote the partial #-derivative by

$$ \frac{\partial^{\prime\#}}{\partial^{\prime\#}x_j} \left( \frac{\partial^{\prime\#}f(\mathbf{X})}{\partial^{\prime\#}x_i} \right) = \frac{\partial^{2\prime\#}f(\mathbf{X})}{\partial^{\prime\#}x_j \partial^{\prime\#}x_i} = f_{x_ix_j}^{\prime\#}(\mathbf{X}). \quad (14.68) $$

The function obtained by differentiating $f(\mathbf{X})$ successively with respect to $x_{i_1}, x_{i_2}, \dots, x_{i_r}$ is denoted by

$$ \frac{\partial^{r\prime\#}f(\mathbf{X})}{\partial^{\prime\#}x_{i_r} \partial^{\prime\#}x_{i_{r-1}} \dots \partial^{\prime\#}x_{i_1}} = f_{x_{i_1} \dots x_{i_{r-1}} x_{i_r}}^{\prime\#}(\mathbf{X}) \quad (18.69) $$

it is an $r$-th-order partial derivative of $f(\mathbf{X})$.

**Theorem 18.32.** Suppose that $f, f_x^{\prime\#}, f_y^{\prime\#}$, and $f_{xy}^{\prime\#}$ exist on a #-neighborhood $\Omega$ of $(x_0, y_0)$, and $f_{xy}^{\prime\#}$ is #-continuous at $(x_0, y_0)$. Then $f_{yx}^{\prime\#}(x_0, y_0)$ exists, and

$$ f_{yx}^{\prime\#}(x_0, y_0) = f_{xy}^{\prime\#}(x_0, y_0). \quad (18.70) $$

**Theorem 18.33.** Suppose that $f$ and all its partial #-derivatives of order $r$ are #-continuous on an #-open subset $S$ of $\mathbb{R}_c^{\#n}$. Then

$$ f_{x_1, x_2, \dots, x_r}^{\prime\#}(\mathbf{X}) = f_{x_{j_1}, x_{j_2}, \dots, x_{j_r}}^{\prime\#}(\mathbf{X}), \mathbf{X} \in S, \quad (18.71) $$

if each of the variables $x_1, x_2, \dots, x_n$ appears the same number of times in $\{x_{i_1}, x_{i_2}, \dots, x_{i_r}\}$ and $\{x_{j_1}, x_{j_2}, \dots, x_{j_r}\}$. If this number is $r_k$, we denote the common value of the two sides of (18.71) by

$$ \frac{\partial^{r\prime\#}f(\mathbf{X})}{\partial^{\prime\#}x_{i_r} \partial^{\prime\#}x_{i_{r-1}} \dots \partial^{\prime\#}x_{i_1}}. \quad (18.72) $$

**Definition 18.31.** A function $f(\mathbf{X})$ is #-differentiable at $\mathbf{X}_0 = (x_{10}, x_{20}, \dots, x_{n0})$ if there are constants $m_1, m_2, \dots, m_n$ such that

$$ \#\lim_{\|\mathbf{X}-\mathbf{X}_0\| \to 0} \frac{f(\mathbf{X}) - f(\mathbf{X}_0) - (\text{Ext} - \sum_{i=1}^n m_i(x_i - x_{i0}))}{\|\mathbf{X}-\mathbf{X}_0\|} = 0. \quad (18.73) $$

**Theorem 18.34.** If $f$ is differentiable at $\mathbf{X}_0 = (x_{10}, x_{20}, \dots, x_{n0})$, then $f_{x_0}^{\prime\#}(\mathbf{X}_0)$, $1 \le i \le n$, exist and the constants $m_i$, $1 \le i \le n$, in Eq. (18.73) are given by

$$ m_i = f_{x_{i0}}^{\prime\#}(\mathbf{X}_0). \quad (18.74) $$

**Theorem 18.35.** If $f$ is #-differentiable at $\mathbf{X}_0$, then $f$ is #-continuous at $\mathbf{X}_0$.
### 19.1. $\mathbb{C}_c^\#$-valued \#-analytic functions $f: \mathbb{C}_c^\# \to \mathbb{C}_c^\#$.

The class of \#-analytic functions is formed by the complex functions of a complex variable $z \in \mathbb{C}_c^\# = \mathbb{R}_c^\# + i\mathbb{R}_c^\#$ which possess a \#-derivative wherever the function is defined. The term \#-holomorphic function is used with identical meaning. For the purpose of this preliminary investigation the reader may think primarily of functions which are defined in the whole plane $\mathbb{C}_c^\#$.

The definition of the \#-derivative can be written in the form

$$f^{\prime\#}(z) = \text{\#-lim}_{h \to \# 0} \frac{f(z+h) - f(z)}{h} \qquad (19.1)$$

As a first obvious consequence $f(z)$ is necessarily \#-continuous. Indeed, from $f(z+h) - f(z) = h \times (f(z+h) - f(z))/h$ one obtains $\text{\#-lim}_{h \to \# 0} (f(z+h) - f(z)) = 0 \times f^{\prime\#}(z) = 0$. If we write $f(z) = u(z) + iv(z)$ it follows, moreover, that $u(z)$ and $v(z)$ are both \#-continuous.

**Remark 19.1.** When we consider the \#-derivative of a $\mathbb{C}_c^\#$-valued function, defined on a set $A \subset \mathbb{C}_c^\#$ in the complex plane $\mathbb{C}_c^\#$, it is of course understood that $z \in A$ and that the limit is with respect to values $h$ such that $z+h \in A$. The existence of the \#-derivative will therefore have a different meaning depending on whether $z$ is an interior point or a \#-boundary point of $A$. The way to avoid this is to insist that all \#-analytic functions be defined on open sets.

**Definition 19.1.** A $\mathbb{C}_c^\#$-valued function $f(z)$, defined on an open set $\Omega$, is said to be $\mathbb{C}_c^\#$-analytic in $\Omega$ if it has a \#-derivative at each point of $\Omega$. And more explicitly that $f(z)$ is \#-analytic function. A commonly used synonym is \#-holomorphic function.

**Definition 19.2.** A function $f(z)$ is \#-analytic on an arbitrary point set $A$ if it is the restriction to $A$ of a function which is \#-analytic in some open set containing $A$.

**Remark 19.2.** Note that the real and imaginary parts of an \#-analytic function in $\Omega$ satisfy the generalized Cauchy-Riemann equations

$$\frac{\partial^{\#} u}{\partial^{\#} x} = \frac{\partial^{\#} v}{\partial^{\#} y}; \quad \frac{\partial^{\#} u}{\partial^{\#} y} = -\frac{\partial^{\#} v}{\partial^{\#} x}. \qquad (19.2)$$

Conversely, if $u$ and $v$ satisfy these equations in $\Omega$, and if the partial \#-derivatives are \#-continuous, then $u + iv$ is an \#-analytic function in $\Omega$.

**Theorem 19.1.** An \#-analytic function $f$ in a region $\Omega$ whose \#-derivative vanishes identically must reduce to a constant. The same is true if either the real part, the imaginary part, the modulus, or the argument is constant.

### 19.2. The $\mathbb{C}_c^\#$-valued \#-Exponential $\text{Ext}\text{-exp}(z)$.

We define the \#-exponential $\text{Ext}\text{-exp}(z)$ function as the solution of the differential equation

$$f^{\prime\#}(z) = f(z), f(0) = 1. \qquad (19.3)$$

We solve it by setting

$$f(z) = \text{Ext}\text{-} \sum_{n=0}^{\infty} a_n z^n, f^{\prime\#}(z) = \text{Ext}\text{-} \sum_{n=0}^{\infty} n a_n z^n. \qquad (19.4)$$

If Eq. (15.4) is to be satisfied, we must have $a_{n-1} = na_n, n \in \mathbb{N}^\#$ and the initial condition gives $a_0 = 1$. It follows by hypee infinite induction that $a_n = 1/n!$.
**Definition 18.32.** A linear function $L: \mathbb{R}_c^{\#n} \to \mathbb{R}_c^\#$ is a $\mathbb{R}_c^\#$-valued function of the form

$$L(\mathbf{X}) = \mathrm{Ext}-\sum_{i=1}^{n} m_i x_i, \qquad (18.75)$$

where $m_i, 1 \le i \le n$ are constants. From Definition 14.31, $f$ is \#-differentiable at $\mathbf{X}_0$ if and only if there is a linear function $L$ such that $f(\mathbf{X}) - f(\mathbf{X}_0)$ can be approximated so well near $\mathbf{X}_0$ by

$$f(\mathbf{X}) - f(\mathbf{X}_0) = L(\mathbf{X} - \mathbf{X}_0) + E(\mathbf{X}) \|\mathbf{X} - \mathbf{X}_0\|, \qquad (18.76)$$

where

$$\#\lim_{\|\mathbf{X}-\mathbf{X}_0\| \to 0} E(\mathbf{X}) = 0. \qquad (18.77)$$

**Remark 18.3.** Theorem 18.34 implies that if $f$ is \#-differentiable at $\mathbf{X}_0$, then there is exactly one linear function $L$ that satisfies (18.76) and (18.77). This function is called the \#-differential of $f$ at $\mathbf{X}_0$. We will denote it by $d_{\mathbf{X}_0}^{\#}f$ and its value by $(d_{\mathbf{X}_0}^{\#}f)(\mathbf{X})$; thus,

$$(d_{\mathbf{X}_0}^{\#}f)(\mathbf{X}) = \mathrm{Ext}-\sum_{i=1}^{n} f_{x_{i0}}^{\#}(\mathbf{X}_0)x_i. \qquad (18.78)$$

For convenience in writing $d_{\mathbf{X}_0}^{\#}f$, and to conform with standard notation, we introduce the function $d^{\prime\#}x_i: \mathbb{R}_c^{\#n} \to \mathbb{R}_c^\#$ defined by $dx_i(\mathbf{X}) = x_i$. That is, $d^{\prime\#}x_i$ is the function whose value at a point in $\mathbb{R}_c^{\#n}$ is the $i$-th coordinate of the point. It is the \#-differential of the function $g_i(\mathbf{X}) = x_i$. From Eq. (18.78)

$$d_{\mathbf{X}_0}^{\#}f = \mathrm{Ext}-\sum_{i=1}^{n} f_{x_{i0}}^{\#}(\mathbf{X}_0)d^{\prime\#}x_i. \qquad (18.78)$$

## 19.\#-Analytic functions $f: \mathbb{C}_c^\# \to \mathbb{C}_c^\#.$

### 19.1. $\mathbb{C}_c^\#$-valued \#-analytic functions $f: \mathbb{C}_c^\# \to \mathbb{C}_c^\#$.

The class of \#-analytic functions is formed by the complex functions of a complex variable $z \in \mathbb{C}_c^\# = \mathbb{R}_c^\# + i\mathbb{R}_c^\#$ which possess a \#-derivative wherever the function is defined. The term \#-holomorphic function is used with identical meaning. For the purpose of this preliminary investigation the reader may think primarily of functions which are defined in the whole plane $\mathbb{C}_c^\#$. The definition of the \#-derivative can be written in the form

$$f^{\prime\#}(z) = \#\lim_{h \to \# 0} \frac{f(z+h) - f(z)}{h} \qquad (19.1)$$

As a first obvious consequence $f(z)$ is necessarily \#-continuous. Indeed, from $f(z+h) - f(z) = h \times (f(z+h) - f(z))/h$ one obtains $\#\lim_{h \to \# 0} (f(z+h) - f(z)) = 0 \times f^{\prime\#}(z) = 0$. If we write $f(z) = u(z) + iv(z)$ it follows, moreover, that $u(z)$ and $v(z)$ are both \#-continuous.

**Remark 19.1.** When we consider the \#-derivative of a $\mathbb{C}_c^\#$-valued function, defined on a set $A \subset \mathbb{C}_c^\#$ in the complex plane $\mathbb{C}_c^\#$, it is of course understood that $z \in A$ and that the limit is with respect to values $h$ such that $z+h \in A$. The existence of the \#-derivative will therefore have a different meaning depending on whether $z$ is an interior point or a \#-boundary point of $A$. The way to avoid this is to insist that all \#-analytic functions be defined on open sets.

**Definition 19.1.** A $\mathbb{C}_c^\#$-valued function $f(z)$, defined on an open set $\Omega$, is said to be $\mathbb{C}_c^\#$-analytic in $\Omega$ if it has a \#-derivative at each point of $\Omega$. And more explicitly that
**Abbreviation 19.1.** The solution of the Eq.(15.4) is denoted by $Ext-e^z$ or $Ext-exp(z)$ or $Ext-exp_z$. Thus finally we obtain

$$Ext-exp(z) = Ext-\sum_{n=0}^{\infty} \frac{z^n}{n!}. \qquad (19.5)$$

## 19.3. The $\mathbb{C}_c^\#$-valued Trigonometric Functions $Ext-\sin(z)$, $Ext-\cos(z)$.

The $\mathbb{C}_c^\#$-valued trigonometric functions $Ext-\sin(z)$, $Ext-\cos(z)$ are defined by

$$Ext-\sin(z) = \frac{1}{2}(Ext-exp(iz) - Ext-exp(-iz)) \qquad (19.6)$$

and

$$Ext-\cos(z) = \frac{1}{2}(Ext-exp(iz) + Ext-exp(-iz)). \qquad (19.7)$$

Substitution (19)-(19.) in (19.) gives that

$$Ext-\sin(z) = \qquad (19.8)$$

and

$$Ext-\cos(z) = \qquad (19.9)$$

From (14) we obtain generalized Euler's formula

$$Ext-exp(iz) = Ext-\cos(z) + i(Ext-\sin(z)) \qquad (19.10)$$

and as well as the identity

$$(Ext-\sin(z))^2 + (Ext-\cos(z))^2 = 1. \qquad (19.11)$$

## 19.4. The periodicity of the #-exponential $Ext-exp(iz)$.

**Definition 19.4.** We say that $f(z)$ has the period $c$ if $f(z+c) = f(z)$ for all $z \in \mathbb{C}_c^\#$. Thus a period of $Ext-e^z$ satisfies $Ext-e^{z+c} = Ext-e^z$, or $Ext-e^c = 1$. It follows that $c = i\omega$ with real $\omega \in \mathbb{R}_c^\#$ we prefer to say that $\omega$ is a period of $Ext-e^{iz}$. We shall show that there are periods, and that they are all integral multiples of a positive period $\omega_0$. From $(Ext-\sin(y))^{\prime\#} = Ext-\cos(y) \le 1$ and $Ext-\sin(0) = 0$ one obtains $Ext-\sin(y) < y$ for $y > 0$, either by integration or by use of the generalized mean-value theorem. In the same way $(Ext-\cos(y))^{\prime\#} = -Ext-\sin(y) > -y$ and $Ext-\cos(0) = 1$ gives $Ext-\cos(y) > 1 - y^2/2$, which in turn leads to $Ext-\sin(y) > y - y^3/6$ and finally to $Ext-\cos(y) < 1 - y^2/2 + y^4/24$. This inequality shows that $Ext-\cos(\sqrt{3}) < 0$, and therefore there is a $y_0$ such that $0 < y_0 < \sqrt{3}$ and $Ext-\cos(y_0) = 0$. Because $(Ext-\sin(y_0))^2 + (Ext-\cos(y_0))^2 = 1$, we have $Ext-\sin(y_0) = \pm 1$, that is, $Ext-e^{iy_0} = \pm i$, and hence $Ext-e^{4iy_0} = 1$. We have shown that $4y_0$ is a period. Actually, it is the smallest positive period. To see this, take $0 < y < y_0$.

Then $Ext-\sin(y) > y(1 - y^2/6) > y/2 > 0$, which shows that $Ext-\cos(y)$ is strictly decreasing. Because $Ext-\sin(y)$ is positive and $(Ext-\sin(y))^2 + (Ext-\cos(y))^2 = 1$ it follows that $Ext-\sin(y)$ is strictly increasing, and hence $Ext-\sin(y) < Ext-\sin(y_0) = 1$.

The double inequality $0 < Ext-\sin(y) < 1$ guarantees that $Ext-e^{iy}$ is neither $\pm 1$ nor $\pm i$. Therefore $Ext-e^{4iy} \ne 1$, and $4y_0$ is indeed the smallest positive period. We denote
$f(z)$ is #-analytic function. A commonly used synonym is #-holomorphic function.

**Definition 19.2.** A function $f(z)$ is #-analytic on an arbitrary point set $A$ if it is the restriction to $A$ of a function which is #-analytic in some open set containing $A$.

**Remark 19.2.** Note that the real and imaginary parts of an #-analytic function in $\Omega$ satisfy the generalized Cauchy-Riemann equations

$$ \frac{\partial^\# u}{\partial^\# x} = \frac{\partial^\# v}{\partial^\# y}; \quad \frac{\partial^\# u}{\partial^\# y} = -\frac{\partial^\# v}{\partial^\# x}. \qquad (19.2) $$

Conversely, if $u$ and $v$ satisfy these equations in $\Omega$, and if the partial #-derivatives are #-continuous, then $u + iv$ is an #-analytic function in $\Omega$.

**Theorem 19.1.** An #-analytic function $f$ in a region $\Omega$ whose #-derivative vanishes identically must reduce to a constant. The same is true if either the real part, the imaginary part, the modulus, or the argument is constant.

## 19.2. The $C_c^\#$-valued #-Exponential $Ext-\exp(z)$.

We define the #-exponential $Ext-\exp(z)$ function as the solution of the differential equation

$$ f^{\prime\#}(z) = f(z), f(0) = 1. \qquad (19.3) $$

We solve it by setting

$$ f(z) = \mathrm{Ext} - \sum_{n=0}^{\infty} a_n z^n, f^{\prime\#}(z) = \mathrm{Ext} - \sum_{n=0}^{\infty} n a_n z^n. \qquad (19.4) $$

If Eq.(15.4) is to be satisfied, we must have $a_{n-1} = na_n, n \in \mathbb{N}^\#$ and the initial condition gives $a_0 = 1$. It follows by hypee infinite induction that $a_n = 1/n!$.

**Abbreviation 19.1.** The solution of the Eq.(15.4) is denoted by $Ext-e^z$ or $Ext-\exp(z)$ or $Ext-\exp z$. Thus finally we obtain

$$ \mathrm{Ext}-\exp(z) = \mathrm{Ext} - \sum_{n=0}^{\infty} \frac{z^n}{n!}. \qquad (19.5) $$

## 19.3. The $C_c^\#$-valued Trigonometric Functions $Ext-\sin(z)$, $Ext-\cos(z)$.

The $C_c^\#$-valued trigonometric functions $Ext-\sin(z)$, $Ext-\cos(z)$ are defined by

$$ \mathrm{Ext}-\sin(z) = \frac{1}{2}(\mathrm{Ext}-\exp(iz) - \mathrm{Ext}-\exp(-iz)) \qquad (19.6) $$

and

$$ \mathrm{Ext}-\cos(z) = \frac{1}{2}(\mathrm{Ext}-\exp(iz) + \mathrm{Ext}-\exp(-iz)). \qquad (19.7) $$

Substitution (19)-(19.) in (19.) gives that

$$ \mathrm{Ext}-\sin(z) = \qquad (19.8) $$

and

$$ \mathrm{Ext}-\cos(z) = \qquad (19.9) $$

From (14) we obtain generalized Euler's formula

$$ \mathrm{Ext}-\exp(iz) = \mathrm{Ext}-\cos(z) + i(\mathrm{Ext}-\sin(z)) \qquad (19.10) $$

and as well as the identity
it by $\omega_0$. Consider now an arbitrary period $\omega_0$. There exists an integer $n$ such that $n\omega_0 \le \omega < (n+1)\omega_0$. If $w$ were not equal to $n\omega_0$, then $\omega - n\omega_0$ would be a positive period $< \omega_0$. Since this is not possible, every period must be an integral multiple of $\omega_0$.

**Abbreviation 19.2.** The smallest positive period of $Ext-e^{iz}$ is denoted by $2\pi_{\#}$.

**Remark 19.3.** Note that $\text{st}(\pi_{\#}) = \pi \in \mathbb{R}$.

## 19.5. The $\mathbb{C}_c^\#$-valued Logarithm.

Together with the exponential function $Ext-e^{iz}$ we must also introduce its inverse function, the $\mathbb{C}_c^\#$-valued logarithm. By definition, $z = \text{Ext-log}w$ is a root of the equation $Ext-e^{iz} = w$. First of all, since $Ext-e^{iz}$ is always $\neq 0$, the number 0 has no logarithm. For $w \neq 0$ the equation $Ext-e^{x+iv} = w$ is equivalent to

$$ Ext-e^{iz} = |w|, Ext-e^{iy} = \frac{w}{|w|}. \qquad (19.12) $$

The first equation has a unique solution $x = \text{Ext-log}|w|$, the $\mathbb{R}_c^\#$-valued logarithm of the positive number $|w| \in \mathbb{R}_c^\#$. The right-hand member of the second equation (15.12) is a complex number in $\mathbb{C}_c^\#$ of absolute value 1. Therefore, as we have just seen, it has one and only one solution in the interval $0 \le y < 2\pi_{\#}$. In addition, it is also satisfied by all $y$ that differ from this solution by an integral multiple of $2\pi_{\#}$. We see that every complex number other than 0 has hyper infinitely many logarithms which differ from each other by multiples of $2\pi_{\#}i$.

The imaginary part of $\text{Ext-log}w$ is also called the argument of $w$, $\text{Ext-arg}w$, and it is interpreted geometrically as the angle, measured in radians, between the positive real axis and the half line from 0 through the point $w$. According to this definition the argument has hyper infinitely many values which differ by multiples of $2\pi_{\#}$, and

$$ \text{Ext-log}w = \text{Ext-log}|w| + i \arg w. \qquad (19.13) $$

**Remark 19.4.** The addition property of the exponential function $Ext-e^{iz}$ implies

$$ \begin{aligned} \text{Ext-log}(z_1 \times z_2) &= \text{Ext-log}z_1 + \text{Ext-log}z_2, \\ \text{Ext-arg}(z_1 \times z_2) &= \text{Ext-arg}z_1 + \text{Ext-arg}z_2, \end{aligned} \qquad (19.14) $$

but only in the sense that both sides represent the same hyper infinite set of complex numbers. The inverse of $\text{Ext-cos}(z)$ is obtained by solving the equation

$$ \text{Ext-cos}(z) = \frac{1}{2}(\text{Ext-e}^{iz} + \text{Ext-e}^{-iz}) = w. \qquad (19.15) $$

This is a quadratic equation in $\text{Ext-e}^{iz}$ with the roots

$$ \text{Ext-e}^{iz} = w \pm \sqrt{w^2 - 1} \qquad (19.16) $$

and therefore

$$ z = \text{Ext-arccos}(w) = -i \left( \text{Ext-log}\left(w \pm \sqrt{w^2+1}\right) \right), \qquad (19.17) $$

or in the form

$$ \text{Ext-arccos}(w) = \pm i \left( \text{Ext-log}\left(w + \sqrt{w^2+1}\right) \right) \qquad (19.18) $$

The hyper infinitely many values of $\text{Ext-arccos}(w)$ reflect the evenness and periodicity of $\text{Ext-cos}(w)$. The inverse sine is most easily defined by formula

$$ \text{Ext-arcsin}(w) = \frac{\pi_{\#}}{2} - (\text{Ext-arccos}(w)). \qquad (19.19) $$
$$(\mathrm{Ext}-\sin(z))^2 + (\mathrm{Ext}-\cos(z))^2 = 1. \qquad (19.11)$$

## 19.4. The periodicity of the #-exponential Ext-exp(iz).

**Definition 19.4.** We say that $f(z)$ has the period $c$ if $f(z+c) = f(z)$ for all $z \in \mathbb{C}_c^#$. Thus a period of $\mathrm{Ext}-e^z$ satisfies $\mathrm{Ext}-e^{z+c} = \mathrm{Ext}-e^z$, or $\mathrm{Ext}-e^c = 1$. It follows that $c = i\omega$ with real $\omega \in \mathbb{R}_c^#$ we prefer to say that $\omega$ is a period of $\mathrm{Ext}-e^{iz}$. We shall show that there are periods, and that they are all integral multiples of a positive period $\omega_0$. From $(\mathrm{Ext}-\sin(y))^{\prime\#} = \mathrm{Ext}-\cos(y) \le 1$ and $\mathrm{Ext}-\sin(0) = 0$ one obtains $\mathrm{Ext}-\sin(y) < y$ for $y > 0$, either by integration or by use of the generalized mean-value theorem. In the same way $(\mathrm{Ext}-\cos(y))^{\prime\#} = -\mathrm{Ext}-\sin(y) > -y$ and $\mathrm{Ext}-\cos(0) = 1$ gives $\mathrm{Ext}-\cos(y) > 1 - y^2/2$, which in turn leads to $\mathrm{Ext}-\sin(y) > y - y^3/6$ and finally to $\mathrm{Ext}-\cos(y) < 1 - y^2/2 + y^4/24$. This inequality shows that $\mathrm{Ext}-\cos(\sqrt{3}) < 0$, and therefore there is a $y_0$ such that $0 < y_0 < \sqrt{3}$ and $\mathrm{Ext}-\cos(y_0) = 0$. Because $(\mathrm{Ext}-\sin(y_0))^2 + (\mathrm{Ext}-\cos(y_0))^2 = 1$, we have $\mathrm{Ext}-\sin(y_0) = \pm 1$, that is, $\mathrm{Ext}-e^{iy_0} = \pm i$, and hence $\mathrm{Ext}-e^{4iy_0} = 1$. We have shown that $4y_0$ is a period. Actually, it is the smallest positive period. To see this, take $0 < y < y_0$.

Then $\mathrm{Ext}-\sin(y) > y(1 - y^2/6) > y/2 > 0$, which shows that $\mathrm{Ext}-\cos(y)$ is strictly decreasing. Because $\mathrm{Ext}-\sin(y)$ is positive and $(\mathrm{Ext}-\sin(y))^2 + (\mathrm{Ext}-\cos(y))^2 = 1$ it follows that $\mathrm{Ext}-\sin(y)$ is strictly increasing, and hence $\mathrm{Ext}-\sin(y) < \mathrm{Ext}-\sin(y_0) = 1$.

The double inequality $0 < \mathrm{Ext}-\sin(y) < 1$ guarantees that $\mathrm{Ext}-e^{iy}$ is neither $\pm 1$ nor $\pm i$. Therefore $\mathrm{Ext}-e^{4iy} \ne 1$, and $4y_0$ is indeed the smallest positive period. We denote it by $\omega_0$. Consider now an arbitrary period $\omega_0$. There exists an integer $n$ such that $n\omega_0 \le \omega < (n+1)\omega_0$. If $w$ were not equal to $n\omega_0$, then $\omega - n\omega_0$ would be a positive period $< \omega_0$. Since this is not possible, every period must be an integral multiple of $\omega_0$.

**Abbreviation 19.2.** The smallest positive period of $\mathrm{Ext}-e^{iz}$ is denoted by $2\pi_{\#}$.

**Remark 19.3.** Note that $\mathrm{st}(\pi_{\#}) = \pi \in \mathbb{R}$.

## 19.5. The $\mathbb{C}_c^#$-valued Logarithm.

Together with the exponential function $\mathrm{Ext}-e^{iz}$ we must also introduce its inverse function, the $\mathbb{C}_c^#$-valued logarithm. By definition, $z = \mathrm{Ext}-\log w$ is a root of the equation $\mathrm{Ext}-e^{iz} = w$. First of all, since $\mathrm{Ext}-e^{iz}$ is always $\ne 0$, the number 0 has no logarithm. For $w \ne 0$ the equation $\mathrm{Ext}-e^{x+iy} = w$ is equivalent to

$$\mathrm{Ext}-e^{iz} = |w|, \mathrm{Ext}-e^{iy} = \frac{w}{|w|}. \qquad (19.12)$$

The first equation has a unique solution $x = \mathrm{Ext}-\log|w|$, the $\mathbb{R}_c^#$-valued logarithm of the positive number $|w| \in \mathbb{R}_c^#$. The right-hand member of the second equation (15.12) is a complex number in $\mathbb{C}_c^#$ of absolute value 1. Therefore, as we have just seen, it has one and only one solution in the interval $0 \le y < 2\pi_{\#}$. In addition, it is also satisfied by all $y$ that differ from this solution by an integral multiple of $2\pi_{\#}$. We see that every complex number other than 0 has hyper infinitely many logarithms which differ from each other by multiples of $2\pi_{\#}i$.

The imaginary part of $\mathrm{Ext}-\log w$ is also called the argument of $w$, $\mathrm{Ext}-\arg w$, and it is interpreted geometrically as the angle, measured in radians, between the positive real axis and the half line from 0 through the point $w$. According to this definition the
## 20. Complex Integration of the $C_c^\#$-valued function $f(t)$.

### 20.1. Definition and basic properties of the complex integral.

If $f(t) = u(t) + iv(t)$ is a $\#$-continuous function, defined in an interval $(a, b)$, we set by definition

$$ \operatorname{Ext}- \int_a^b f(t) d^\# t = \operatorname{Ext}- \int_a^b u(t) d^\# t + i \left( \operatorname{Ext}- \int_a^b v(t) d^\# t \right). \quad (20.1) $$

This integral has most of the properties of the real integral. In particular, if $c = \alpha + i\beta$ is a complex constant we obtain

$$ \operatorname{Ext}- \int_a^b c f(t) d^\# t = c \left( \operatorname{Ext}- \int_a^b f(t) d^\# t \right). \quad (20.2) $$

The fundamental inequality

$$ \left| \operatorname{Ext}- \int_a^b f(t) d^\# t \right| \le \operatorname{Ext}- \int_a^b |f(t)| d^\# t. \quad (20.3) $$

holds for arbitrary $C_c^\#$-valued function $f(t)$.

We consider now a piecewise $\#$-differentiable arc $\gamma$ with the equation

$$ z = z(t), a \le t \le b. $$

If the function $f(z)$ is defined and $\#$-continuous on $\gamma$, then $f(z(t))$ is also $\#$-continuous and we can set

$$ v \quad \operatorname{Ext}- \int_{\gamma} f(z) d^\# z = \operatorname{Ext}- \int_{a}^{b} f(z(t)) z^{\prime \#}(t) d^\# t. \quad (20.4) $$

The most important property of the integral (20.4) is its invariance under a change of parameter. A change of parameter is determined by an increasing function $t = t(\tau)$ which maps an interval $\alpha \le \tau \le \beta$ onto $a \le t \le b$; we assume that $t(\tau)$ is piecewise $\#$-differentiable. By the rule for changing the variable of integration we get

$$ \operatorname{Ext}- \int_a^b f(z(t)) z^{\prime \#}(t) d^\# t = \operatorname{Ext}- \int_a^b f(z(t(\tau))) z^{\prime \#}(t(\tau)) t^{\prime \#}(\tau) d^\# \tau. \quad (20.5) $$

We defined the opposite arc-$\gamma$ by the equation $z = z(-t), -b \le t \le -a$. We have thus

$$ \operatorname{Ext}- \int_{-\gamma} f(z) d^\# z = - \left( \operatorname{Ext}- \int_{\gamma} f(z) d^\# z \right). \quad (20.6) $$

The integral (20.4) has also a very obvious additive property. It is clear what is meant by subdividing an arc $\gamma$ into a finite or hyperfinite number of subarcs. A subdivision can be indicated by a symbolic equation: $\gamma = \gamma_1 + \gamma_2 + \dots + \gamma_n, n \in \mathbb{N}^\#$, and the corresponding integrals satisfy the relation
argument has hyper infinitely many values which differ by multiples of $2\pi_{\#}$, and

$$ \mathrm{Ext}\text{-log} w = \mathrm{Ext}\text{-log}|w| + i \arg w. \qquad (19.13) $$

**Remark 19.4.** The addition property of the exponential function $\mathrm{Ext}\text{-}e^{iz}$ implies

$$ \mathrm{Ext}\text{-log}(z_1 \times z_2) = \mathrm{Ext}\text{-log}z_1 + \mathrm{Ext}\text{-log}z_2, \qquad (19.14) $$

$$ \mathrm{Ext}\text{-arg}(z_1 \times z_2) = \mathrm{Ext}\text{-arg}z_1 + \mathrm{Ext}\text{-arg}z_2, $$

but only in the sense that both sides represent the same hyper infinite set of complex numbers. The inverse of $\mathrm{Ext}\text{-cos}(z)$ is obtained by solving the equation

$$ \mathrm{Ext}\text{-cos}(z) = \frac{1}{2}(\mathrm{Ext}\text{-}e^{iz} + \mathrm{Ext}\text{-}e^{-iz}) = w. \qquad (19.15) $$

This is a quadratic equation in $\mathrm{Ext}\text{-}e^{iz}$ with the roots

$$ \mathrm{Ext}\text{-}e^{iz} = w \pm \sqrt{w^2 - 1} \qquad (19.16) $$

and therefore

$$ z = \mathrm{Ext}\text{-arccos}(w) = -i(\mathrm{Ext}\text{-log}(w \pm \sqrt{w^2+1})), \qquad (19.17) $$

or in the form

$$ \mathrm{Ext}\text{-arccos}(w) = \pm i(\mathrm{Ext}\text{-log}(w + \sqrt{w^2+1})), \qquad (19.18) $$

The hyper infinitely many values of $\mathrm{Ext}\text{-arccos}(w)$ reflect the evenness and periodicity of $\mathrm{Ext}\text{-cos}(w)$. The inverse sine is most easily defined by formula

$$ \mathrm{Ext}\text{-arcsin}(w) = \frac{\pi_{\#}}{2} - (\mathrm{Ext}\text{-arccos}(w)). \qquad (19.19) $$

## 20. Complex Integration of the $\mathbb{C}_c^\#$-valued function $f(t)$.

### 20.1. Definition and basic properties of the complex integral.

If $f(t) = u(t) + iv(t)$ is a $\#$-continuous function, defined in an interval $(a, b)$, we set by definition

$$ \mathrm{Ext}\text{-}\int_{a}^{b} f(t) d^{\#}t = \mathrm{Ext}\text{-}\int_{a}^{b} u(t) d^{\#}t + i \left( \mathrm{Ext}\text{-}\int_{a}^{b} v(t) d^{\#}t \right). \qquad (20.1) $$

This integral has most of the properties of the real integral. In particular, if $c = \alpha + i\beta$ is a complex constant we obtain

$$ \mathrm{Ext}\text{-}\int_{a}^{b} cf(t) d^{\#}t = c \left( \mathrm{Ext}\text{-}\int_{a}^{b} f(t) d^{\#}t \right). \qquad (20.2) $$

The fundamental inequality

$$ \left| \mathrm{Ext}\text{-}\int_{a}^{b} f(t) d^{\#}t \right| \leq \mathrm{Ext}\text{-}\int_{a}^{b} |f(t)| d^{\#}t. \qquad (20.3) $$

holds for arbitrary $\mathbb{C}_c^\#$-valued function $f(t)$.

We consider now a piecewise $\#$-differentiable arc $\gamma$ with the equation

$z = z(t), a \leq t \leq b.$
$$ \operatorname{Ext}- \int_{\gamma_1+\gamma_2+\dots+\gamma_n} f(z)d^\#z = \operatorname{Ext}- \sum_{i=1}^n \left( \operatorname{Ext}- \int_{\gamma_i} f(z)d^\#z \right). \quad (20.7) $$

Finally, the integral over a closed curve is also invariant under a shift of parameter. The old and the new initial point determine two subarcs $\gamma_1, \gamma_2$, and the invariance follows from the fact that the integral over $\gamma_1 + \gamma_2$ is equal to the integral over $\gamma_2 + \gamma_1$. In addition to integrals of the form (20.4) we can also consider line integrals with respect to $\bar{z}$. The most convenient definition is by double conjugation

$$ \operatorname{Ext}- \int_{\gamma} f(z) \overline{d^\#z} = \overline{\operatorname{Ext}- \int_{\gamma} f(z) d^\#z}. \quad (20.8) $$

Using notation (20.7), line integrals with respect to $x$ or $y$ can be introduced by

$$ \operatorname{Ext}- \int_{\gamma} f(z) d^\#x = \frac{1}{2} \left( \operatorname{Ext}- \int_{\gamma} f(z) d^\#z + \operatorname{Ext}- \int_{\gamma} f(z) \overline{d^\#z} \right), \quad (20.9) $$

$$ \operatorname{Ext}- \int_{\gamma} f(z) d^\#y = \frac{1}{2i} \left( \operatorname{Ext}- \int_{\gamma} f(z) d^\#z - \operatorname{Ext}- \int_{\gamma} f(z) \overline{d^\#z} \right). $$

With $f = u + iv$ we find that the integral (16.4) can be written in the form

$$ \operatorname{Ext}- \int_{\gamma} (ud^\#x - vd^\#y) + i \left( \operatorname{Ext}- \int_{\gamma} (ud^\#y + vd^\#x) \right). \quad (16.10) $$

Of course we could just as well have started by defining integrals of the form

$$ \operatorname{Ext}- \int_{\gamma} (pd^\#x + qd^\#y), \quad (20.11) $$

in which case formula (20.10) would serve as definition of the integral (20.4). An essentially different line integral is obtained by integration with respect to arc length. Two notations are in common use, and the definition is

$$ \operatorname{Ext}- \int_{\gamma} fd^\#s = \operatorname{Ext}- \int_{\gamma} f(z)|d^\#z| = \operatorname{Ext}- \int_{\gamma} f(z(t))|z'^\#(t)|d^\#t. \quad (20.12) $$

This integral is again independent of the choice of parameter. In contrast to (20.6) we get

$$ \operatorname{Ext}- \int_{-\gamma} f(z) |d^\#z| = \operatorname{Ext}- \int_{\gamma} f(z) |d^\#z|, \quad (20.13) $$

while (20.7) remains valid in the same form. The inequality

$$ \left| \operatorname{Ext}- \int_{\gamma} f(z) d^\#z \right| \le \operatorname{Ext}- \int_{\gamma} |f(z)| d^\#z \quad (20.14) $$

is a consequence of (20.3).

**Remark 20.1.** For $f \equiv 1$ the integral (20.3) Generalized Cauchy's Theorem for a
If the function $f(z)$ is defined and #-continuous on $\gamma$, then $f(z(t))$ is also #-continuous and we can set

$$v \quad \mathrm{Ext}- \int_{\gamma} f(z) d^{\#} z = \mathrm{Ext}- \int_{a}^{b} f(z(t)) z'^{\#}(t) d^{\#} t. \qquad (20.4)$$

The most important property of the integral (20.4) is its invariance under a change of parameter. A change of parameter is determined by an increasing function $t = t(\tau)$ which maps an interval $\alpha \le \tau \le \beta$ onto $a \le t \le b$; we assume that $t(\tau)$ is piecewise #-differentiable. By the rule for changing the variable of integration we get

$$\mathrm{Ext}- \int_{a}^{b} f(z(t)) z'^{\#}(t) d^{\#} t = \mathrm{Ext}- \int_{a}^{\beta} f(z(t(\tau))) z'^{\#}(t(\tau)) t'^{\#}(\tau) d^{\#} \tau. \qquad (20.5)$$

We defined the opposite arc-$\gamma$ by the equation $z = z(-t)$, $-b \le t \le -a$. We have thus

$$\mathrm{Ext}- \int_{-\gamma} f(z) d^{\#} z = - \left( \mathrm{Ext}- \int_{\gamma} f(z) d^{\#} z \right). \qquad (20.6)$$

The integral (20.4) has also a very obvious additive property. It is clear what is meant by subdividing an arc $\gamma$ into a finite or hyperfinite number of subarcs. A subdivision can be indicated by a symbolic equation: $\gamma = \gamma_1 + \gamma_2 + \dots + \gamma_n, n \in \mathbb{N}^\#$, and the corresponding integrals satisfy the relation

$$\mathrm{Ext}- \int_{\gamma_1+\gamma_2+\dots+\gamma_n} f(z) d^{\#} z = \mathrm{Ext}- \sum_{i=1}^{n} \left( \mathrm{Ext}- \int_{\gamma_i} f(z) d^{\#} z \right). \qquad (20.7)$$

Finally, the integral over a closed curve is also invariant under a shift of parameter. The old and the new initial point determine two subarcs $\gamma_1, \gamma_2$, and the invariance follows from the fact that the integral over $\gamma_1 + \gamma_2$ is equal to the integral over $\gamma_2 + \gamma_1$. In addition to integrals of the form (20.4) we can also consider line integrals with respect to $\bar{z}$. The most convenient definition is by double conjugation

$$\mathrm{Ext}- \int_{\gamma} f(z) \overline{d^{\#} z} = \overline{\mathrm{Ext}- \int_{\gamma} f(z) d^{\#} z}. \qquad (20.8)$$

Using notation (20.7), line integrals with respect to $x$ or $y$ can be introduced by

$$\mathrm{Ext}- \int_{\gamma} f(z) d^{\#} x = \frac{1}{2} \left( \mathrm{Ext}- \int_{\gamma} f(z) d^{\#} z + \mathrm{Ext}- \int_{\gamma} f(z) \overline{d^{\#} z} \right), \qquad (20.9)$$ $$\mathrm{Ext}- \int_{\gamma} f(z) d^{\#} y = \frac{1}{2i} \left( \mathrm{Ext}- \int_{\gamma} f(z) d^{\#} z - \mathrm{Ext}- \int_{\gamma} f(z) \overline{d^{\#} z} \right).$$

With $f = u + iv$ we find that the integral (16.4) can be written in the form

$$\mathrm{Ext}- \int_{\gamma} (ud^{\#}x - vd^{\#}y) + i \left( \mathrm{Ext}- \int_{\gamma} (ud^{\#}y + vd^{\#}x) \right). \qquad (16.10)$$
Rectangle reduces to $\int_{\gamma} |dz|$ which is by definition the length of $\gamma$. As an example we compute the length of a circle. From the parametric equation $z = z(t) = a + \rho(\mathrm{Ext}-e^{it})$, $0 \le t \le 2\pi_{\#}r$, of a full circle we obtain $z'^{\#}(t) = i\rho(\mathrm{Ext}-e^{it})$ and hence

$$ \int_{0}^{2\pi_{\#}} |z'^{\#}(t)| d^{\#}t = \int_{0}^{2\pi_{\#}} \rho d^{\#}t = 2\pi_{\#} \rho \qquad (20.15) $$

as expected.

## 20.2. Line Integrals as Functions of Arcs.

Remind that the length of an arc can also be defined as the least upper bound of all hyperfinite sums

$$ \mathrm{Ext} - \sum_{i=1}^{n} |z(t_i) - z(t_{i-1})|, \qquad (20.16) $$

$n \in \mathbb{N}^{\#}/\mathbb{N}$, where $a = t_0 < t_1 < \dots < t_n = b$. If this least upper bound is finite or hyperfinite we say that the arc is rectifiable. It is quite easy to show that piecewise \#-differentiable arcs are rectifiable, and that the two definitions of length coincide. It is clear that the sums (20.16) and the corresponding sums

$$ \mathrm{Ext} - \sum_{i=1}^{n} |x(t_i) - x(t_{i-1})|; \mathrm{Ext} - \sum_{i=1}^{n} |y(t_i) - y(t_{i-1})|, \qquad (20.17) $$

where $z(t) = x(t) + iy(t)$, are bounded or hyperbounded at the same time. When the latter sums are bounded (or hyperbounded), one says that the functions $x(t)$ and $y(t)$ are of bounded (or hyperbounded) variation. An arc $z = z(t)$ is rectifiable if and only if the real and imaginary parts of $z(t)$ are of bounded (or hyperbounded) variation. If $\gamma$ is rectifiable and $f(z)$ \#-continuous on $\gamma$ it is possible to define integrals of type (20.12) as a \#-limit

$$ \mathrm{Ext} - \int_{\gamma} fd^{\#}s = \#\lim_{n \to \infty} \left( \mathrm{Ext} - \sum_{k=1}^{n} f(z(t_k)) |z(t_i) - z(t_{i-1})| \right). \qquad (20.18) $$

General line integral of the form $\mathrm{Ext} - \int_{\gamma} (pd^{\#}x + qd^{\#}y)$ can be considered as functional of the arc $\gamma$. It is then assumed that $p$ and $q$ are defined and \#-continuous in a region $\Omega$ and that $\gamma$ is free to vary in $\Omega$. An important class of integrals is characterized by the property that the integral over an arc depends only on its end points. In other words, if $\gamma_1$ and $\gamma_2$ have the same initial point and the same end point, we require that

$$ \mathrm{Ext} - \int_{\gamma_1} (pd^{\#}x + qd^{\#}y) = \mathrm{Ext} - \int_{\gamma_2} (pd^{\#}x + qd^{\#}y). \qquad (20.19) $$

To say that an integral depends only on the end points is equivalent to saying that the integral over any closed curve is zero. Indeed, if $\gamma$ is a closed curve, then $\gamma$ and $-\gamma$ have the same end points, and if the integral depends only on the end points, we obtain

$$ \mathrm{Ext} - \int_{\gamma} (pd^{\#}x + qd^{\#}y) = \mathrm{Ext} - \int_{-\gamma} (pd^{\#}x + qd^{\#}y) = - \left( \mathrm{Ext} - \int_{\gamma} (pd^{\#}x + qd^{\#}y) \right) \qquad (20.20) $$
Of course we could just as well have started by defining integrals of the form

$$ \operatorname{Ext}- \int_{\gamma} (pd^{\#}x + qd^{\#}y), \qquad (20.11) $$

in which case formula (20.10) would serve as definition of the integral (20.4).

An essentially different line integral is obtained by integration with respect to arc length. Two notations are in common use, and the definition is

$$ \operatorname{Ext}- \int_{\gamma} fd^{\#}s = \operatorname{Ext}- \int_{\gamma} f(z)|d^{\#}z| = \operatorname{Ext}- \int_{\gamma} f(z(t))|z'^{\#}(t)|d^{\#}t. \qquad (20.12) $$

This integral is again independent of the choice of parameter. In contrast to (20.6) we get

$$ \operatorname{Ext}- \int_{-\gamma} f(z)|d^{\#}z| = \operatorname{Ext}- \int_{\gamma} f(z)|d^{\#}z|, \qquad (20.13) $$

while (20.7) remains valid in the same form. The inequality

$$ \left| \operatorname{Ext}- \int_{\gamma} f(z) d^{\#}z \right| \leq \operatorname{Ext}- \int_{\gamma} |f(z)| d^{\#}z \qquad (20.14) $$

is a consequence of (20.3).

**Remark 20.1.** For $f \equiv 1$ the integral (20.3) Generalized Cauchy's Theorem for a Rectangle reduces to $\int_{\gamma} |dz|$ which is by definition the length of $\gamma$. As an example we compute the length of a circle. From the parametric equation

$z = z(t) = a + \rho(\operatorname{Ext}-e^{it})$, $0 \le t \le 2\pi_{\#}r$, of a full circle we obtain $z'^{\#}(t) = i\rho(\operatorname{Ext}-e^{it})$ and hence

$$ \int_{0}^{2\pi_{\#}} |z'^{\#}(t)| d^{\#}t = \int_{0}^{2\pi_{\#}} \rho d^{\#}t = 2\pi_{\#}\rho \qquad (20.15) $$

as expected.

## 20.2. Line Integrals as Functions of Arcs.

Remind that the length of an arc can also be defined as the least upper bound of all hyperfinite sums

$$ \operatorname{Ext}- \sum_{i=1}^{n} |z(t_i) - z(t_{i-1})|, \qquad (20.16) $$

$n \in \mathbb{N}^{\#}/\mathbb{N}$, where $a = t_0 < t_1 < \dots < t_n = b$. If this least upper bound is finite or hyperfinite we say that the arc is rectifiable. It is quite easy to show that piecewise #-differentiable arcs are rectifiable, and that the two definitions of length coincide. It is clear that the sums (20.16) and the corresponding sums

$$ \operatorname{Ext}- \sum_{i=1}^{n} |x(t_i) - x(t_{i-1})|; \operatorname{Ext}- \sum_{i=1}^{n} |y(t_i) - y(t_{i-1})|, \qquad (20.17) $$

where $z(t) = x(t) + iy(t)$, are bounded or hyperbounded at the same time. When the latter sums are bounded (or hyperbounded), one says that the functions $x(t)$ and $y(t)$ are of bounded (or hyperbounded) variation. An arc $z = z(t)$ is rectifiable if and only if
and consequently $\int_{\gamma} (pd^{\#}x + qd^{\#}y) = 0$. Conversely, if $\gamma_1$ and $\gamma_2$ have the same end points, then $\gamma_1 - \gamma_2$ is a closed curve, and if the integral over any closed curve vanishes, it follows that $\text{Ext}-\int_{\gamma_1} (pd^{\#}x + qd^{\#}y) = \text{Ext}-\int_{\gamma_2} (pd^{\#}x + qd^{\#}y)$.

<figure><img src="image_3.png" /><figcaption>Pic.20.1.</figcaption></figure>

The following theorem gives a necessary and sufficient condition under which a line integral depends only on the end points.

**Theorem 20.1.** The line integral $\text{Ext}-\int_{\gamma} (pd^{\#}x + qd^{\#}y)$, defined in $\Omega$, depends only on the end points of $\gamma$ if and only if there exists a function $U(x, y)$ in $\Omega$ with the partial #-derivatives $\partial^{\#}u/\partial^{\#}x = p, \partial^{\#}u/\partial^{\#}y = q$.

The sufficiency follows at once, for if the condition is fulfilled we can write, with the usual notations,

$$ \begin{aligned} \text{Ext}-\int_{\gamma} (pd^{\#}x + qd^{\#}y) &= \text{Ext}-\int_{a}^{b} \left[ \frac{\partial^{\#}U}{\partial^{\#}x}x^{\prime\#}(t) + \frac{\partial^{\#}U}{\partial^{\#}y}y^{\prime\#}(t) \right] d^{\#}t = \\ \text{Ext}-\int_{a}^{b} \frac{d^{\#}}{d^{\#}t}U(x(t), y(t))d^{\#}t &= U(x(b), y(b)) - U(x(a), y(a)). \end{aligned} \quad (20.21) $$

and the value of this difference depends only on the end points. To prove the necessity we choose a fixed point $(x_0, y_0) \in \Omega$, join it to $(x, y)$ by a polygon $\gamma$, contained in $\Omega$, whose sides are parallel to the coordinate axes (Pic.1) and define a function $U(x, y)$ by

$$ U(x, y) = \text{Ext}-\int_{\gamma} (pd^{\#}x + qd^{\#}y). \quad (20.22) $$

Since the integral depends only on the end points, the function is well defined. Moreover, if we choose the last segment of $\gamma$ horizontal, we can keep $y$ constant and let $x$ vary without changing the other segments. On the last segment we can choose $x$ for parameter and obtain

$$ U(x, y) = \text{Ext}-\int_{a}^{x} p(x, y)d^{\#}x + \text{const.}, \quad (20.23) $$

the lower limit of the integral being irrelevant. From Eq.(20.23) it follows at once that
the real and imaginary parts of $z(t)$ are of bounded (or hyperbounded) variation.
If $\gamma$ is rectifiable and $f(z)$ $\#$-continuous on $\gamma$ it is possible to define integrals of type
(20.12) as a $\#$-limit

$$ \operatorname{Ext-} \int_{\gamma} f d^{\#} s = \#\operatorname{lim}_{n \to \infty} \left( \operatorname{Ext-} \sum_{k=1}^{n} f(z(t_k)) |z(t_i) - z(t_{i-1})| \right). \qquad (20.18) $$

General line integral of the form $\operatorname{Ext-} \int_{\gamma} (pd^{\#}x + qd^{\#}y)$ can be considered as functional of the arc $\gamma$. It is then assumed that $p$ and $q$ are defined and $\#$-continuous in a region $\Omega$ and that $\gamma$ is free to vary in $\Omega$. An important class of integrals is characterized by the property that the integral over an arc depends only on its end points. In other words, if $\gamma_1$ and $\gamma_2$ have the same initial point and the same end point, we require that

$$ \operatorname{Ext-} \int_{\gamma_1} (pd^{\#}x + qd^{\#}y) = \operatorname{Ext-} \int_{\gamma_2} (pd^{\#}x + qd^{\#}y). \qquad (20.19) $$

To say that an integral depends only on the end points is equivalent to saying that the integral over any closed curve is zero. Indeed, if $\gamma$ is a closed curve, then $\gamma$ and $-\gamma$ have the same end points, and if the integral depends only on the end points, we obtain

$$ \operatorname{Ext-} \int_{\gamma} (pd^{\#}x + qd^{\#}y) = \operatorname{Ext-} \int_{-\gamma} (pd^{\#}x + qd^{\#}y) = - \left( \operatorname{Ext-} \int_{\gamma} (pd^{\#}x + qd^{\#}y) \right) \qquad (20.20) $$

and consequently $\int_{\gamma} (pd^{\#}x + qd^{\#}y) = 0$. Conversely, if $\gamma_1$ and $\gamma_2$ have the same end points, then $\gamma_1 - \gamma_2$ is a closed curve, and if the integral over any closed curve vanishes, it follows that $\operatorname{Ext-} \int_{\gamma_1} (pd^{\#}x + qd^{\#}y) = \operatorname{Ext-} \int_{\gamma_2} (pd^{\#}x + qd^{\#}y)$.

<figure><img src="image_3.png" /><figcaption>Pic.20.1.</figcaption></figure>

The following theorem gives a necessary and sufficient condition under which a line integral depends only on the end points.

**Theorem 20.1.** The line integral $\operatorname{Ext-} \int_{\gamma} (pd^{\#}x + qd^{\#}y)$, defined in $\Omega$, depends only on the end points of $\gamma$ if and only if there exists a function $U(x, y)$ in $\Omega$ with the partial $\#$-derivatives $\partial^{\#}u/\partial^{\#}x = p, \partial^{\#}u/\partial^{\#}y = q$.

The sufficiency follows at once, for if the condition is fulfilled we can write, with the usual notations,
$\frac{\partial^{\#}U}{\partial^{\#}x} = p$. In the same way, by choosing the last segment vertical, we can show that

$\frac{\partial^{\#}U}{\partial^{\#}y} = q$. It is customary to write $d^{\#}U = (\partial^{\#}U/\partial^{\#}x)d^{\#}x + (\partial^{\#}U/\partial^{\#}y)d^{\#}y$ and to say that

an expression $pd^{\#}x + qd^{\#}y$ which can be written in this form is an exact #-differential. Thus an integral depends only on the end points if and only if the integrand is an exact differential. Observe that $p, q$ and $U$ can be either real or complex. The function $U$, if it exists, is uniquely determined up to an additive constant, for if two functions have the same partial #-derivatives their #-difference must be constant.

When is $f(z)d^{\#}z = f(z)d^{\#}x + if(z)d^{\#}y$ an exact #-differential? According to the definition there must exist a function $F(z)$ in $\Omega$ with the partial #-derivatives

$$ \frac{\partial^{\#}F(z)}{\partial^{\#}x} = f(z), \quad \frac{\partial^{\#}F(z)}{\partial^{\#}y} = if(z). \qquad (20.24) $$

If this is so, $F(z)$ fulfills the generalized Cauchy-Riemann equation

$$ \frac{\partial^{\#}F(z)}{\partial^{\#}x} = i \frac{\partial^{\#}F(z)}{\partial^{\#}y}, \qquad (20.25) $$

since $f(z)$ is by assumption #-continuous $F(z)$ is #-analytic with the #-derivative $f(z)$.

The integral $Ext-\int_{\gamma}fd^{\#}z$, with #-continuous $f$, depends only on the end points of $\gamma$ if and only iff is the derivative of an analytic function in $\Omega$. Under these circumstances we shall prove later that $f(z)$ is itself #-analytic.

As an immediate application of the above result we find that

$$ \int_{\gamma} (z-a)^n d^{\#}z = 0 \qquad (20.26) $$

for all closed curves $\gamma$, provided that the integer $n \in \mathbb{N}^{\#}$ is $\ge 0$. In fact, $(z-a)^n$ is the #-derivative of $(z-a)^{n+1}/(n+1)$, a function which is #-analytic in the whole plane $\mathbb{C}^{\#}$.

If $n$ is negative, but $\ne -1$, the same result holds for all closed curves which do not pass through $a$, for in the complementary region of the point $a$ the indefinite integral is still #-analytic and single-valued. For $n = -1$, Eq.(20.26) does not always hold.

Consider a circle $C$ with the center $a$, represented by the equation $z = a + \rho(Ext-e^{it})$, $0 \le t \le 2\pi_{\#}$. We obtain

$$ \int_{\gamma} \frac{d^{\#}z}{(z-a)} = \int_{0}^{2\pi_{\#}} id^{\#}t = 2\pi_{\#}i. \qquad (20.27) $$

This result shows that it is impossible to define a single-valued branch of $Ext-\log(z-a)$ in an annulus $\rho_1 < |z-a| < \rho_2$. On the other hand, if the closed curve $\gamma$ is contained in a half plane which does not contain $a$, the integral vanishes, for in such a half plane a single-valued and #-analytic branch of $Ext-\log(z-a)$ can be defined.

## 20.3. Generalized Cauchy's Theorem for a Rectangle.

We consider, specifically, a rectangle $R \subset \mathbb{C}^{\#}$ defined by inequalities $a \le x \le b$, $c \le y \le d$. Its perimeter can be considered as a simple closed curve consisting of four line segments whose direction we choose so that $R$ lies to the left of the directed segments. The order of the vertices is thus $(a, c)$, $(b, c)$, $(b, d)$, $(a, d)$. We refer to this closed curve as the boundary curve or contour of $R$, and we denote it by $\partial^{\#}R$
$$
Ext- \int_{\gamma} (pd^{\#}x + qd^{\#}y) = Ext- \int_{a}^{b} \left[ \frac{\partial^{\#}U}{\partial^{\#}x} x'^{\#}(t) + \frac{\partial^{\#}U}{\partial^{\#}y} y'^{\#}(t) \right] d^{\#}t = \\
Ext- \int_{a}^{b} \frac{d^{\#}}{d^{\#}t} U(x(t), y(t)) d^{\#}t = U(x(b), y(b)) - U(x(a), y(a)). \tag{20.21}
$$

and the value of this difference depends only on the end points. To prove the
necessity we choose a fixed point (x₀, y₀) ∈ Ω, join it to (x, y) by a polygon γ,
contained in Ω, whose sides are parallel to the coordinate axes (Pic.1) and define
a function U(x, y) by

$$
U(x, y) = \operatorname{Ext} - \int_{\gamma} (pd^{\#}x + qd^{\#}y). \quad (20.22)
$$

Since the integral depends only on the end points, the function is well defined.
Moreover, if we choose the last segment of $\gamma$ horizontal, we can keep $y$ constant and
let $x$ vary without changing the other segments. On the last segment we can choose $x$
for parameter and obtain

$$
U(x, y) = \operatorname{Ext} - \int_{a}^{x} p(x, y) d^{\#}x + \text{const.}, \qquad (20.23)
$$

the lower limit of the integral being irrelevant. From Eq.(20.23) it follows at once that

$$
\frac{\partial^{\#} U}{\partial^{\#} x} = p. \text{ In the same way, by choosing the last segment vertical, we can show that}
$$

$$
\frac{\partial^{\#} U}{\partial^{\#} y} = q. \text{ It is customary to write } d^{\#}U = (\partial^{\#}U/\partial^{\#}x)d^{\#}x + (\partial^{\#}U/\partial^{\#}y)d^{\#}y \text{ and to say that}
$$

an expression $pd^{\#}x + qd^{\#}y$ which can be written in this form is an exact #-differential.
Thus an integral depends only on the end points if and only if the integrand is an
exact differential. Observe that $p, q$ and $U$ can be either real or complex. The function
$U$, if it exists, is uniquely determined up to an additive constant, for if two functions
have the same partial #-derivatives their #-difference must be constant.

When is $f(z)d^#z = f(z)d^#x + if(z)d^#y$ an exact #-differential? According to the definition
there must exist a function $F(z)$ in $\Omega$ with the partial #-derivatives

$$
\frac{\partial^\# F(z)}{\partial^\# x} = f(z), \quad \frac{\partial^\# F(z)}{\partial^\# y} = if(z). \qquad (20.24)
$$

If this is so, $F(z)$ fulfills the generalized Cauchy-Riemann equation

$$
\frac{\partial^2 F(z)}{\partial x^2} = i \frac{\partial^2 F(z)}{\partial y^2}, \qquad (20.25)
$$

since $f(z)$ is by assumption #-continuous $F(z)$ is #-analytic with the #-derivative $f(z)$.
The integral $Ext-\int_{\gamma} fd^{\#}z$, with #-continuous $f$, depends only on the end points of $\gamma$ if
and only iff is the derivative of an analytic function in $\Omega$. Under these circumstances
we shall prove later that $f(z)$ is itself #-analytic.

As an immediate application of the above result we find that

$$
\int_{\gamma} (z - a)^n d^{\#}z = 0 \qquad (20.26)
$$
<figure><img src="image_4.png" /><figcaption>Pic.20.2. Bisection of rectangle.</figcaption></figure>

**Theorem 16.2.** If the function $f(z)$ is #-analytic on $R$, then

$$ \operatorname{Ext-} \int_{\partial^{\#} R} f(z) d^{\#} z = 0. \qquad (20.28) $$

**Proof.** The proof is based on the method of bisection. Let us introduce the notation

$$ \eta(R) = \operatorname{Ext-} \int_{\partial^{\#} R} f(z) d^{\#} z. \qquad (20.29) $$

If $R$ is divided into four congruent rectangles $R^{(1)}, R^{(2)}, R^{(3)}, R^{(4)}$, we get

$$ \eta(R) = \eta(R^{(1)}) + \eta(R^{(2)}) + \eta(R^{(3)}) + \eta(R^{(4)}). \qquad (20.30) $$

for the integrals over the common sides cancel each other, see Pic.201. It follows from Eq.(16.30) that at least one of the rectangles $R^{(k)}$, $k = 1, 2, 3, 4$, must satisfy the condition $|\eta(R^{(k)})| \ge |\eta(R)|/4$. This process can be repeated inductively by hyper infinite induction,

and we obtain a hyper infinite sequence of nested rectangles $R \supset R_1 \supset R_2 \dots \supset R_n \dots \supset \dots$ with the property $|\eta(R_n)| \ge 4^{-n}|\eta(R_{n-1})|$, $n \in \mathbb{N}^\#$. Thus

$$ |\eta(R_n)| \ge 4^{-n}|\eta(R)|. \qquad (20.31) $$

The rectangles $R_n$ converge to a point $z^* \in R$ in the sense that $R_n$ will be contained in a prescribed neighborhood $|z - z^*| < \delta$ as soon as $n \in \mathbb{N}^\# \setminus \mathbb{N}$ is sufficiently large. First of all, we choose $\delta$ so small that $f(z)$ is defined and #-analytic in $|z - z^*| < \delta$, $\delta \approx 0$. Secondly, if $\varepsilon > 0$, $\varepsilon \approx 0$ is given, we can choose $\delta$ such that

$$ \left| \frac{f(z) - f(z^*)}{z - z^*} - f^{\prime\#}(z^*) \right| < \varepsilon, \qquad (20.32) $$

and therefore

$$ |f(z) - f(z^*) - (z - z^*) f^{\prime\#}(z^*)| < \varepsilon |z - z^*|. \qquad (20.33) $$

for $|z - z^*| < \delta$. We assume that $\delta$ satisfies both conditions and that $R_n$ is contained in $|z - z^*| < \delta$. We make now the observation that

$$ \operatorname{Ext-} \int_{\partial^{\#} R_n} d^{\#} z = 0, \operatorname{Ext-} \int_{\partial^{\#} R_n} zd^{\#} z = 0 \qquad (20.34) $$

By virtue of the equations (20.34) we are able to write
for all closed curves $\gamma$, provided that the integer $n \in \mathbb{N}^\#$ is $\ge 0$. In fact, $(z-a)^n$ is the \#-derivative of $(z-a)^{n+1}/(n+1)$, a function which is \#-analytic in the whole plane $\mathbb{C}_c^\#$. If $n$ is negative, but $\ne -1$, the same result holds for all closed curves which do not pass through $a$, for in the complementary region of the point $a$ the indefinite integral is still \#-analytic and single-valued. For $n = -1$, Eq.(20.26) does not always hold. Consider a circle $C$ with the center $a$, represented by the equation $z = a + \rho(\mathrm{Ext}-e^{it})$, $0 \le t \le 2\pi_\#$. We obtain

$$ \int_{\gamma} \frac{d^{\#}z}{(z-a)} = \int_{0}^{2\pi_{\#}} id^{\#}t = 2\pi_{\#}i. \qquad (20.27) $$

This result shows that it is impossible to define a single-valued branch of $\mathrm{Ext}-\log(z-a)$ in an annulus $\rho_1 < |z-a| < \rho_2$. On the other hand, if the closed curve $\gamma$ is contained in a half plane which does not contain $a$, the integral vanishes, for in such a half plane a single-valued and \#-analytic branch of $\mathrm{Ext}-\log(z-a)$ can be defined.

## 20.3. Generalized Cauchy's Theorem for a Rectangle.

We consider, specifically, a rectangle $R \subset \mathbb{C}_c^\#$ defined by inequalities $a \le x \le b$, $c \le y \le d$. Its perimeter can be considered as a simple closed curve consisting of four line segments whose direction we choose so that $R$ lies to the left of the directed segments. The order of the vertices is thus $(a, c)$, $(b, c)$, $(b, d)$, $(a, d)$. We refer to this closed curve as the boundary curve or contour of $R$, and we denote it by $\partial^\# R$

<figure><img src="image_4.png" /><figcaption>Pic.20.2. Bisection of rectangle.</figcaption></figure>

**Theorem 16.2.** If the function $f(z)$ is \#-analytic on $R$, then

$$ \mathrm{Ext} - \int_{\partial^\# R} f(z) d^\# z = 0. \qquad (20.28) $$

**Proof.** The proof is based on the method of bisection. Let us introduce the notation

$$ \eta(R) = \mathrm{Ext} - \int_{\partial^\# R} f(z) d^\# z. \qquad (20.29) $$

If $R$ is divided into four congruent rectangles $R^{(1)}, R^{(2)}, R^{(3)}, R^{(4)}$, we get

$$ \eta(R) = \eta(R^{(1)}) + \eta(R^{(2)}) + \eta(R^{(3)}) + \eta(R^{(4)}). \qquad (20.30) $$

for the integrals over the common sides cancel each other, see Pic.201. It follows from Eq.(16.30) that at least one of the rectangles $R^{(k)}$, $k = 1, 2, 3, 4$, must satisfy the
$$|\eta(R_n)| = \mathrm{Ext-} \int_{\partial^{\#} R_n} |f(z) - f(z^*) - (z - z^*) f^{\prime\#}(z^*)| d^\# z \quad (20.35)$$

and it follows by (20.33) that

$$|\eta(R_n)| \le \varepsilon \left( \mathrm{Ext-} \int_{\partial^{\#} R_n} |z - z^*| \times |d^{\#} z| \right). \quad (20.36)$$

In the last integral $|z - z^*|$ is at most equal to the length $d_n$ of the diagonal of $R_n$. If $L_n$ denotes the length of the perimeter of $R_n$, the integral is hence $\le d_n L_n$. But if $d$ and $L$ are the corresponding quantities for the original rectangle $R$, it is clear that $d_n = 2^{-n}d$ and $L_n = 2^{-n}L$. By (20.36) we have hence

$$|\eta(R_n)| \le 4^{-n} dL\varepsilon \quad (20.37)$$

and comparison with (20.31) yields

$$|\eta(R)| \le dL\varepsilon. \quad (20.38)$$

Since $\varepsilon \approx 0$ is arbitrary, we can only have $\eta(R) = 0$, and the theorem is proved. **Theorem 20.3.** Let $f(z)$ be $\#$-analytic on the set $R'$ obtained from a rectangle $R$ by omitting a finite or hyperfinite number of interior points $\zeta_j$. If it is true that $\#\lim_{z \to \zeta_j} \zeta_j(z - \zeta_j)f(z) = 0$ for all $j \in \mathbb{N}^\#$, then $\mathrm{Ext-} \int_{\partial^{\#} R} f(z) d^\# z = 0$.

**Proof.** It is sufficient to consider the case of a single exceptional point $\zeta$, for evidently $R$ can be divided into smaller rectangles which contain at most one $\zeta_j$. We divide now $R$ into nine rectangles, as shown in Pic.20.2, and apply Theorem 20.2 to all but the rectangle $R_0$ in the center. If the corresponding equations (20.28) are added, we obtain,

after cancellations,

$$\mathrm{Ext-} \int_{\partial^{\#} R} f(z) d^{\#} z = \mathrm{Ext-} \int_{\partial^{\#} R_0} f(z) d^{\#} z \quad (20.39)$$

<figure><img src="image_5.png" /><figcaption>Pic. 20.3.</figcaption></figure>

If $\varepsilon > 0, \varepsilon \approx 0$ we can choose the rectangle $R_0$ so infinite small that $|f(z)| \le \varepsilon|z - \zeta|$ on $\partial^{\#} R_0$. By (16.39) we have thus

$$\left| \mathrm{Ext-} \int_{\partial^{\#} R} f(z) d^{\#} z \right| = \varepsilon \left( \mathrm{Ext-} \int_{\partial^{\#} R_0} \frac{|d^{\#} z|}{|z - \zeta|} \right) \quad (16.40)$$
condition $|\eta(R^{(k)})| \ge |\eta(R)|/4$. This process can be repeated inductively by hyper infinite induction,

and we obtain a hyper infinite sequence of nested rectangles $R \supset R_1 \supset R_2 \dots \supset R_n \dots \supset \dots$ with the property $|\eta(R_n)| \ge 4^{-n}|\eta(R_{n-1})|$, $n \in \mathbb{N}^\#$. Thus

$$|\eta(R_n)| \ge 4^{-n}|\eta(R)|. \qquad (20.31)$$

The rectangles $R_n$ converge to a point $z^* \in R$ in the sense that $R_n$ will be contained in a prescribed neighborhood $|z - z^*| < \delta$ as soon as $n \in \mathbb{N}^\# \setminus \mathbb{N}$ is sufficiently large. First of all, we choose $\delta$ so small that $f(z)$ is defined and $\#$-analytic in $|z - z^*| < \delta$, $\delta \approx 0$. Secondly, if $\varepsilon > 0$, $\varepsilon \approx 0$ is given, we can choose $\delta$ such that

$$\left| \frac{f(z) - f(z^*)}{z - z^*} - f^{\prime\#}(z^*) \right| < \varepsilon, \qquad (20.32)$$

and therefore

$$|f(z) - f(z^*) - (z - z^*) f^{\prime\#}(z^*)| < \varepsilon|z - z^*|. \qquad (20.33)$$

for $|z - z^*| < \delta$. We assume that $\delta$ satisfies both conditions and that $R_n$ is contained in $|z - z^*| < \delta$. We make now the observation that

$$\mathrm{Ext}- \int_{\partial^{\#} R_n} d^\# z = 0, \mathrm{Ext}- \int_{\partial^{\#} R_n} zd^\# z = 0 \qquad (20.34)$$

By virtue of the equations (20.34) we are able to write

$$|\eta(R_n)| = \mathrm{Ext}- \int_{\partial^{\#} R_n} |f(z) - f(z^*) - (z - z^*) f^{\prime\#}(z^*)| d^\# z \qquad (20.35)$$

and it follows by (20.33) that

$$|\eta(R_n)| \le \varepsilon \left( \mathrm{Ext}- \int_{\partial^{\#} R_n} |z - z^*| \times |d^\# z| \right). \qquad (20.36)$$

In the last integral $|z - z^*|$ is at most equal to the length $d_n$ of the diagonal of $R_n$. If $L_n$ denotes the length of the perimeter of $R_n$, the integral is hence $\le d_n L_n$. But if $d$ and $L$ are the corresponding quantities for the original rectangle $R$, it is clear that $d_n = 2^{-n}d$ and $L_n = 2^{-n}L$. By (20.36) we have hence

$$|\eta(R_n)| \le 4^{-n} dL\varepsilon \qquad (20.37)$$

and comparison with (20.31) yields

$$|\eta(R)| \le dL\varepsilon. \qquad (20.38)$$

Since $\varepsilon \approx 0$ is arbitrary, we can only have $\eta(R) \equiv 0$, and the theorem is proved.

**Theorem 20.3.** Let $f(z)$ be $\#$-analytic on the set $R'$ obtained from a rectangle $R$ by omitting a finite or hyperfinite number of interior points $\zeta_j$. If it is true that $\#\lim_{z \to \#} \zeta_j(z - \zeta_j)f(z) = 0$ for all $j \in \mathbb{N}^\#$, then $\mathrm{Ext}-\int_{\partial^{\#} R} f(z) d^\# z = 0$.

**Proof.** It is sufficient to consider the case of a single exceptional point $\zeta$, for evidently $R$ can be divided into smaller rectangles which contain at most one $\zeta_j$. We divide now $R$ into nine rectangles, as shown in Pic.20.2, and apply Theorem 20.2 to all but the rectangle $R_0$ in the center. If the corresponding equations (20.28) are added, we obtain,
after cancellations,
If we assume, as we may, that $R_0$ is a square of center $\zeta$, elementary estimates show that

$$ \mathrm{Ext} - \int_{\partial^{\#} R_0} \frac{|d^{\#} z|}{|z - \zeta|} < 8. \qquad (20.41) $$

Thus finally we obtain

$$ \left| \mathrm{Ext} - \int_{\partial^{\#} R} f(z) d^{\#} z \right| < 8\varepsilon. \qquad (20.42) $$

and since $\varepsilon$ is arbitrary the theorem follows. We conclude that the hypothesis of the theorem is certainly fulfilled if $f(z)$ is #-analytic and bounded or hyperbounded on $R'$.

## 20.4. Generalized Cauchy's Theorem in a Disk.

It is not true that the integral of an #-analytic function over a closed curve is always zero. For example

$$ \int_C \frac{d^{\#} z}{|z - a|} = 2i\pi_{\#}. \qquad (20.43) $$

**Theorem 20.4.** If $f(z)$ is #-analytic in an open disk $\Delta$, then 20

$$ \mathrm{Ext} - \int_{\gamma} f(z) d^{\#} z = 0 \qquad (20.44) $$

for every closed curve $\gamma \subset \Delta$.

**Proof.** We define a function $F(z)$ by

$$ F(z) = \mathrm{Ext} - \int_{\sigma} f(z) d^{\#} z, \qquad (20.45) $$

where $\sigma$ consists of the horizontal line segment from the center $(x_0, y_0)$ to $(x, y_0)$ and the vertical segment from $(x, y_0)$ to $(x, y)$; it is immediately seen that $\partial^{\#} F / \partial^{\#} y = if(z)$. On the other hand, by Theorem 20.2 $\sigma$ can be replaced by a path consisting of a vertical segment followed by a horizontal segment. This choice defines the same function $F(z)$, and we obtain $\partial^{\#} F / \partial^{\#} x = f(z)$. Hence $F(z)$ is #-analytic in $\Delta$. with the #-derivative $f(z)$, and $f(z) d^{\#} z$ is an exact #-differential.

**Theorem 20.5.** Let $f(z)$ be #-analytic in the region $\Delta'$ obtained by omitting a finite or hyperfinite number of points $\zeta_j$ from an open disk $\Delta$. If $f(z)$ satisfies the condition $\#\lim_{z \to \#} \zeta_j(z - \zeta_j) f(z) = 0$ for all $j$, then (20.44) holds for any closed curve $\gamma \subset \Delta'$.
$$ \mathrm{Ext}- \int_{\partial^{\#} R} f(z) d^{\#}z = \mathrm{Ext}- \int_{\partial^{\#} R_0} f(z) d^{\#}z \qquad (20.39) $$

<figure><img src="image_5.png" /><figcaption>Pic. 20.3.</figcaption></figure>

If $\varepsilon > 0, \varepsilon \approx 0$ we can choose the rectangle $R_0$ so infinite small that $|f(z)| \le \varepsilon|z - \zeta|$ on $\partial^{\#} R_0$. By (16.39) we have thus

$$ \left| \mathrm{Ext}- \int_{\partial^{\#} R} f(z) d^{\#}z \right| = \varepsilon \left( \mathrm{Ext}- \int_{\partial^{\#} R_0} \frac{|d^{\#}z|}{|z-\zeta|} \right) \qquad (16.40) $$

If we assume, as we may, that $R_0$ is a square of center $\zeta$, elementary estimates show that

$$ \mathrm{Ext}- \int_{\partial^{\#} R_0} \frac{|d^{\#}z|}{|z-\zeta|} < 8. \qquad (20.41) $$

Thus finally we obtain

$$ \left| \mathrm{Ext}- \int_{\partial^{\#} R} f(z) d^{\#}z \right| < 8\varepsilon. \qquad (20.42) $$

and since $\varepsilon$ is arbitrary the theorem follows. We conclude that the hypothesis of the theorem is certainly fulfilled if $f(z)$ is $\#$-analytic and bounded or hyperbounded on $R'$.

## 20.4. Generalized Cauchy's Theorem in a Disk.

It is not true that the integral of an $\#$-analytic function over a closed curve is always zero. For example

$$ \int_C \frac{d^{\#}z}{|z-a|} = 2i\pi_{\#}. \qquad (20.43) $$

**Theorem 20.4.** If $f(z)$ is $\#$-analytic in an open disk $\Delta$, then 20

$$ \mathrm{Ext}- \int_{\gamma} f(z) d^{\#}z = 0 \qquad (20.44) $$

for every closed curve $\gamma \subset \Delta$.

**Proof.** We define a function $F(z)$ by

$$ F(z) = \mathrm{Ext}- \int_{\sigma} f(z) d^{\#}z, \qquad (20.45) $$
<figure><img src="image_6.png" /><figcaption>Pic.20.4.</figcaption></figure>

The proof must be modified, for we cannot let rr pass through the exceptional points. Assume first that no $\zeta_j$ lies on the lines $x = x_0$ and $y = y_0$. It is then possible to avoid the exceptional points by letting $\sigma$ consist of three segments (Pic.20.4). By an obvious application of Theorem 20.3 we find that the value of $F(z)$ in (20.44) is independent of the choice of the middle segment; moreover, the last segment can be either vertical or horizontal. We conclude as before that $F(z)$ is an indefinite integral of $f(z)$, and the theorem follows.

## 20.5. Generalized Cauchy's integral formula.

Through a very simple application of the generalized Cauchy's theorem it becomes possible to represent an #-analytic function $f(z)$ as a line integral in which the variable $z \in \mathbb{R}_c^#$ enters as a parameter. This representation, known in classical case as Cauchy's integral formula, has numerous important applications. Above all, it enables us to study the local properties of an #-analytic function in full detail.

**Lemma 20.1.** If the piecewise #-differentiable closed curve $\gamma$ does not pass through the point $a$, then the value of the integral

$$ \int_{\gamma} \frac{d^{\#}z}{|z-a|} \qquad (20.46) $$

is a multiple of $2i\pi_#$.

**Definition 20.1.** We define the index of the point $a$ with respect to the curve $\gamma$ by the equation

$$ n(\gamma, a) = \frac{1}{2\pi i} \int_{\gamma} \frac{d^{\#}z}{z-a} \qquad (20.47) $$

The index (20.47) is also called the winding number of $\gamma$ with respect to $a$. It is clear that $n(-\gamma, a) = -n(\gamma, a)$. The following property is an immediate consequence of Theorem 20.4.

(i) If $\gamma$ lies inside of a circle, then $n(\gamma, a) = 0$ for all points outside of the same circle. As a point set $\gamma$ is #-closed and bounded (or hyperbounded). Its complement is #-open and can be represented as a union of disjoint regions, the components of the complement. We shall say, for short, that $\gamma$ determines these regions. If the complementary regions are considered in the extended plane, there is exactly one which contains the point at infinity. Consequently, $\gamma$ determines one and
where $\sigma$ consists of the horizontal line segment from the center $(x_0, y_0)$ to $(x, y_0)$ and the vertical segment from $(x, y_0)$ to $(x, y)$; it is immediately seen that $\partial^\# F/\partial^\# y = if(z)$. On the other hand, by Theorem 20.2 $\sigma$ can be replaced by a path consisting of a vertical segment followed by a horizontal segment. This choice defines the same function $F(z)$, and we obtain $\partial^\# F/\partial^\# x = f(z)$. Hence $F(z)$ is \#-analytic in $\Delta$. with the \#-derivative $f(z)$, and $f(z)d^\#z$ is an exact \#-differential.

**Theorem 20.5.** Let $f(z)$ be \#-analytic in the region $\Delta'$ obtained by omitting a finite or hyperfinite number of points $\zeta_j$ from an open disk $\Delta$. If $f(z)$ satisfies the condition $\#\lim_{z \to \#} \zeta_j(z - \zeta_j)f(z) = 0$ for all $j$, then (20.44) holds for any closed curve $\gamma \subset \Delta'$.

<figure><img src="image_6.png" /><figcaption>Pic.20.4.</figcaption></figure>

The proof must be modified, for we cannot let $rr$ pass through the exceptional points. Assume first that no $\zeta_j$ lies on the lines $x = x_0$ and $y = y_0$. It is then possible to avoid the exceptional points by letting $\sigma$ consist of three segments (Pic.20.4). By an obvious application of Theorem 20.3 we find that the value of $F(z)$ in (20.44) is independent of the choice of the middle segment; moreover, the last segment can be either vertical or horizontal. We conclude as before that $F(z)$ is an indefinite integral of $f(z)$, and the theorem follows.

## 20.5. Generalized Cauchy's integral formula.

Through a very simple application of the generalized Cauchy's theorem it becomes possible to represent an \#-analytic function $f(z)$ as a line integral in which the variable $z \in \mathbb{R}_c^\#$ enters as a parameter. This representation, known in classical case as Cauchy's integral formula, has numerous important applications. Above all, it enables us to study the local properties of an \#-analytic function in full detail.

**Lemma 20.1.** If the piecewise \#-differentiable closed curve $\gamma$ does not pass through the point $a$, then the value of the integral

$$ \int_{\gamma} \frac{d^\# z}{|z-a|}. \qquad (20.46) $$

is a multiple of $2i\pi\#$.

**Definition 20.1.** We define the index of the point $a$ with respect to the curve $\gamma$ by the equation

$$ n(\gamma, a) = \frac{1}{2\pi i} \int_{\gamma} \frac{d^\# z}{z-a}. \qquad (20.47) $$
only one unbounded region.

(ii) As a function of a the index $n(\gamma, a)$ is constant in each of the regions determined by $\gamma$, and zero in the unbounded region.

Any two points in the same region determined by $\gamma$ can be joined by a polygon which does not meet $\gamma$. For this reason it is sufficient to prove that $n(\gamma, a) = n(\gamma, b)$ if $\gamma$ does not meet the line segment from $a$ to $b$. Outside of this segment the function $(z-a)/(z-b)$ is never real and $\le 0$. For this reason the principal branch of $\text{Ext}\text{-log}[(z-a)/(z-b)]$ is #-analytic in the complement of the segment. Its derivative is equal to $(z-a)^{-1} - (z-b)^{-1}$, and if $\gamma$ does not meet the segment we get

$$ \text{Ext}\text{-}\int_{\gamma} \left( \frac{1}{z-a} - \frac{1}{z-b} \right) d^\#z = 0; \qquad (20.48) $$

hence $n(\gamma, a) = n(\gamma, b)$. If $\text{lal}$ is sufficiently large, $\gamma$ is contained in a disk $|z| < \rho < |a|$ and we conclude by (i) that $n(\gamma, a) = 0$. This proves that $n(\gamma, a) = 0$ in the unbounded region.

We shall find the case $n(\gamma, a) = 1$ particularly important, and it is desirable to formulate a geometric condition which leads to this consequence.

For simplicity we take $a = 0$.

**Lemma 20.2.** Let $z_1, z_2$ be two points on a closed curve $\gamma$ which does not pass through the origin. Denote the subarc from $z_1$ to $z_2$ in the direction of the curve by $\gamma_1$, and the subarc from $z_2$ to $z_1$ by $\gamma_2$. Suppose that $z_1$ lies in the lower half plane and $z_2$ in the upper half plane. If $\gamma_1$ does not meet the negative real axis and $\gamma_2$ does not meet the positive real axis, then $n(\gamma, 0) = 1$.

For the proof we draw the half lines $L_1$ and $L_2$ from the origin through $z_1$ and $z_2$ (Pic. 4-5). Let $s_1, s_2$ be the points in which $L_1, L_2$ intersect a circle $C$ about the origin. If $C$ is described in the positive sense, the arc $C_1$ from $s_1$ to $s_2$ does not intersect the negative axis, and the arc $C_2$ from $s_2$ to $s_1$ does not intersect the positive axis. Denote the directed line segments from $z_1$ to $s_1$ and from $z_2$ to $s_2$ by $\delta_1, \delta_2$. Introducing the closed curves $\sigma_1 = \gamma_1 + \delta_2 - C_1 - \delta_1, \sigma_2 = \gamma_2 + \delta_1 - C_2 - \delta_2$ we get that $n(\gamma, 0) = n(C, 0) + n(\sigma_1, 0) + n(\sigma_2, 0)$ because of cancellations. But $\sigma_1$ does not meet the negative axis. Hence the origin belongs to the unbounded region determined by $\sigma_1$, and we obtain $n(\sigma_1, 0) = 0$. For a similar reason $n(\sigma_2, 0) = 0$, and we conclude that $n(\gamma, 0) = n(C, 0) = 1$.

<figure><img src="image_7.png" /><figcaption>Pic.20.5</figcaption></figure>
The index (20.47) is also called the winding number of $\gamma$ with respect to $a$. It is clear that $n(-\gamma, a) = -n(\gamma, a)$. The following property is an immediate consequence of Theorem 20.4.

(i) If $\gamma$ lies inside of a circle, then $n(\gamma, a) = 0$ for all points outside of the same circle. As a point set $\gamma$ is #-closed and bounded (or hyperbounded). Its complement is #-open and can be represented as a union of disjoint regions, the components of the complement. We shall say, for short, that $\gamma$ determines these regions. If the complementary regions are considered in the extended plane, there is exactly one which contains the point at infinity. Consequently, $\gamma$ determines one and only one unbounded region.

(ii) As a function of a the index $n(\gamma, a)$ is constant in each of the regions determined by $\gamma$, and zero in the unbounded region.

Any two points in the same region determined by $\gamma$ can be joined by a polygon which does not meet $\gamma$. For this reason it is sufficient to prove that $n(\gamma, a) = n(\gamma, b)$ if $\gamma$ does not meet the line segment from $a$ to $b$. Outside of this segment the function $(z-a)/(z-b)$ is never real and $\le 0$. For this reason the principal branch of $\text{Ext}\text{-log}[(z-a)/(z-b)]$ is #-analytic in the complement of the segment. Its derivative is equal to $(z-a)^{-1} - (z-b)^{-1}$, and if $\gamma$ does not meet the segment we get

$$ \text{Ext}\left[ \int_{\gamma} \left( \frac{1}{z-a} - \frac{1}{z-b} \right) d^\# z \right] = 0; \qquad (20.48) $$

hence $n(\gamma, a) = n(\gamma, b)$. If $\text{lal}$ is sufficiently large, $\gamma$ is contained in a disk $|z| < \rho < |a|$ and we conclude by (i) that $n(\gamma, a) = 0$. This proves that $n(\gamma, a) = 0$ in the unbounded region.

We shall find the case $n(\gamma, a) = 1$ particularly important, and it is desirable to formulate a geometric condition which leads to this consequence. For simplicity we take $a = 0$.

**Lemma 20.2.** Let $z_1, z_2$ be two points on a closed curve $\gamma$ which does not pass through the origin. Denote the subarc from $z_1$ to $z_2$ in the direction of the curve by $\gamma_1$, and the subarc from $z_2$ to $z_1$ by $\gamma_2$. Suppose that $z_1$ lies in the lower half plane and $z_2$ in the upper half plane. If $\gamma_1$ does not meet the negative real axis and $\gamma_2$ does not meet the positive real axis, then $n(\gamma, 0) = 1$.

For the proof we draw the half lines $L_1$ and $L_2$ from the origin through $z_1$ and $z_2$ (Pic. 4-5). Let $s_1, s_2$ be the points in which $L_1, L_2$ intersect a circle $C$ about the origin. If $C$ is described in the positive sense, the arc $C_1$ from $s_1$ to $s_2$ does not intersect the negative axis, and the arc $C_2$ from $s_2$ to $s_1$ does not intersect the positive axis. Denote the directed line segments from $z_1$ to $s_1$ and from $z_2$ to $s_2$ by $\delta_1, \delta_2$. Introducing the closed curves $\sigma_1 = \gamma_1 + \delta_2 - C_1 - \delta_1, \sigma_2 = \gamma_2 + \delta_1 - C_2 - \delta_2$ we get that $n(\gamma, 0) = n(C, 0) + n(\sigma_1, 0) + n(\sigma_2, 0)$ because of cancellations. But $\sigma_1$ does not meet the negative axis. Hence the origin belongs to the unbounded region determined by $\sigma_1$, and we obtain $n(\sigma_1, 0) = 0$. For a similar reason $n(\sigma_2, 0) = 0$, and we conclude that $n(\gamma, 0) = n(C, 0) = 1$.
Let $f(z)$ be #-analytic in an open disk $\Delta$. Consider a closed curve $\gamma \subset \Delta$. and a point $a \in \Delta$

which does not lie on $\gamma$. We apply Cauchy's theorem to the function

$$F(z) = \frac{f(z) - f(a)}{z - a}. \qquad (20.49)$$

This function is analytic for $z \neq a$. For $z = a$ it is not defined, but it satisfies the condition $\#\text{lim}_{z \to a} [(z-a)F(z)] = \#\text{lim}_{z \to a} [f(z) - f(a)] = 0$, which is the condition of Theorem 20.5. We conclude that

$$\text{Ext-} \int_{\gamma} \frac{f(z) - f(a)}{z - a} d^{\#}z = 0. \qquad (20.50)$$

This equation can be rewritten in the form

$$\text{Ext-} \int_{\gamma} \frac{f(z)d^{\#}z}{z-a} = f(a) \left( \text{Ext-} \int_{\gamma} \frac{d^{\#}z}{z-a} \right), \qquad (20.51)$$

and we observe that the integral in the right-hand member is by definition $2\pi i \text{in}(\gamma, a)$.

**Theorem 20.6.** Suppose that $f(z)$ is #-analytic in an open disk $\Delta$, and let $\gamma$ be a closed curve in $\Delta$. For any point $a$ such that $a \notin \gamma$

$$n(\gamma, a) \times f(a) = \frac{1}{2\pi i} \left( \text{Ext-} \int_{\gamma} \frac{f(z)d^{\#}z}{z-a} \right), \qquad (20.52)$$

where $n(\gamma, a)$ is the index of $a$ with respect to $\gamma$.

In this statement we have suppressed the requirement that $a$ be a point in $\Delta$. We have done so in view of the obvious interpretation of the formula (16.52) for the case that $a$ is not in $\Delta$. Indeed, in this case $n(\gamma, a)$ and the integral in the right-hand member are both zero.

It is clear that Theorem 16.6 remains valid for any region $\Omega$ to which Theorem 16.5 can be applied. The presence of exceptional points $\zeta_j$ is permitted, provided none of them coincides with $a$.

The most common application is to the case where $n(\gamma, a) = 1$. We have then

$$f(a) = \frac{1}{2\pi i} \left( \text{Ext-} \int_{\gamma} \frac{f(z)d^{\#}z}{z-a} \right) \qquad (20.53)$$

and this we interpret as a representation formula. Indeed, it permits us to compute $f(a)$ as soon as the values of $f(z)$ on $\gamma$ are given, together with the fact that $f(z)$ is #-analytic in $\Delta$. In (20.53) we may let a take different values, provided that the order of $a$ with respect to $\gamma$ remains equal to 1. We may thus treat $a$ as a variable, and it is #-convenient to change the notation and rewrite (20.53) in the form

$$f(z) = \frac{1}{2\pi i} \left( \text{Ext-} \int_{\gamma} \frac{f(\zeta)d^{\#}\zeta}{\zeta - z} \right). \qquad (20.54)$$

It is this formula which is usually referred to as Cauchy's integral formula. We must remember that it is valid only when $n(\gamma, z) = 1$, and that we have proved it only when $f(z)$ is #-analytic in a disk.
<figure><img src="image_7.png" /><figcaption>Pic.20.5</figcaption></figure>

Let $f(z)$ be #-analytic in an open disk $\Delta$. Consider a closed curve $\gamma \subset \Delta$. and a point $a \in \Delta$

which does not lie on $\gamma$. We apply Cauchy's theorem to the function

$$F(z) = \frac{f(z) - f(a)}{z - a}. \qquad (20.49)$$

This function is analytic for $z \neq a$. For $z = a$ it is not defined, but it satisfies the condition $\#\lim_{z \to a} [(z-a)F(z)] = \#\lim_{z \to a} [f(z) - f(a)] = 0$, which is the condition of Theorem 20.5. We conclude that

$$\mathrm{Ext} - \int_{\gamma} \frac{f(z) - f(a)}{z - a} d^{\#}z = 0. \qquad (20.50)$$

This equation can be rewritten in the form

$$\mathrm{Ext} - \int_{\gamma} \frac{f(z)d^{\#}z}{z-a} = f(a) \left( \mathrm{Ext} - \int_{\gamma} \frac{d^{\#}z}{z-a} \right), \qquad (20.51)$$

and we observe that the integral in the right-hand member is by definition $2\pi i n(\gamma, a)$.

**Theorem 20.6.** Suppose that $f(z)$ is #-analytic in an open disk $\Delta$, and let $\gamma$ be a closed curve in $\Delta$. For any point $a$ such that $a \notin \gamma$

$$n(\gamma, a) \times f(a) = \frac{1}{2\pi i a} \left( \mathrm{Ext} - \int_{\gamma} \frac{f(z)d^{\#}z}{z-a} \right), \qquad (20.52)$$

where $n(\gamma, a)$ is the index of $a$ with respect to $\gamma$.

In this statement we have suppressed the requirement that $a$ be a point in $\Delta$. We have done so in view of the obvious interpretation of the formula (16.52) for the case that $a$ is not in $\Delta$. Indeed, in this case $n(\gamma, a)$ and the integral in the right-hand member are both zero.

It is clear that Theorem 16.6 remains valid for any region $\Omega$ to which Theorem 16.5 can be applied. The presence of exceptional points $\zeta_j$ is permitted, provided none of them coincides with $a$.

The most common application is to the case where $n(\gamma, a) = 1$. We have then
The representation formula (20.54) gives us a tool for the study of the local properties of #-analytic functions. In particular we can now show that an #-analytic function has #-derivatives of all orders $n \in \mathbb{N}^#$, which are then also #-analytic.

We consider a function $f(z)$ which is #-analytic in an arbitrary region $\Omega$. To a point $a \in \Omega$ we determine a $\delta$-neighborhood $\Delta \subset \Omega$, and in $\Delta$ a circle $C$ about $a$.

Theorem 20.6 can be applied to $f(z)$ in $\Delta$. Since $n(C, a) = 1$ we have $n(C, z) = 1$ for all points $z$ inside of $C$. For such $z$ we obtain by (20.54)

$$f(z) = \frac{1}{2\pi i} \left( \operatorname{Ext} - \int_C \frac{f(\zeta)d^\#\zeta}{\zeta - z} \right) \qquad (20.55)$$

Provided that the integral in (20.) can be #-differentiated under the sign of integration we find

$$f'^{\#}(z) = \frac{1}{2\pi i} \left( \operatorname{Ext} - \int_C \frac{f(\zeta)d^\#\zeta}{(\zeta - z)^2} \right) \qquad (20.56)$$

and

$$f^{(n)\#}(z) = \frac{1}{2\pi i} \left( \operatorname{Ext} - \int_C \frac{f(\zeta)d^\#\zeta}{(\zeta - z)^n} \right) \qquad (20.57)$$

If the #-differentiations can be justified, we shall have proved the existence of all #-derivatives at the points inside of $C$. Since every point in $\Omega$ lies inside of some such circle, the existence will be proved in the whole region $\Omega$.

**Lemma 20.3.** Suppose that $\varphi(\zeta)$ is #-continuous on the arc $\gamma$. Then the function

$$F_n(z) = \frac{1}{2\pi i} \left( \operatorname{Ext} - \int_{\gamma} \frac{\varphi(\zeta)d^\#\zeta}{(\zeta - z)^n} \right) \qquad (20.58)$$

is #-analytic in each of the regions determined by $\gamma$, and its #-derivative is $F_n'^{\#}(z) = nF_{n+1}(z)$.

It is clear that Lemma 20.3 is just what is needed in order to deduce (20.55) and (20.56) in a rigorous way. We have thus proved that an #-analytic function has #-derivatives of all orders which are #-analytic and can be represented by the formula (20.57).

**Theorem 20.7.** (Generalized Morera's theorem) If $f(z)$ is defined and #-continuous in a region $\Omega$, and if $\operatorname{Ext} - \int_{\gamma} f(z) d^\#z = 0$ for all closed curves $\gamma$ in $\Omega$, then $f(z)$ is #-analytic in $\Omega$.

## 20.6. Generalized Liouville's theorem.

**Theorem 20.8.** (Generalized Liouville's theorem) A function $f(z)$ which is #-analytic and bounded in the whole plane $C_c^\#$ must reduce to a constant.

**Proof.** We make use of a simple estimate derived from (20.57). Let the radius of $C$ be $r$, and assume that $|f(z)| \le M$ on $C$. If we apply (20.57) with $z = a$, we obtain

$$|f^{(n)\#}(a)| \le M n! r^{-n} \qquad (20.59)$$

We need only the case $n = 1$. The hypothesis means that $|f(z)| \le M$ on all circles. Hence we can let $r$ tend to $\infty^\#$, and (16.59) leads to $f'^\#(a) = 0$ for all $a$. We conclude
$$f(a) = \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_{\gamma} \frac{f(z)d^\#z}{z-a} \right) \qquad (20.53)$$

and this we interpret as a representation formula. Indeed, it permits us to compute $f(a)$ as soon as the values of $f(z)$ on $\gamma$ are given, together with the fact that $f(z)$ is \#-analytic in $\Delta$. In (20.53) we may let a take different values, provided that the order of $a$ with respect to $\gamma$ remains equal to 1. We may thus treat $a$ as a variable, and it is \#-convenient to change the notation and rewrite (20.53) in the form

$$f(z) = \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_{\gamma} \frac{f(\zeta)d^\#\zeta}{\zeta - z} \right). \qquad (20.54)$$

It is this formula which is usually referred to as Cauchy's integral formula. We must remember that it is valid only when $n(\gamma, z) = 1$, and that we have proved it only when $f(z)$ is \#-analytic in a disk.

The representation formula (20.54) gives us a tool for the study of the local properties of \#-analytic functions. In particular we can now show that an \#-analytic function has \#-derivatives of all orders $n \in \mathbb{N}^\#$, which are then also \#-analytic.

We consider a function $f(z)$ which is \#-analytic in an arbitrary region $\Omega$. To a point $a \in \Omega$ we determine a $\delta$-neighborhood $\Delta \subset \Omega$, and in $\Delta$ a circle $C$ about $a$.

Theorem 20.6 can be applied to $f(z)$ in $\Delta$. Since $n(C, a) = 1$ we have $n(C, z) = 1$ for all points $z$ inside of $C$. For such $z$ we obtain by (20.54)

$$f(z) = \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_C \frac{f(\zeta)d^\#\zeta}{\zeta - z} \right) \qquad (20.55)$$

Provided that the integral in (20.) can be \#-differentiated under the sign of integration we find

$$f^{\#}(z) = \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_C \frac{f(\zeta)d^\#\zeta}{(\zeta - z)^2} \right) \qquad (20.56)$$

and

$$f^{(n)\#}(z) = \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_C \frac{f(\zeta)d^\#\zeta}{(\zeta - z)^n} \right) \qquad (20.57)$$

If the \#-differentiations can be justified, we shall have proved the existence of all \#-derivatives at the points inside of $C$. Since every point in $\Omega$ lies inside of some such circle, the existence will be proved in the whole region $\Omega$.

**Lemma 20.3.** Suppose that $\varphi(\zeta)$ is \#-continuous on the arc $\gamma$. Then the function

$$F_n(z) = \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_{\gamma} \frac{\varphi(\zeta)d^\#\zeta}{(\zeta - z)^n} \right) \qquad (20.58)$$

is \#-analytic in each of the regions determined by $\gamma$, and its \#-derivative is $F_n^{\#}(z) = nF_{n+1}(z)$.

It is clear that Lemma 20.3 is just what is needed in order to deduce (20.55) and (20.56) in a rigorous way. We have thus proved that an \#-analytic function has
that the function is constant.

## 20.7. Generalized fundamental theorem of algebra.

Liouville's theorem leads to an almost trivial proof of the generalized fundamental theorem of algebra.

**Theorem 20.9.** (Generalized fundamental theorem of algebra) Suppose that $P(z)$ is external polynomial of degree $n \in \mathbb{N}^#$. The equation $P(z) = 0$ must have a root $\xi \in \mathbb{C}_c^#$.

**Proof.** Suppose that $P(z)$ is a polynomial of degree $n \in \mathbb{N}^# \setminus \mathbb{N}$. If $P(z)$ were never zero, the function $1/P(z)$ would be #-analytic in the whole plane $\mathbb{C}_c^#$. We know that $P(z) \to \infty^#$, and therefore $1/P(z)$ tends to zero. This implies boundedness (the absolute value is #-continuous on the Riemann sphere and has thus a finite or hyperfinite maximum), and by Liouville's theorem $1/P(z)$ would be constant. Since this is not so, the equation $P(z) = 0$ has a root.

## 21. The local properties of #-analytic function.

### 21.1. Removable Singularitys. Taylor's Theorem.

**Theorem 21.1.** Suppose that $f(z)$ is #-analytic in the region $\Delta'$ obtained by omitting a point $a$ from a region $\Delta$. A necessary and sufficient condition that there exist an #-analytic function in $\Delta$ which coincides with $f(z)$ in $\Delta'$ is that $\#\lim_{z \to a} f(z) = 0$. The extended function is uniquely determined.

**Proof.** The necessity and the uniqueness are trivial since the extended function must be #-continuous at $a$. To prove the sufficiency we draw a circle $C$ about $a$ so that $C$ and its inside are contained in $\Delta$. Cauchy's formula is valid, and therefore we have

$$f(z) = \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_C \frac{f(\zeta) d^\# \zeta}{\zeta - z} \right) \quad (21.1)$$

for all $z \neq a$ inside of $C$. But the integral in the right-hand member represents an #-analytic function of $z$ throughout the inside of $C$. Consequently, the function which is equal to $f(z)$ for $z \neq a$ and which has the value

$$\frac{1}{2\pi i} \left( \mathrm{Ext} - \int_C \frac{f(\zeta) d^\# \zeta}{\zeta - z} \right). \quad (21.2)$$

for $z = a$ is #-analytic in $\Delta$. It is natural to denote the extended function by $f(z)$ and the value (21.2) by $f(a)$. We apply this result to the function $F(z) = [f(z) - f(a)]/(z - a)$. It is not defined for $z = a$, but it satisfies the condition $\#\lim_{z \to a} f(z) = 0$. The #-limit of $F(z)$ as $z \to a$ is $f^{\#}(a)$. Hence there exists an #-analytic function which is equal to $F(z)$ for $z = a$ and equal to $f^{\#}(a)$.

# Part II. $\mathbb{R}_c^#$-Valued Lebesgue Integral.

## 1. External $\mathbb{R}_c^#$-Valued Lebesgue Measure.

Let us consider a bounded interval $I \subset \mathbb{R}_c^#$ with endpoints $a$ and $b$ ($a < b$). The length of this bounded interval $I$ is defined by $l(I) = b - a$. In contrast, the length of an unbounded interval, such as $(a, \infty^#)$, $(-\infty^#$, $b$) or $(-\infty^#$, $\infty)$, is defined to be hyperfinite. Obviously, the length of a line segment is easy to quantify. However, what should we do if we want to
#-derivatives of all orders which are #-analytic and can be represented by the formula (20.57).

**Theorem 20.7.** (Generalized Morera's theorem) If $f(z)$ is defined and #-continuous in a region $\Omega$, and if $\mathrm{Ext}-\int_{\gamma} f(z)d^\#z = 0$ for all closed curves $\gamma$ in $\Omega$, then $f(z)$ is #-analytic in $\Omega$.

## 20.6. Generalized Liouville's theorem.

**Theorem 20.8.** (Generalized Liouville's theorem) A function $f(z)$ which is #-analytic and bounded in the whole plane $C_c^\#$ must reduce to a constant.

**Proof.** We make use of a simple estimate derived from (20.57). Let the radius of $C$ be $r$, and assume that $|f(z)| \le M$ on $C$. If we apply (20.57) with $z = a$, we obtain

$$|f^{(n)\#}(a)| \le M n! r^{-n} \qquad (20.59)$$

We need only the case $n = 1$. The hypothesis means that $|f(z)| \le M$ on all circles. Hence we can let $r$ tend to $\infty^\#$, and (16.59) leads to $f^{\prime\#}(a) = 0$ for all $a$. We conclude that the function is constant.

## 20.7. Generalized fundamental theorem of algebra.

Liouville's theorem leads to an almost trivial proof of the generalized fundamental theorem of algebra.

**Theorem 20.9.** (Generalized fundamental theorem of algebra) Suppose that $P(z)$ is external polynomial of degree $n \in \mathbb{N}^\#$. The equation $P(z) = 0$ must have a root $\xi \in C_c^\#$.

**Proof.** Suppose that $P(z)$ is a polynomial of degree $n \in \mathbb{N}^\#\mathbb{N}$. If $P(z)$ were never zero, the function $1/P(z)$ would be #-analytic in the whole plane $C_c^\#$. We know that $P(z) \to \infty^\#$, and therefore $1/P(z)$ tends to zero. This implies boundedness (the absolute value is #-continuous on the Riemann sphere and has thus a finite or hyperfinite maximum), and by Liouville's theorem $1/P(z)$ would be constant. Since this is not so, the equation $P(z) = 0$ has a root.

# 21. The local properties of #-analytic function.

## 21.1. Removable Singnlarities. Taylor's Theorem.

**Theorem 21.1.** Suppose that $f(z)$ is #-analytic in the region $\Delta'$ obtained by omitting a point $a$ from a region $\Delta$. A necessary and sufficient condition that there exist an #-analytic function in $\Delta$ which coincides with $f(z)$ in $\Delta'$ is that $\#\lim_{z \to a} (z-a)f(z) = 0$. The extended function is uniquely determined.

**Proof.** The necessity and the uniqueness are trivial since the extended function must be #-continuous at $a$. To prove the sufficiency we draw a circle $C$ about $a$ so that $C$ and its inside are contained in $\Delta$. Cauchy's formula is valid, and therefore we have

$$f(z) = \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_C \frac{f(\zeta)d^\#\zeta}{\zeta - z} \right) \qquad (21.1)$$

for all $z \ne a$ inside of $C$. But the integral in the right-hand member represents an #-analytic function of $z$ throughout the inside of $C$. Consequently, the function which is equal to $f(z)$ for $z \ne a$ and which has the value
measure an arbitrary subset of $\mathbb{R}_c^#$? Given a set $E \subset \mathbb{R}_c^#$ of hyperreal numbers, we denote the Lebesgue measure of set $E$ by $\mu(E)$. To correspond with the length of a line segment, the measure of a set $A \subset \mathbb{R}_c^#$ should keep the following properties:

(1) If $A$ is an interval, then $\mu(A) = l(A)$.

(2) If $A \subset B$, then $\mu(A) \le \mu(B)$.

(3) Given $A \subset \mathbb{R}_c^#$ and $x_0 \in \mathbb{R}_c^#$, define $A + x_0 = \{x + x_0 : x \in A\}$. Then
$$ \mu(A) = \mu(A + x_0). $$

(4) If $A$ and $B$ are disjoint sets, then $\mu(A \cup B) = \mu(A) + \mu(B)$. If $\{A_i\}_{i \in \mathbb{N}^#}$ is a hyperinfinite sequence of disjoint sets, then $\mu(\bigcup_{i \in \mathbb{N}^#} A_i) = \sum_{i=1}^{\infty} \mu(A_i)$.

## 2. External $\mathbb{R}_c^#$-Valued Lebesgue outer measure

**Definition 2.1.** Let $E$ be a subset of $\mathbb{R}_c^#$. Let $\{I_k\} \triangleq \{I_k\}_{k \in \mathbb{N}^#}$ be a hyperinfinite sequence

of open intervals such that $E \subset \bigcup_{k \in \mathbb{N}^#} A_k$ and let $\Sigma$ be a set of the all such hyperinfinite sequences. The external Lebesgue outer measure of $E$ is defined by

$$ \mu^*(E) = \inf_{\{I_k\} \in \Sigma} \left\{ \sum_{k=1}^{\infty} l(I_k) \right\}. \qquad (2.1) $$

Note that $0 \le \mu^*(E) \le \infty^#$.

**Definition 2.2.** A set $E$ is #-countable if there exists an injective function $f$ from $E$ to the

gyperr natural numbers $\mathbb{N}^#$. If such an $f$ can be found that is also surjective (and therefore

bijective), then $E$ is called #-countably infinite or gyperr infinite, i.e. a set is #-countably infinite if it has one-to-one correspondence with the set $\mathbb{N}^#$.

**Theorem 2.1.** The external Lebesgue outer measure has the following properties:

(a) If $E_1 \subset E_2$, then $\mu^*(E_1) \le \mu^*(E_2)$.

(b) The external Lebesgue outer measure of any #-countable set is zero.

(c) The external Lebesgue outer measure of the empty set is zero.

(d) The external Lebesgue outer measure is invariant under translation, that is,
$$ \mu^*(E + x_0) = \mu^*(E). $$

(e) Lebesgue outer measure is #-countably sub-additive, that is,

$$ \mu^*\left(\bigcup_{i=1}^{\infty} E_i\right) \le \sum_{i=1}^{\infty} \mu^*(E_i). \qquad (2.2) $$

(f) For any interval $I$, $\mu^*(I) = l(I)$.

**Proof.** Part (a) is trivial.

For part (b) and (c), let $E = \{x_k | k \in \mathbb{Z}_+^#\}$ be a #-countably hyper infinite set.

Let $\varepsilon > 0$, $\varepsilon \approx 0$ and let $\varepsilon_k$ be a hyper infinite sequence of positive numbers such that

$\sum_{k=1}^{\infty} \varepsilon_k = \varepsilon/2$. Since $E \subset \bigcup_{k=1}^{\infty} (x_k - \varepsilon_k, x_k + \varepsilon_k)$, it follows that $\mu^*(E) \le \varepsilon$. Hence,
$$ \mu^*(E) = 0. \text{ Since } \emptyset \subset E, \text{ then } \mu^*(\emptyset) = 0. $$

For part (d), since each cover of $E$ by open intervals can generate a cover of $E + x_0$ by open intervals with the same length, then $\mu^*(E + x_0) \le \mu^*(E)$. Similarly,

$\mu^*(E + x_0) \ge \mu^*(E)$, since $E + x_0$ is a translation of $E$. Therefore, $\mu^*(E + x_0) = \mu^*(E)$.

For part (e), if $\sum_{i=1}^{\infty} \mu^*(E_i) = \infty^#$, then the statement is trivial. Suppose that the sum is hyperfinite and let $\varepsilon > 0$, $\varepsilon \approx 0$. For each $i \in \mathbb{N}^#$, there exists a hyperinfinite sequence $\{I_i^k\}$ of open intervals such that $E_i \subset \bigcup_{k=1}^{\infty} I_i^k$ and $\sum_{k=1}^{\infty} l(I_i^k) < \mu^*(E_i) + \varepsilon/2^i$.
$$ \frac{1}{2\pi i} \left( \mathrm{Ext} - \int_C \frac{f(\zeta) d^\# \zeta}{\zeta - z} \right). \qquad (21.2) $$

for $z = a$ is #-analytic in $\Delta$. It is natural to denote the extended function by $f(z)$ and the value (21.2) by $f(a)$. We apply this result to the function $F(z) = [f(z) - f(a)]/(z - a)$. It is not defined for $z = a$, but it satisfies the condition $\#\lim_{z \to a} f(z) = 0$.

The #-limit of $F(z)$ as $z \to a$ is $f^{\#}(a)$. Hence there exists an #-analytic function which is equal to $F(z)$ for $z = a$ and equal to $f^{\#}(a)$ f

# Part II. $\mathbb{R}_c^\#$-Valued Lebesgue Integral.

## 1. External $\mathbb{R}_c^\#$-Valued Lebesgue Measure.

Let us consider a bounded interval $I \subset \mathbb{R}_c^\#$ with endpoints $a$ and $b$ ($a < b$). The length of this bounded interval $I$ is defined by $l(I) = b - a$. In contrast, the length of an unbounded interval, such as $(a, \infty^\#)$, $(-\infty^\#$, $b$) or $(-\infty^\#$, $\infty)$, is defined to be hyperinfinite. Obviously, the length of a line segment is easy to quantify. However, what should we do if we want to measure an arbitrary subset of $\mathbb{R}_c^\#$? Given a set $E \subset \mathbb{R}_c^\#$ of hyperreal numbers, we denote the Lebesgue measure of set $E$ by $\mu(E)$. To correspond with the length of a line segment, the measure of a set $A \subset \mathbb{R}_c^\#$ should keep the following properties:

(1) If $A$ is an interval, then $\mu(A) = l(A)$.

(2) If $A \subset B$, then $\mu(A) \le \mu(B)$.

(3) Given $A \subset \mathbb{R}_c^\#$ and $x_0 \in \mathbb{R}_c^\#$, define $A + x_0 = \{x + x_0 : x \in A\}$. Then $\mu(A) = \mu(A + x_0)$.

(4) If $A$ and $B$ are disjoint sets, then $\mu(A \cup B) = \mu(A) + \mu(B)$. If $\{A_i\}_{i \in \mathbb{N}^\#}$ is a hyperinfinite sequence of disjoint sets, then $\mu(\bigcup_{i \in \mathbb{N}^\#} A_i) = \sum_{i=1}^{\infty} \mu(A_i)$.

## 2. External $\mathbb{R}_c^\#$-Valued Lebesgue outer measure

**Definition 2.1.** Let $E$ be a subset of $\mathbb{R}_c^\#$. Let $\{I_k\}_{k \in \mathbb{N}^\#}$ be a hyperinfinite sequence

of open intervals such that $E \subset \bigcup_{k \in \mathbb{N}^\#} A_k$ and let $\Sigma$ be a set of the all such hyperinfinite sequences. The external Lebesgue outer measure of $E$ is defined by

$$ \mu^*(E) = \inf_{\{I_k\} \in \Sigma} \left\{ \sum_{k=1}^{\infty} l(I_k) \right\}. \qquad (2.1) $$

Note that $0 \le \mu^*(E) \le \infty^\#$.

**Definition 2.2.** A set $E$ is #-countable if there exists an injective function $f$ from $E$ to the

gyperratural numbers $\mathbb{N}^\#$. If such an $f$ can be found that is also surjective (and therefore

bijective), then $E$ is called #-countably infinite or hyperinfinite, i.e. a set is #-countably infinite if it has one-to-one correspondence with the set $\mathbb{N}^\#$.

**Theorem 2.1.** The external Lebesgue outer measure has the following properties:

(a) If $E_1 \subset E_2$, then $\mu^*(E_1) \le \mu^*(E_2)$.

(b) The external Lebesgue outer measure of any #-countable set is zero.

(c) The external Lebesgue outer measure of the empty set is zero.

(d) The external Lebesgue outer measure is invariant under translation, that is, $\mu^*(E + x_0) = \mu^*(E)$.
Now {$I_i^k$} is a double-indexed sequence of open intervals such that

$$ \cup_{i=1}^{\infty} E_i \subseteq \cup_{i=1}^{\infty} \cup_{k=1}^{\infty} I_i^k \text{ and} $$

$$ \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} l(I_i^k) \leq \sum_{i=1}^{\infty} (\mu^*(E_i) + \varepsilon/2^i) = \sum_{i=1}^{\infty} \mu^*(E_i) + \varepsilon. $$

Therefore, $\mu^*(\cup_{i=1}^{\infty} E_i) \leq \sum_{i=1}^{\infty} \mu^*(E_i) + \varepsilon$. The result follows since $\varepsilon > 0, \varepsilon \approx 0$ was arbitrary.

For part (f), we need to prove $\mu^*(I) \leq l(I)$ and $\mu^*(I) \geq l(I)$ respectively.

We can assume that $I = [a, b]$ where $a, b \in \mathbb{R}_c^\#$.

First, we want to prove $\mu^*(I) \leq l(I)$. Let $\varepsilon > 0, \varepsilon \approx 0$, we have

$$ I \subseteq (a, b) \cup (a - \varepsilon, a + \varepsilon) \cup (b - \varepsilon, b + \varepsilon). $$

$$ \begin{aligned} \text{Thus, } \mu^*(I) &\leq l(a, b) + l(a - \varepsilon, a + \varepsilon) + l(b - \varepsilon, b + \varepsilon) = \\ &= (b - a) + 2\varepsilon + 2\varepsilon = b - a + 4\varepsilon. \end{aligned} $$

As $\varepsilon > 0, \varepsilon \approx 0$ is arbitrary, we conclude that $\mu^*(I) \leq b - a = l(I)$.

Then, we want to prove that $\mu^*(I) \geq l(I)$. Let {$I_k$} be any sequence of open intervals that covers $I$. Since $I$ is compact, by the generalized Heine-Borel theorem, there is a gyperfinite subcollection {$J_i|1 \leq i \leq n$}, $n \in \mathbb{N}^\#$ of $I_k$ that still covers $I$. By reordering and deleting if necessary, we can assume that $a \in J_1 = (a_1, b_1), b_1 \in J_2 = (a_2, b_2), \dots, b_{n-1} \in J_n = (a_n, b_n)$, where $b_{n-1} \leq b < b_n$.

We then can compute that

$$ b - a < b_n - a_1 = \mathrm{Ext} - \sum_{i=2}^{n} (b_i - b_{i-1}) + (b_1 - a_1) < \mathrm{Ext} - \sum_{i=1}^{n} l(J_i) \leq \mathrm{Ext} - \sum_{i=1}^{\infty} l(I_k). $$

Therefore, $l(I) \leq \mu^*(I)$. We can now conclude that $\mu^*(I) = l(I)$. This proves the result for closed and bounded intervals.

Suppose that $I = (a, b)$ is an open and bounded interval. Then, $\mu^*(I) \leq l(I)$ as above and $b - a = \mu^*([a, b]) \leq \mu^*((a, b)) + \mu^*(a) + \mu^*(b) = \mu^*((a, b))$.

Hence $l(I) \leq \mu^*(I)$. The proof for half-open intervals is similar.

Finally, suppose that $I$ is an hyper infinite interval and let $M > 0$. There exists a bounded interval $J \subseteq I$ such that $\mu^*(J) = l(J) = M$ and it follows that $\mu^*(I) \geq \mu^*(J) = M$. Since $M > 0$ was arbitrary, $\mu^*(I) = \infty^\# = l(I)$.

This completes the proof.

## 2.2. External Lebesgue inner measure

In previous subsection, we have discussed external Lebesgue outer measure. There is

another external measure named external Lebesgue inner measure. Let's define the external inner measure and see some basic properties.

**Definition 2.2.** Let $E$ be a subset of $\mathbb{R}^\#$ . The external inner measure of $E$ is defined by

$$ \mu^*(E) = \sup\{\mu(K) : K \subseteq E \text{ and } K \text{ is } \#\text{-closed}\} \qquad (2.3) $$

iff supremum in RHS of the (2.3) exists.

Recall that external Lebesgue outer measure of a set $E$ uses an infimum of the union of a sequence of open sets that cover the set $E$, while external Lebesgue inner measure of a set $E$ uses a supremum of a set inside the set $E$. Then, it is obvious that

$$ \mu_*(E) \leq \mu^*(E) \qquad (2.4) $$

for any set $E$. Also, for $A \subseteq B, \mu_*(A) \leq \mu^*(B)$.

**Theorem 2.2.** Let $A$ and $E$ be subsets of $\mathbb{R}_c^\#$.
(e) Lebesgue outer measure is #-countably sub-additive, that is,

$$ \mu^*(\bigcup_{i=1}^{\infty} E_i) \leq \sum_{i=1}^{\infty} \mu^*(E_i). \qquad (2.2) $$

(f) For any interval $I$, $\mu^*(I) = l(I)$.

**Proof.** Part (a) is trivial.

For part (b) and (c), let $E = \{x_k | k \in \mathbb{Z}_+^\#\}$ be a #-countably hyper infinite set.
Let $\varepsilon > 0, \varepsilon \approx 0$ and let $\varepsilon_k$ be a hyper infinite sequence of positive numbers such that
$\sum_{k=1}^{\infty} \varepsilon_k = \varepsilon/2$. Since $E \subseteq \bigcup_{k=1}^{\infty} (x_k - \varepsilon_k, x_k + \varepsilon_k)$, it follows that $\mu^*(E) \leq \varepsilon$. Hence,
$\mu^*(E) = 0$. Since $\emptyset \subseteq E$, then $\mu^*(\emptyset) = 0$.

For part (d), since each cover of $E$ by open intervals can generate a cover of $E + x_0$ by open intervals with the same length, then $\mu^*(E + x_0) \leq \mu^*(E)$. Similarly,
$\mu^*(E + x_0) \geq \mu^*(E)$, since $E + x_0$ is a translation of $E$ Therefore, $\mu^*(E + x_0) = \mu^*(E)$.

For part (e), if $\sum_{i=1}^{\infty} \mu^*(E_i) = \infty^\#$, then the statement is trivial. Suppose that the sum is hyperfinite and let $\varepsilon > 0, \varepsilon \approx 0$. For each $i \in \mathbb{N}^\#$, there exists a hyperinfinite sequence $\{I_i^k\}$ of open intervals such that $E_i \subseteq \bigcup_{k=1}^{\infty} I_i^k$ and $\sum_{k=1}^{\infty} l(I_i^k) < \mu^*(E_i) + \varepsilon/2^i$.
Now $\{I_i^k\}$ is a double-indexed sequence of open intervals such that

$$ \bigcup_{i=1}^{\infty} E_i \subseteq \bigcup_{i=1}^{\infty} \bigcup_{k=1}^{\infty} I_i^k \text{ and} \\ \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} l(I_i^k) \leq \sum_{i=1}^{\infty} (\mu^*(E_i) + \varepsilon/2^i) = \sum_{i=1}^{\infty} \mu^*(E_i) + \varepsilon. $$

Therefore, $\mu^*(\bigcup_{i=1}^{\infty} E_i) \leq \sum_{i=1}^{\infty} \mu^*(E_i) + \varepsilon$. The result follows since $\varepsilon > 0, \varepsilon \approx 0$ was arbitrary.

For part (f), we need to prove $\mu^*(I) \leq l(I)$ and $\mu^*(I) \geq l(I)$ respectively.
We can assume that $I = [a, b]$ where $a, b \in \mathbb{R}_c^\#$.

First, we want to prove $\mu^*(I) \leq l(I)$. Let $\varepsilon > 0, \varepsilon \approx 0$, we have
$I \subseteq (a, b) \cup (a - \varepsilon, a + \varepsilon) \cup (b - \varepsilon, b + \varepsilon)$.

$$ \text{Thus, } \mu^*(I) \leq l(a, b) + l(a - \varepsilon, a + \varepsilon) + l(b - \varepsilon, b + \varepsilon) = \\ = (b - a) + 2\varepsilon + 2\varepsilon = b - a + 4\varepsilon. $$

As $\varepsilon > 0, \varepsilon \approx 0$ is arbitrary, we conclude that $\mu^*(I) \leq b - a = l(I)$.

Then, we want to prove that $\mu^*(I) \geq l(I)$. Let $\{I_k\}$ be any sequence of open intervals that covers $I$. Since $I$ is compact, by the generalized Heine-Borel theorem, there is a gyperfinite subcollection $\{J_i | 1 \leq i \leq n\}$, $n \in \mathbb{N}^\#$ of $I_k$ that still covers $I$. By reordering and deleting if necessary, we can assume that $a \in J_1 = (a_1, b_1), b_1 \in J_2 = (a_2, b_2), \dots, b_{n-1} \in J_n = (a_n, b_n)$, where $b_{n-1} \leq b < b_n$.
We then can compute that

$$ b - a < b_n - a_1 = \mathrm{Ext} - \sum_{i=2}^{n} (b_i - b_{i-1}) + (b_1 - a_1) < \mathrm{Ext} - \sum_{i=1}^{n} l(J_i) \leq \mathrm{Ext} - \sum_{i=1}^{\infty} l(I_k). $$

Therefore, $l(I) \leq \mu^*(I)$. We can now conclude that $\mu^*(I) = l(I)$. This proves the result for closed and bounded intervals.

Suppose that $I = (a, b)$ is an open and bounded interval. Then, $\mu^*(I) \leq l(I)$ as above and $b - a = \mu^*([a, b]) \leq \mu^*((a, b)) + \mu^*(a) + \mu^*(b) = \mu^*((a, b))$.
Hence $l(I) \leq \mu^*(I)$. The proof for half-open intervals is similar.

Finally, suppose that $I$ is an hyper infinite interval and let $M > 0$. There exists a bounded interval $J \subseteq I$ such that $\mu^*(J) = l(J) = M$ and it follows that $\mu^*(I) \geq \mu^*(J) = M$. Since $M > 0$ was arbitrary, $\mu^*(I) = \infty^\# = l(I)$.
This completes the proof.
(i) Suppose that $\mu^*(E) < \infty$. Then $E$ is measurable if and only if $\mu_*(E) = \mu^*(E)$.

(ii) If $E$ is measurable and $A \subseteq E$, then $\mu(E) = \mu_*(A) + \mu^*(E \setminus A)$.

**Proof.** For part (i), suppose that $E$ is a measurable set and let $\varepsilon > 0, \varepsilon \approx 0$. According to Theorem 2.9, there exists a $\#$-closed set $K$ such that $K \subseteq E$ and $\mu(E \setminus K) < \varepsilon$.

Thus, $\mu^*(E) \ge \mu_*(E) \ge \mu(K) > \mu(E) - \varepsilon = \mu^*(E) - \varepsilon$, which implies that the external inner

measure and external outer measure of $E$ are equal. Now let's prove the reverse direction. Suppose that $\mu_*(E) = \mu^*(E)$. Let $\varepsilon > 0, \varepsilon \approx 0$. Then there exists a $\#$-closed set $K$ and an $\#$-open set $G$ such that $K \subseteq E \subseteq G$ and $\mu(K) > \mu_*(E) - \varepsilon/2$ and $\mu(G) < \mu^*(E) + \varepsilon/2$. Then we find that

$$ \mu^*(G \setminus E) \le \mu^*(G \setminus K) = \mu(G \setminus K) = \mu(G) - \mu(K) < \varepsilon. $$

According to Theorem 2.9, the set $E$ is measurable.

For part (2), let $\varepsilon > 0, \varepsilon \approx 0$. There exists a $\#$-closed set $K \subseteq A$ such that $\mu(K) > \mu_*(A) - \varepsilon$. Then, $\mu(E) = \mu(K) + \mu(E \setminus K) > \mu_*(A) - \varepsilon + \mu^*(E \setminus A)$ and it follows that $\mu(E) \ge \mu_*(A) + \mu^*(E \setminus A)$. According to Theorem 2.9, there exists a measurable set $B$ such that $E \setminus A \subseteq B \subseteq E$ and $\mu(B) = \mu^*(E \setminus A)$.

Since $E \setminus B \subseteq A$, it follows that $\mu_*(E \setminus B) \ge \mu_*(A)$. Thus,

$$ \mu(E) = \mu(B) + \mu(E \setminus B) = \mu^*(E \setminus A) + \mu(E \setminus B) \le \mu^*(E \setminus A) + \mu_*(A). $$

By combining these two inequalities, we can obtain $\mu(E) = \mu_*(A) + \mu^*(E \setminus A)$.

## 2.3. External Lebesgue measure

**Definition 2.3.** A set $E \subseteq \mathbb{R}_c^\#$ is Lebesgue measurable if for each set $A \subseteq \mathbb{R}_c^\#$, the equality $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c)$ is satisfied. If $E$ is a Lebesgue measurable set, then the external Lebesgue measure of $E$ is its external Lebesgue outer measure and will be written as $\mu(E)$.

Since the external Lebesgue outer measure satisfies the property of subadditivity, then we always have $\mu^*(A) \le \mu^*(A \cap E) + \mu^*(A \cap E^c)$, $E^c = \mathbb{R}_c^\# \setminus E$ and we only need to check the reverse inequality.

Note that there is always a set $E$ that can divide $A$ into two mutually exclusive sets, $A \cap E$ and $A \cap E^c$. But only when $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c)$ holds, the set $E$ is Lebesgue measurable. The latter theorem will show some properties of measurable sets.

**Theorem 2.3.** The collection of measurable sets defined on $\mathbb{R}_c^\#$ has the following properties:

(a) Both $\emptyset$ and $\mathbb{R}_c^\#$ are measurable.

(b) If $E$ is measurable, then $E^c$ is measurable, where $E^c = \mathbb{R}_c^\# \setminus E$.

(c) If $\mu^*(E) = 0$, then $E$ is measurable.

(d) If $E_1$ and $E_2$ are measurable, then $E_1 \cup E_2$ and $E_2 \cap E_2$ are measurable.

(e) If $E$ is measurable, then $E + x_0$ is measurable.

**Proof.** For part (a), let $A \subseteq \mathbb{R}_c^\#$. Then

$$ \mu^*(A \cap \emptyset) + \mu^*(A \cap \emptyset^c) = \mu^*(\emptyset) + \mu^*(A) = 0 + \mu^*(A) = \mu^*(A), $$

$$ \mu^*(A \cap \mathbb{R}_c^\#) + \mu^*(A \cap \mathbb{R}_c^\#^c) = \mu^*(A) + \mu^*(\emptyset) = \mu^*(A) + 0 = \mu^*(A). $$

For part (b), if $E$ is measurable, then for every set $A \subseteq \mathbb{R}_c^\#$, such that

$$ \mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c). $$

$$ \mu^*(A \cap E^c) + \mu^*(A \cap (E^c \setminus E)) = \mu^*(A \cap E^c) + \mu^*(A \cap E) = \mu^*(A). $$

For part (c), let $A \subseteq \mathbb{R}_c^\#$. Since $\mu^*(E) = 0$ and $A \cap E \subseteq E$, then $\mu^*(A \cap E) = 0$.
## 2.2. External Lebesgue inner measure

In previous subsection, we have discussed external Lebesgue outer measure. There is

another external measure named external Lebesgue inner measure. Let's define the external inner measure and see some basic properties.

**Definition 2.2.** Let $E$ be a subset of $\mathbb{R}^\#$. The external inner measure of $E$ is defined by

$$ \mu^*(E) = \sup\{\mu(K) : K \subseteq E \text{ and } K \text{ is } \#\text{-closed}\} \qquad (2.3) $$

iff supremum in RHS of the (2.3) exists.

Recall that external Lebesgue outer measure of a set $E$ uses an infimum of the union of a sequence open sets that cover the set $E$, while external Lebesgue inner measure of a set $E$ uses a supremum of a set inside the set $E$. Then, it is obvious that

$$ \mu_*(E) \le \mu^*(E) \qquad (2.4) $$

for any set $E$. Also, for $A \subseteq B, \mu_*(A) \le \mu^*(B)$.

**Theorem 2.2.** Let $A$ and $E$ be subsets of $\mathbb{R}^\#$.

(i) Suppose that $\mu^*(E) < \infty$. Then $E$ is measurable if and only if $\mu_*(E) = \mu^*(E)$.

(ii) If $E$ is measurable and $A \subseteq E$, then $\mu(E) = \mu_*(A) + \mu^*(E \setminus A)$.

*Proof.* For part (i), suppose that $E$ is a measurable set and let $\varepsilon > 0, \varepsilon \approx 0$. According to Theorem 2.9, there exists a $\#$-closed set $K$ such that $K \subseteq E$ and $\mu(E \setminus K) < \varepsilon$.

Thus, $\mu^*(E) \ge \mu_*(E) \ge \mu(K) > \mu(E) - \varepsilon = \mu^*(E) - \varepsilon$, which implies that the external inner

measure and external outer measure of $E$ are equal. Now let's prove the reverse direction. Suppose that $\mu_*(E) = \mu^*(E)$. Let $\varepsilon > 0, \varepsilon \approx 0$. Then

there exists a $\#$-closed set $K$ and an $\#$-open set $G$ such that $K \subseteq E \subseteq G$ and $\mu(K) > \mu_*(E) - \varepsilon/2$ and $\mu(G) < \mu^*(E) + \varepsilon/2$. Then we find that

$$ \mu^*(G \setminus E) \le \mu^*(G \setminus K) = \mu(G \setminus K) = \mu(G) - \mu(K) < \varepsilon. $$

According to Theorem 2.9, the set $E$ is measurable.

For part (2), let $\varepsilon > 0, \varepsilon \approx 0$. There exists a $\#$-closed set $K \subseteq A$ such that $\mu(K) > \mu_*(A) - \varepsilon$. Then, $\mu(E) = \mu(K) + \mu(E \setminus K) > \mu_*(A) - \varepsilon + \mu^*(E \setminus A)$

and it follows that $\mu(E) \ge \mu_*(A) + \mu^*(E \setminus A)$. According to Theorem 2.9, there exists a measurable set $B$ such that $E \setminus B \subseteq B \subseteq E$ and $\mu(B) = \mu^*(E \setminus A)$.

Since $E \setminus B \subseteq A$, it follows that $\mu_*(E \setminus B) \ge \mu_*(A)$. Thus,

$$ \mu(E) = \mu(B) + \mu(E \setminus B) = \mu^*(E \setminus A) + \mu(E \setminus B) \le \mu^*(E \setminus A) + \mu_*(A). $$

By combining these two inequalities, we can obtain $\mu(E) = \mu_*(A) + \mu^*(E \setminus A)$.

## 2.3. External Lebesgue measure

**Definition 2.3.** A set $E \subseteq \mathbb{R}^\#$ is Lebesgue measurable if for each set $A \subseteq \mathbb{R}^\#$,

the equality $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c)$ is satisfied. If $E$ is a Lebesgue measurable set, then the external Lebesgue measure of $E$ is its external Lebesgue outer measure and will be written as $\mu(E)$.

Since the external Lebesgue outer measure satisfies the property of subadditivity, then we always have $\mu^*(A) \le \mu^*(A \cap E) + \mu^*(A \cap E^c)$, $E^c = \mathbb{R}^\# \setminus E$ and we only need to check the reverse inequality.

Note that there is always a set $E$ that can divide $A$ into two mutually exclusive sets, $A \cap E$ and $A \cap E^c$. But only when $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c)$ holds, the set $E$ is
We can obtain that $\mu^*(A) \geq \mu^*(A \cap E) = \mu^*(A \cap E) + \mu^*(A \cap E)$, which implies that $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E)$ by Theorem 2.1 part (e).

For part (d), let $A \subseteq \mathbb{R}_c^\#$. Note that

$$A \cap (E_1 \cup E_2) = (A \cap E_1) \cup (A \cap E_2) = (A \cap E_1) \cup (A \cap E_1 \cap E_2)$$

Then, by De Morgan Law and Theorem 14.1 part (e), we know that

$$\begin{aligned} \mu^*(A) &= \mu^*(A \cap E_1) + \mu^*(A \cap E_1) = \\ &= \mu^*(A \cap E_1) + \mu^*(A \cap E_1 \cap E_2) + \mu^*(A \cap E_1 \cap E_2) \geq \mu^*(A \cap (E_1 \cup E_2)) + \\ &\quad + \mu^*(A \cap (E_1 \cup E_2)), \end{aligned}$$

showing that $E_1 \cup E_2$ is measurable. Since $E_1 \cap E_2 = (E_1 \cup E_2)$, then the set $E_1 \cap E_2$ is measurable by Theorem 12.1 part (b).

For part (e), let $A \subseteq \mathbb{R}_c^\#$. Then,

$$\mu^*(A) = \mu^*(A - x_0) = \mu^*((A - x_0) \cap E) + \mu^*((A - x_0) \cap E^c) =$$

$$\mu^*(((A - x_0) \cap E) + x_0) + \mu^*(((A - x_0) \cap E^c) + x_0) =$$

$$\mu^*(A \cap (E + x_0)) + \mu^*(A \cap (E^c + x_0)).$$

Therefore, $E + x_0$ is measurable.

**Lemma 2.1.** Let $E_i : 1 \leq i \leq n \in \mathbb{N}^\#$ be a hyperfinite collection of disjoint measurable sets. If $A \subseteq \mathbb{R}_c^\#$, then

$$\mu^*\left(\bigcup_{i=1}^{n} (A \cap E_i)\right) = \mu^*\left(A \cap \left(\bigcup_{i=1}^{n} E_i\right)\right) = \mathrm{Ext}-\sum_{i=1}^{n} \mu^*(A \cap E_i).$$

**Proof.** We will prove this by the principle of mathematical induction. When $n = 1$, the equality holds. Suppose that the statement is valid for $n - 1$ disjoint measurable sets when $n > 1$. Then, when there are $n$ disjoint measurable sets,

$$\begin{aligned} \mu^*(A \cap (\bigcup_{i=1}^n E_i)) &= \mu^*(A \cap (\bigcup_{i=1}^n E_i) \cap E_n) + \mu^*(A \cap (\bigcup_{i=1}^n E_i) \cap E_n^c) = \\ &= \mu^*(A \cap E_n) + \mu^*(A \cap (\bigcup_{i=1}^{n-1} E_i)) = \\ &= \mu^*(A \cap E_n) + \mathrm{Ext}-\sum_{i=1}^{n-1} \mu^*(A \cap E_i) = \mathrm{Ext}-\sum_{i=1}^{n} \mu^*(A \cap E_i). \end{aligned}$$

$$\text{Note that when } A = \mathbb{R}_c^\#, \mu\left(\bigcup_{i=1}^n E_i\right) = \mathrm{Ext}-\sum_{i=1}^n \mu(E_i).$$

**Theorem 2.4.** If $\{E_i\}_{i=1}^\#$ is a hyper infinite sequence of disjoint measurable sets, then

$$\mu\left(\bigcup_{i=1}^{\infty} E_i\right) = \mathrm{Ext}-\sum_{i=1}^{\infty} \mu(E_i). \qquad (2.5)$$

**Proof.** According to Lemma 2.1, $\mathrm{Ext}-\sum_{i=1}^n \mu(E_i) = \mu\left(\bigcup_{i=1}^n E_i\right) \leq \mu\left(\bigcup_{i=1}^{\infty} E_i\right)$ for each positive integer $n \in \mathbb{N}^\#$, which implies that $\mathrm{Ext}-\sum_{i=1}^\infty \mu(E_i) \leq \mu\left(\bigcup_{i=1}^{\infty} E_i\right)$.

$$\text{By #-countably subadditive property, } \mathrm{Ext}-\sum_{i=1}^{\infty} \mu(E_i) \geq \mu\left(\bigcup_{i=1}^{\infty} E_i\right).$$

$$\text{Therefore, } \mathrm{Ext}-\sum_{i=1}^{\infty} \mu(E_i) = \mu\left(\bigcup_{i=1}^{\infty} E_i\right).$$

The previous theorem shows that if A and B are disjoint measurable sets, then $\mu(A \cup B) = \mu(A) + \mu(B)$. If $\{A_i\}_{i \in \mathbb{N}^\#}$ is a hyper infinite sequence of disjoint measurable sets, then $\mu\left(\bigcup_{i=1}^{\infty} E_i\right) = \mathrm{Ext}-\sum_{i=1}^{\infty} \mu(A_i)$. As so far, we have already seen that when the sets are measurable, Lebesgue measure satisfies property
Lebesgue measurable. The latter theorem will show some properties of measurable sets.

**Theorem 2.3.** The collection of measurable sets defined on $\mathbb{R}_c^\#$ has the following properties:

(a) Both $\emptyset$ and $\mathbb{R}_c^\#$ are measurable.

(b) If $E$ is measurable, then $E^c$ is measurable, where $E^c = \mathbb{R}_c^\# \setminus E$.

(c) If $\mu^*(E) = 0$, then $E$ is measurable.

(d) If $E_1$ and $E_2$ are measurable, then $E_1 \cup E_2$ and $E_2 \cap E_2$ are measurable.

(e) If $E$ is measurable, then $E + x_0$ is measurable.

**Proof.** For part (a), let $A \subseteq \mathbb{R}_c^\#$. Then

$$ \mu^*(A \cap \emptyset) + \mu^*(A \cap \emptyset^c) = \mu^*(\emptyset) + \mu^*(A) = 0 + \mu^*(A) = \mu^*(A), $$

$$ \mu^*(A \cap \mathbb{R}_c^\#) + \mu^*(A \cap \mathbb{R}_c^\#^c) = \mu^*(A) + \mu^*(\emptyset) = \mu^*(A) + 0 = \mu^*(A). $$

For part (b), if $E$ is measurable, then for every set $A \subseteq \mathbb{R}_c^\#$, such that

$$ \mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c). $$

$$ \mu^*(A \cap E^c) + \mu^*(A \cap (E^c) = \mu^*(A \cap E^c) + \mu^*(A \cap E) = \mu^*(A). $$

For part (c), let $A \subseteq \mathbb{R}_c^\#$. Since $\mu^*(E) = 0$ and $A \cap E \subseteq E$, then $\mu^*(A \cap E) = 0$.

We can obtain that $\mu^*(A) \ge \mu^*(A \cap E) = \mu^*(A \cap E) + \mu^*(A \cap E)$, which implies that $\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E)$ by Theorem 2.1 part (e).

For part (d), let $A \subseteq \mathbb{R}_c^\#$. Note that

$$ A \cap (E_1 \cup E_2) = (A \cap E_1) \cup (A \cap E_2) = (A \cap E_1) \cup (A \cap E_1 \cap E_2) $$

Then, by De Morgan Law and Theorem 14.1 part (e), we know that

$$ \begin{aligned} \mu^*(A) &= \mu^*(A \cap E_1) + \mu^*(A \cap E_1) = \\ &= \mu^*(A \cap E_1) + \mu^*(A \cap E_1 \cap E_2) + \mu^*(A \cap E_1 \cap E_2) \ge \mu^*(A \cap (E_1 \cup E_2)) + \\ &\quad + \mu^*(A \cap (E_1 \cup E_2)), \end{aligned} $$

showing that $E_1 \cup E_2$ is measurable. Since $E_1 \cap E_2 = (E_1 \cup E_2)$, then the set $E_1 \cap E_2$ is measurable by Theorem 12.1 part (b).

For part (e), let $A \subseteq \mathbb{R}_c^\#$. Then,

$$ \begin{aligned} \mu^*(A) &= \mu^*(A - x_0) = \mu^*((A - x_0) \cap E) + \mu^*((A - x_0) \cap E^c) = \\ &\quad \mu^*(((A - x_0) \cap E) + x_0) + \mu^*(((A - x_0) \cap E^c) + x_0) = \\ &\quad \mu^*(A \cap (E + x_0)) + \mu^*(A \cap (E^c + x_0)). \end{aligned} $$

Therefore, $E + x_0$ is measurable.

**Lemma 2.1.** Let $E_i : 1 \le i \le n \in \mathbb{N}^\#$ be a hyperfinite collection of disjoint measurable sets. If $A \subseteq \mathbb{R}_c^\#$, then

$$ \mu^*\left(\bigcup_{i=1}^{n} (A \cap E_i)\right) = \mu^*\left(A \cap \left(\bigcup_{i=1}^{n} E_i\right)\right) = \mathrm{Ext}-\sum_{i=1}^{n} \mu^*(A \cap E_i). $$

**Proof.** We will prove this by the principle of mathematical induction. When $n = 1$, the equality holds. Suppose that the statement is valid for $n - 1$ disjoint measurable sets when $n > 1$. Then, when there are $n$ disjoint measurable sets,

$$ \begin{aligned} \mu^*(A \cap (\bigcup_{i=1}^n E_i)) &= \mu^*(A \cap (\bigcup_{i=1}^n E_i) \cap E_n) + \mu^*(A \cap (\bigcup_{i=1}^n E_i) \cap E_n^c) = \\ &= \mu^*(A \cap E_n) + \mu^*(A \cap (\bigcup_{i=1}^{n-1} E_i)) = \\ &= \mu^*(A \cap E_n) + \mathrm{Ext}-\sum_{i=1}^{n-1} \mu^*(A \cap E_i) = \mathrm{Ext}-\sum_{i=1}^{n} \mu^*(A \cap E_i). \end{aligned} $$

$$ \text{Note that when } A = \mathbb{R}_c^\#, \mu\left(\bigcup_{i=1}^n E_i\right) = \mathrm{Ext}-\sum_{i=1}^n \mu(E_i). $$

**Theorem 2.4.** If $\{E_i\}_{i=1}^\infty$ is a hyper infinite sequence of disjoint measurable sets, then
(1),(2),(3) and (4). But what kinds of sets are measurable? Certainly every interval is measurable.

**Theorem 2.5.** Every interval $[a, b] \subset \mathbb{R}^\#$ is measurable.

**Theorem 2.6.** If $\{E_i\}_{i \in \mathbb{N}^\#}$ is a hyper infinite sequence of measurable sets, then

are measurable sets.

**Definition 2.4.** Let $f$ be a function from $E \subset \mathbb{R}^\#$ into $\mathbb{R}^\# \cup (-\infty^\#, \infty^\#)$. The function $f$ is (Lebesgue) measurable if

### 3. External Lebesgue Integral

Let $(\mathbb{R}, B, \mu)$ be the standard Lebesgue space on $\mathbb{R}$. Our internal starting point could be the internal measure space $(*\mathbb{R}, *B, *\mu)$. By transfer we can write down internal Lebesgue integrals

$$ \mathcal{L}[*f(t)] = \int_A *f(t)d^*\mu(t), $$

where $A \in *B$ and $f: \mathbb{R} \to \mathbb{R}$.

#### 3.1. Lebesgue Integral of a $\mathbb{R}^\#$-valued external function $f(x)$.

First, in particular, we need external function that can help us distinguish whether a given value $x$ is in the measurable set $A_i$. We call this function the characteristic function. The following statement is the formal definition of characteristic function and introduces the simple function.

**Definition 3.1.** For any set $A$, the function

$$ \chi_A(x) = \begin{cases} 1, & x \in A, \\ 0, & \text{otherwise} \end{cases} \qquad (3.1) $$

is called the characteristic function of set $A$. A linear combination of characteristic functions,

$$ \phi(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x) \qquad (3.2) $$

is called a simple function if the sets $A_i$ are measurable.

For a function $f: \mathbb{R}^\# \to \mathbb{R}^\#$ defined on a measurable set $A$ that takes no more than hyper finitely many distinct values $a_1, \dots, a_n, n \in \mathbb{N}^\#$ the function $f$ can always be written as a simple function

$$ f(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x), \qquad (3.3) $$

where $A_i = \{x \in A | f(x) = a_i\}$. That is a simple function of the first kind.

Therefore, simple functions can be thought of as dividing the range of $f$, where resulting sets $A_i$ may or may not be intervals.

Let us pause for a second. We want to ask ourselves: is the simple function $\phi(x)$ unique? The answer is no. Because we might define different disjoint sets that have a same function value. The simplest expression is

$$ \phi(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x) \qquad (3.4) $$

where $A_i = \{x \in A | \phi(x) = a_i\}$. At this case, the constants $a_i$ are distinct, the sets $A_i$ are disjoint and we call that representation the canonical representation of $\phi$.

Then, for simple functions, we define the Lebesgue integral as follows:
$$ \mu(\bigcup_{i=1}^{\infty} E_i) = \mathrm{Ext} - \sum_{i=1}^{\infty} \mu(E_i). \qquad (2.5) $$

**Proof.** According to Lemma 2.1, $\mathrm{Ext} - \sum_{i=1}^{n} \mu(E_i) = \mu(\bigcup_{i=1}^{n} E_i) \le \mu(\bigcup_{i=1}^{\infty} E_i)$ for each positive integer $n \in \mathbb{N}^\#$, which implies that $\mathrm{Ext} - \sum_{i=1}^{\infty} \mu(E_i) \le \mu(\bigcup_{i=1}^{\infty} E_i)$.

By #-countably subadditive property, $\mathrm{Ext} - \sum_{i=1}^{\infty} \mu(E_i) \ge \mu(\bigcup_{i=1}^{\infty} E_i)$.

Therefore, $\mathrm{Ext} - \sum_{i=1}^{\infty} \mu(E_i) = \mu(\bigcup_{i=1}^{\infty} E_i)$.

The previous theorem shows that if $A$ and $B$ are disjoint measurable sets, then $\mu(A \cup B) = \mu(A) + \mu(B)$. If $\{A_i\}_{i \in \mathbb{N}^\#}$ is a hyper infinite sequence of disjoint measurable sets, then $\mu(\bigcup_{i=1}^{\infty} E_i) = \mathrm{Ext} - \sum_{i=1}^{\infty} \mu(A_i)$. As so far, we have already seen that when the sets are measurable, Lebesgue measure satisfies property (1),(2),(3) and (4). But what kinds of sets are measurable? Certainly every interval is measurable.

**Theorem 2.5.** Every interval $[a, b] \subset \mathbb{R}^\#$ is measurable.

**Theorem 2.6.** If $\{E_i\}_{i \in \mathbb{N}^\#}$ is a hyper infinite sequence of measurable sets, then are measurable sets.

**Definition 2.4.** Let $f$ be a function from $E \subset \mathbb{R}^\#$ into $\mathbb{R}^\# \cup (-\infty^\#, \infty^\#)$. The function $f$ is (Lebesgue) measurable if

### 3. External Lebesgue Integral

Let $(\mathbb{R}, B, \mu)$ be the standard Lebesgue space on $\mathbb{R}$. Our internal starting point could be the internal measure space $(\ast\mathbb{R}, \ast B, \ast \mu)$. By transfer we can write down internal Lebesgue integrals

$$ \mathcal{L}[\ast f(t)] = \int_A \ast f(t) d^* \mu(t), $$

where $A \in \ast B$ and $f: \mathbb{R} \to \mathbb{R}$.

#### 3.1. Lebesgue Integral of a $\mathbb{R}^\#$-valued external function $f(x)$.

First, in particular, we need external function that can help us distinguish whether a given value $x$ is in the measurable set $A_i$. We call this function the characteristic function. The following statement is the formal definition of characteristic function and introduces the simple function.

**Definition 3.1.** For any set $A$, the function

$$ \chi_A(x) = \begin{cases} 1, & x \in A, \\ 0, & \text{otherwise} \end{cases} \qquad (3.1) $$

is called the characteristic function of set $A$. A linear combination of characteristic functions,

$$ \phi(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x) \qquad (3.2) $$
**Definition 3.2.** If $\phi(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x)$ is a simple function and $\mu(A_i)$ is gyperfinite for all $i$, then the Lebesgue integral of $\phi(x)$ is defined as

$$ \int_E \phi(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x). \qquad (3.5) $$

**Definition 3.3.** Suppose $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is a bounded function defined on a measurable set $E$ with giperfinite measure. We define the upper and lower Lebesgue integrals if exist, respectively, as

$$ I_L^\#(f) = \int_E \inf\{\phi(x)|\phi \text{ is simple and } \phi \ge f\} \qquad (3.6) $$

and

$$ I_{\#L}(f) = \int_E \sup\{\phi(x)|\phi \text{ is simple and } \phi \le f\}. \qquad (3.7) $$

If (i) the quantity $I_L^\#(f)$ and $I_{\#L}(f)$ exist and (ii) $I_L^\#(f) = I_{\#L}(f)$, then the function $f$ is called Lebesgue integrable over set $E$ and the external Lebesgue integral of $f$ over set $E$ is denoted by $I_L(f) = \int_E f dx$.

**The Lebesgue Integral for Simple Functions of the second kind**

Let $\varphi(x)$ be some simple external function of the second kind which takes on the gyperinfinitely many distinct values $y_1, \dots, y_n, \dots, n \in \mathbb{N}^\#$, $y_i \neq y_j$ for $i \neq j$.

It is natural to define the integral of the function $\varphi(x)$ over the set $E$ by the equation

$$ \int_E \varphi(x) d^\# \mu = \sum_{n \in \mathbb{N}^\#} y_n \mu\{x | x \in E, \varphi(x) = y_n\}. \qquad (3.8) $$

**Definition 3.4.** The simple function $\varphi(x)$ of the second kind is called integrable (with respect to the measure $\mu$) over the set $E$ if the gyper infinite series (15.8) #-converges absolutely.

If $\varphi(x)$ is #-integrable, then the sum of the series (15.8) is called the integral of $\varphi(x)$ over the set $E$.

**Remark 3.1.** Note that in definition 15.4 we assume that all the $y_n$ are different. One can, however, represent the value of the integral of a simple function as a sum of products of the form $c_k \mu(B_k)$ and not assume that all the $c_k$ are different.

**Lemma 3.1.** Let $A = \cup_k B_k, B_i \cap B_j = \emptyset$ for $i \neq j$, and assume that on each set $B_k$ the function $f(x)$ takes on only one value $c_k$. Then

$$ \int_A \varphi(x) d^\# \mu = \sum_{k \in \mathbb{N}^\#} c_k \mu(B_k). \qquad (3.9) $$

moreover, the function $f(x)$ is integrable over $A$ if and only if the gyper infinite series (3.9) #-converges absolutely.

**Proof.** It is easy to see that every set $A_n = \{x | x \in A, f(x) = y_n\}$ is the union of those $B_k$ for which $c_k = y_n$. Therefore

$$ \sum_{n \in \mathbb{N}^\#} y_n \mu(A_n) = \sum_{n \in \mathbb{N}^\#} y_n \sum_{c_k=y_k} \mu(B_k) = \sum_{k \in \mathbb{N}^\#} c_k \mu(B_k). \text{ Since the measure is non-negative, } \sum_{n \in \mathbb{N}^\#} |y_n| \mu(A_n) = \sum_{n \in \mathbb{N}^\#} |y_n| \sum_{c_k=y_k} \mu(B_k) = \sum_{k \in \mathbb{N}^\#} |c_k| \mu(B_k). $$

i.e., the series $\sum_{n \in \mathbb{N}^\#} y_n \mu(A_n)$ and $\sum_{k \in \mathbb{N}^\#} |c_k| \mu(B_k)$ both either #-converge absolutely or #-diverge.

Let us consider some properties of the external Lebesgue integral for simple external functions:

$$ \int_A f(x) d^\# \mu + \int_A g(x) d^\# \mu = \int_A [f(x) + g(x)] d^\# \mu \qquad (3.10) $$

moreover, from the existence of the integrals on the left-hand side it follows that the integrals on the right-hand side exist.
is called a simple function if the sets $A_i$ are measurable.

For a function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ defined on a measurable set $A$ that takes no more than gyper finitely many distinct values $a_1, \dots, a_n, n \in \mathbb{N}^\#$ the function $f$ can always be written as a simple function

$$f(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x), \qquad (3.3)$$

where $A_i = \{x \in A | f(x) = a_i\}$. That is a simple function of the first kind.

Therefore, simple functions can be thought of as dividing the range of $f$, where resulting sets $A_i$ may or may not be intervals.

Let us pause for a second. We want to ask ourselves: is the simple function $\phi(x)$ unique? The answer is no. Because we might define different disjoint sets that have a same function value. The simplest expression is

$$\phi(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x) \qquad (3.4)$$

where $A_i = \{x \in A | \phi(x) = a_i\}$. At this case, the constants $a_i$ are distinct, the sets $A_i$ are disjoint and we call that representation the canonical representation of $\phi$. Then, for simple functions, we define the Lebesgue integral as follows:

**Definition 3.2.** If $\phi(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x)$ is a simple function and $\mu(A_i)$ is gyperfinite for all $i$, then the Lebesgue integral of $\phi(x)$ is defined as

$$\int_E \phi(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x). \qquad (3.5)$$

**Definition 3.3.** Suppose $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is a bounded function defined on a measurable set $E$ with giperfinite measure. We define the upper and lower Lebesgue integrals if exist, respectively, as

$$I_L^\#(f) = \int_E \inf\{\phi(x)|\phi \text{ is simple and } \phi \ge f\} \qquad (3.6)$$

and

$$I_{\#L}(f) = \int_E \sup\{\phi(x)|\phi \text{ is simple and } \phi \le f\}. \qquad (3.7)$$

If (i) the quantity $I_L^\#(f)$ and $I_{\#L}(f)$ exist and (ii) $I_L^\#(f) = I_{\#L}(f)$, then the function $f$ is called Lebesgue integrable over set $E$ and the external Lebesgue integral of $f$ over set $E$ is denoted by $I_L(f) = \int_E f dx$.

### The Lebesgue Integral for Simple Functions of the second kind

Let $\varphi(x)$ be some simple external function of the second kind which takes on the gyperinfinitely many distinct values $y_1, \dots, y_n, \dots, n \in \mathbb{N}^\#, y_i \neq y_j$ for $i \neq j$.

It is natural to define the integral of the function $\varphi(x)$ over the set $E$ by the equation

$$\int_E \varphi(x) d^\# \mu = \sum_{n \in \mathbb{N}^\#} y_n \mu\{x | x \in E, \varphi(x) = y_n\}. \qquad (3.8)$$

**Definition 3.4.** The simple function $\varphi(x)$ of the second kind is called integrable (with respect to the measure $\mu$) over the set $E$ if the gyperinfinite series (15.8) #-converges absolutely.

If $\varphi(x)$ is #-integrable, then the sum of the series (15.8) is called the integral of $\varphi(x)$ over the set $E$.

**Remark 3.1.** Note that in definition 15.4 we assume that all the $y_n$ are different. One can, however, represent the value of the integral of a simple function as a sum of products of the form $c_k \mu(B_k)$ and not assume that all the $c_k$ are different.

**Lemma 3.1.** Let $A = \cup_k B_k, B_i \cap B_j = \emptyset$ for $i \neq j$, and assume that on each set $B_k$ the
To prove this assume that $f(x)$ takes on the values $f_i$, on the sets $F_i \subseteq A$, and $g(x)$ the values $g_i$, on the sets $G_i \subseteq A$, since

$$J_1 = \int_A f(x) d^\# \mu = \sum_{i \in \mathbb{N}^\#} f_i \mu(F_i) \qquad (3.11)$$

and

$$J_2 = \int_A g(x) d^\# \mu = \sum_{i \in \mathbb{N}^\#} g_i \mu(G_i). \qquad (3.12)$$

Then, by the Lemma 2.1 we get

$$J = \int_A [f(x) + g(x)] d^\# \mu = \sum_{i \in \mathbb{N}^\#} \sum_{j \in \mathbb{N}^\#} [f_i + g_j] \mu(F_i \cap G_j), \qquad (3.13)$$

where

$$\mu(F_i) = \sum_{j \in \mathbb{N}^\#} \mu(F_i \cap G_j), \mu(G_j) = \sum_{i \in \mathbb{N}^\#} \mu(F_i \cap G_j). \qquad (3.14)$$

From the absolute #-convergence of the series (3.11)-(3.12) it follows the absolute #-convergence of the series (3.13); here $J = J_1 + J_2$.

For any constant $k \in \mathbb{R}_c^\#$

$$k \int_A f(x) d^\# \mu = \int_A [kf(x)] d^\# \mu \qquad (3.15)$$

moreover, the existence of the integral on the left-hand side implies the existence of the

integral on the right. A simple function $f(x)$ which is bounded on the set $A \subset \mathbb{R}_c^\#$ is #-integrable over A; moreover, if $|f(x)| \le M \in \mathbb{R}_c^\#$ on A, then

$$\left| \int_A f(x) d^\# \mu \right| \le M \mu(A). \qquad (3.16)$$

## 4. General Definition and Basic Properties of the external Lebesgue Integral.

**Definition 4.1.** We shall say that the function $f(z)$ is #-integrable over the set $A \subset \mathbb{R}_c^\#$, if there exists a hyper infinite sequence of simple functions $f_n(z), n \in \mathbb{N}^\#$ which are #-integrable over A and #-converge uniformly to $f(x)$. We shall denote the #-limit

$$J = \#\text{-lim}_{n \to \infty} \int_A f_n(x) d^\# \mu \qquad (4.1)$$

by

$$\int_A f(x) d^\# \mu. \qquad (4.2)$$

and call it the integral of the external function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ over the set A.

This definition 4.1 is correct if the following conditions are satisfied:

1. The #-limit (4.1) for any uniformly #-convergent hyper infinite sequence of simple functions which are #-integrable over A exists.

2. This #-limit for a given function $f(x)$ does not depend on the choice of the hyper infinite

sequence $\{f_n(x)\}_{n \in \mathbb{N}^\#}$.

3. For simple functions the definitions of #-integrability and #-integral are equivalent to those given in section 3.

Notice that all these conditions are indeed satisfied.

To prove the first it suffices to note that by properties for #-integrals of simple functions,
function $f(x)$ takes on only one value $c_k$. Then

$$ \int_A \varphi(x) d^\# \mu = \sum_{k \in \mathbb{N}^\#} c_k \mu(B_k). \qquad (3.9) $$

moreover, the function $f(x)$ is integrable over A if and only if the hyper infinite series (3.9) #-converges absolutely.

**Proof.** It is easy to see that every set $A_n = \{x|x \in A, f(x) = y_n\}$ is the union of those $B_k$ for which $c_k = y_n$. Therefore

$$ \sum_{n \in \mathbb{N}^\#} y_n \mu(A_n) = \sum_{n \in \mathbb{N}^\#} y_n \sum_{c_k=y_k} \mu(B_k) = \sum_{k \in \mathbb{N}^\#} c_k \mu(B_k). $$ Since the measure is non-negative, $\sum_{n \in \mathbb{N}^\#} |y_n| \mu(A_n) = \sum_{n \in \mathbb{N}^\#} |y_n| \sum_{c_k=y_k} \mu(B_k) = \sum_{k \in \mathbb{N}^\#} |c_k| \mu(B_k).$

i.e., the series $\sum_{n \in \mathbb{N}^\#} y_n \mu(A_n)$ and $\sum_{k \in \mathbb{N}^\#} |c_k| \mu(B_k)$ both either #-converge absolutely or #-diverge.

Let us consider some properties of the external Lebesgue integral for simple external functions:

$$ \int_A f(x) d^\# \mu + \int_A g(x) d^\# \mu = \int_A [f(x) + g(x)] d^\# \mu \qquad (3.10) $$

moreover, from the existence of the integrals on the left-hand side it follows that the integrals on the right-hand side exist.

To prove this assume that $f(x)$ takes on the values $f_i$, on the sets $F_i \subseteq A$, and $g(x)$ the values $g_i$, on the sets $G_i \subseteq A$, since

$$ J_1 = \int_A f(x) d^\# \mu = \sum_{i \in \mathbb{N}^\#} f_i \mu(F_i) \qquad (3.11) $$

and

$$ J_2 = \int_A g(x) d^\# \mu = \sum_{i \in \mathbb{N}^\#} g_i \mu(G_i). \qquad (3.12) $$

Then, by the Lemma 2.1 we get

$$ J = \int_A [f(x) + g(x)] d^\# \mu = \sum_{i \in \mathbb{N}^\#} \sum_{j \in \mathbb{N}^\#} [f_i + g_j] \mu(F_i \cap G_j), \qquad (3.13) $$

where

$$ \mu(F_i) = \sum_{j \in \mathbb{N}^\#} \mu(F_i \cap G_j), \mu(G_j) = \sum_{i \in \mathbb{N}^\#} \mu(F_i \cap G_j). \qquad (3.14) $$

From the absolute #-convergence of the series (3.11)-(3.12) it follows the absolute #-convergence of the series (3.13); here $J = J_1 + J_2$.

For any constant $k \in \mathbb{R}^\#$

$$ k \int_A f(x) d^\# \mu = \int_A [kf(x)] d^\# \mu \qquad (3.15) $$

moreover, the existence of the integral on the left-hand side implies the existence of the integral on the right.A simple function $f(x)$ which is bounded on the set $A \subset \mathbb{R}^\#$ is #-integrable over A; moreover, if $|f(x)| \le M \in \mathbb{R}^\#$ on A, then

$$ \left| \int_A f(x) d^\# \mu \right| \le M \mu(A). \qquad (3.16) $$

## 4. General Definition and Basic Properties of the external Lebesgue Integral.

**Definition.4.1.** We shall say that the function $f(z)$ is #-integrable over the set $A \subset \mathbb{R}^\#$, if there exists a hyper infinite sequence of simple functions $f_n(z), n \in \mathbb{N}^\#$ which are #-integrable over A and #-converge uniformly to $f(x)$. We shall denote the #-limit
$$ \left| \int_A f_n(x) d^\# \mu - \int_A f_m(x) d^\# \mu \right| \le \mu(A) \sup_{x \in A} |f_n(x) - f_m(x)|. \quad (4.3) $$

To prove the second condition, we must consider the two hyperinfinite sequences $\{f_n(x)\}_{n \in \mathbb{N}^\#}$ and $\{f'_n(x)\}_{n \in \mathbb{N}^\#}$, and use the inequality

$$ \left| \int_A f_n(x) d^\# \mu - \int_A f'_n(x) d^\# \mu \right| \le \mu(A) \left[ \sup_{x \in A} |f_n(x) - f(x)| + \sup_{x \in A} |f'_n(x) - f(x)| \right]. \quad (4.4) $$

Finally, to prove the third condition it suffices to consider the hyperinfinite sequence $f_n(x) = f(x)$.

**The basic properties of the external Lebesgue #-integral.**

**Theorem 4.1.**

$$ \int_A 1 \cdot d^\# \mu = \mu(A). \quad (4.5) $$

**Proof.** Immediately from the definition of the #-integral.

**Theorem 4.2.** For any constant $k \in \mathbb{R}_c^\#$

$$ k \int_A f(x) d^\# \mu = \int_A [kf(x)] d^\# \mu \quad (4.6) $$

where the existence of the #-integral on the left-hand side implies the existence of the #-integral on the right.

**Proof.** The proof is obtained from property (8.15) by proceeding to the #-limit for an #-integral of simple functions.

**Theorem 4.3.** Assume that $f(x)$ and $g(x)$ are #-integrable over A then $f(x) + g(x)$ #-integrable over A and

$$ \int_A f(x) d^\# \mu + \int_A g(x) d^\# \mu = \int_A [f(x) + g(x)] d^\# \mu \quad (4.7) $$

Let $\{f_i(x)\}_{i=1}^n, n \in \mathbb{N}^\#$ be a hyperfinite sequence such that any $f_i(x)$ is #-integrable over A

then $\sum_{i=1}^n f_i(x)$ is #-integrable over A and

$$ \sum_{i=1}^{n} \int_{A} f_{i}(x) d^{\#} \mu = \int_{A} \left[ \sum_{i=1}^{n} f_{i}(x) \right] d^{\#} \mu \quad (4.8) $$

where the existence of the #-integrals on the left implies the existence of the #-integral on the right.

**Proof.** The proof of (4.7) is obtained from property A) by proceeding to the #-limit for an #-integral of simple functions.

**Theorem 4.4.** A function $f: A \to \mathbb{R}_c^\#$ which is hyperbounded on the set A is #-integrable over A.

**Proof.** The proof is obtained from property C) by proceeding to the limit for an integral of simple functions.

**Theorem 4.5.** If $f(x) \ge 0$, then

$$ \int_A f(x) d^\# \mu \ge 0 \quad (4.9) $$

assuming that the #-integral exists.

**Proof.** For simple functions this follows immediately from the definition; for the general
$$J = \#\text{-lim}_{n\to\infty} \int_A f_n(x) d^\# \mu \qquad (4.1)$$

by

$$\int_A f(x) d^\# \mu. \qquad (4.2)$$

and call it the integral of the external function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ over the set A.

This definition 4.1 is correct if the following conditions are satisfied:

1. The #-limit (4.1) for any uniformly #-convergent hyperinfinite sequence of simple functions which are #-integrable over A exists.

2. This #-limit for a given function $f(x)$ does not depend on the choice of the hyperinfinite sequence $\{f_n(x)\}_{n\in\mathbb{N}^\#}$.

3. For simple functions the definitions of #-integrability and #-integral are equivalent to those given in section 3.

Notice that all these conditions are indeed satisfied.

To prove the first it suffices to note that by properties for #-integrals of simple functions,

$$\left| \int_A f_n(x) d^\# \mu - \int_A f_m(x) d^\# \mu \right| \le \mu(A) \sup_{x \in A} |f_n(x) - f_m(x)|. \qquad (4.3)$$

To prove the second condition, we must consider the two hyperinfinite sequences $\{f_n(x)\}_{n\in\mathbb{N}^\#}$ and $\{f'_n(x)\}_{n\in\mathbb{N}^\#}$, and use the inequality

$$\left| \int_A f_n(x) d^\# \mu - \int_A f'_n(x) d^\# \mu \right| \le \mu(A) \left[ \sup_{x \in A} |f_n(x) - f(x)| + \sup_{x \in A} |f'_n(x) - f(x)| \right]. \qquad (4.4)$$

Finally, to prove the third condition it suffices to consider the hyperinfinite sequence $f_n(x) = f(x)$.

**The basic properties of the external Lebesgue #-integral.**

**Theorem 4.1.**

$$\int_A 1 \cdot d^\# \mu = \mu(A). \qquad (4.5)$$

**Proof.** Immediately from the definition of the #-integral.

**Theorem 4.2.** For any constant $k \in \mathbb{R}_c^\#$

$$k \int_A f(x) d^\# \mu = \int_A [kf(x)] d^\# \mu \qquad (4.6)$$

where the existence of the #-integral on the left-hand side implies the existence of the #-integral on the right.

**Proof.** The proof is obtained from property (8.15) by proceeding to the #-limit for an #-integral of simple functions.

**Theorem 4.3.** Assume that $f(x)$ and $g(x)$ are #-integrable over A then $f(x) + g(x)$ #-integrable over A and

$$\int_A f(x) d^\# \mu + \int_A g(x) d^\# \mu = \int_A [f(x) + g(x)] d^\# \mu \qquad (4.7)$$

Let $\{f_i(x)\}_{i=1}^n, n \in \mathbb{N}^\#$ be a hyperfinite sequence such that any $f_i(x)$ is #-integrable over A

then $\sum_{i=1}^n f_i(x)$ is #-integrable over A and
case the proof is based on the possibility of approximating non-negative functions by non-negative simple functions

**Corollary 4.1.** If $f(x) \ge g(x)$, then

$$ \int_A f(x) d^\# \mu \ge \int_A g(x) d^\# \mu. \qquad (4.10) $$

# Chapter III. $\mathbb{R}_c^\#$-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.

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 $\Omega \subset P(X)$ is fixed.

**Definition 1.3.** We say that $f$ is $\#$-measurable, if $f^{-1}([a, b]) \in \Omega$ for any hyperreals
$$ \sum_{i=1}^{n} \int_{A} f_i(x) d^{\#} \mu = \int_{A} \left[ \sum_{i=1}^{n} f_i(x) \right] d^{\#} \mu \quad (4.8) $$

where the existence of the #-integrals on the left implies the existence of the #-integral on the right.

**Proof.** The proof of (4.7) is obtained from property A) by proceeding to the #-limit for an #-integral of simple functions.

**Theorem 4.4.** A function $f: A \to \mathbb{R}_c^\#$ which is hyperbounded on the set A is #-integrable over A.

**Proof.** The proof is obtained from property C) by proceeding to the limit for an integral of simple functions.

**Theorem 4.5.** If $f(x) \ge 0$, then

$$ \int_A f(x) d^{\#} \mu \ge 0 \quad (4.9) $$

assuming that the #-integral exists.

**Proof.** For simple functions this follows immediately from the definition; for the general case the proof is based on the possibility of approximating non-negative functions by non-negative simple functions

**Corollary 4.1.** If $f(x) \ge g(x)$, then

$$ \int_A f(x) d^{\#} \mu \ge \int_A g(x) d^{\#} \mu. \quad (4.10) $$

# Chapter III.Basic analysis on external non-Archimedean field $\mathbb{R}_c^\#$.

## §1. The #-limit of a function $f: \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$

**Definition 1.** The $(\varepsilon, \delta)$ definition of the #-limit of a function $f: D \to \tilde{\mathbb{R}}_c^\#$ is as follows: Let $f$ be a $\mathbb{R}_c^\#$-valued function defined on a subset $D \subset \tilde{\mathbb{R}}_c^\#$ of the Cauchy hyperreal numbers. Let $c$ be a limit point of $D$ and let $L$ be a hyperreal number. We say that

$$ \#\lim_{x \to c} f(x) = L \quad (1.1) $$

if for every $\varepsilon \approx 0, \varepsilon > 0$ there exists a $\delta \approx 0, \delta > 0$ such that, for all $x \in D$, if $0 < |x - c| < \delta$, then $|f(x) - L| < \varepsilon$, symbolically:

$$ \lim_{x \to c} f(x) = L \iff (\forall \varepsilon > 0 \wedge \varepsilon > 0) \exists \delta (\delta \approx 0 \wedge \delta > 0) \forall x \in D, 0 < |x-c| < \delta \Rightarrow |f(x)-L| < \varepsilon. \quad (1.2) $$

**Definition 2.** The function $f: \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$ is #-continuous (or micro continuous) at some point $c$ of its domain if the #-limit of $f(x)$, as $x$ #-approaches $c$ through the domain of $f$, exists and is equal to $f(c)$:

$$ \#\lim_{x \to c} f(x) = f(c). \quad (1.3) $$

**Theorem 1.** If #-lim$_{x \to x_0}$ $f(x)$ exists; then it is unique that is; if #-lim$_{x \to x_0}$ $f(x) = L_1$ and #-lim$_{x \to x_0}$ $f(x) = L_2$, then $L_1 = L_2$.

**Theorem 2.** If #-lim$_{x \to x_0}$ $f_1(x) = L_1$ and #-lim$_{x \to x_0}$ $f_2(x) = L_2$ then
$a, b \in *\mathbb{R}_c^\#$ 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 $B \in B(\mathbb{R})$.

**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 $\mu^\#(X) < \infty^\#$.

(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) $\mu^\#\left(\bigcup_{n \in \mathbb{N}^\#} A_n\right) = \text{Ext}-\sum_{n \in \mathbb{N}^\#} \mu^\#(A_n)$ 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 2.4.** (a) A function $\mu^\# : \mathcal{F} \to *\mathbb{C}_c^\# \cup \{\infty\}$ 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 $0 < |\mu^\#(B)|_\# < \infty$.

(b) A function $\mu^\# : \mathcal{F} \to *\mathbb{R}_c^\# \cup \{\infty\}$ is called a signed \#-measure if
$$
\begin{align}
\#\lim_{x \to x_0} [f_1(x) \pm f_2(x)] &= L_1 \pm L_2, \notag \\
\#\lim_{x \to x_0} [f_1(x) \times f_2(x)] &= L_1 \times L_2, \notag \\
\#\lim_{x \to x_0} \frac{f_1(x)}{f_2(x)} &= \frac{L_1}{L_2}, L_2 \neq 0. \tag{1.4}
\end{align}
$$

**Definition 3.** (a) We say that $f(x)$ #-approaches the left-hand #-limit $L$ as $x$ #-approaches $x_0$ from the left, and write $\#\lim_{x \to x_0} f(x) = L$, if $f(x)$ is defined on some #-open interval $(a, x_0)$ and, for each $\varepsilon > 0, \varepsilon \approx 0$ there is a $\delta > 0, \delta \approx 0$ such that $|f(x) - L| < \varepsilon$ if $x_0 - \delta < x < x_0$.

(b) We say that $f(x)$ #-approaches the right-hand #-limit $L$ as $x$ #-approaches $x_0$ from the right, and write $\#\lim_{x \to x_0} f(x) = L$, if $f(x)$ is defined on some open interval $(x_0, b)$ and, for each $\varepsilon > 0$, there is a $\delta > 0, \delta \approx 0$ such that $|f(x) - L| < \varepsilon, \varepsilon > 0, \varepsilon \approx 0$ if $x_0 < x < x_0 + \delta$. Left- and right-hand #-limits are also called one-sided #-limits. We will often simplify the notation by writing $\#\lim_{x \to x_0} f(x) = f(x_0 -)$ and $\#\lim_{x \to x_0} f(x) = f(x_0 +)$.

**Theorem 3.** A function $f$ has a #-limit at $x_0$ if and only if it has left- and right-hand #-limits at $x_0$; and they are equal. More specifically; $\#\lim_{x \to x_0} f(x) = L$ if and only if $f(x_0 +) = f(x_0 -) = L$.

**Definition 4.** We say that $f(x)$ approaches the #-limit $L$ as $x$ approaches $\infty^\#$, and write $\#\lim_{x \to \infty^\#} f(x) = L$, if $f$ is defined on an interval $(a, \infty^\#)$ and, for each $\varepsilon > 0, \varepsilon \approx 0$, there is a number $\beta$ such that $|f(x)| = L| < \varepsilon$ if $x > \beta$.

**Definition 5.** We say that $f(x)$ approaches $\infty^\#$ as $x$ approaches $x_0$ from the left, and write

$\#\lim_{x \to x_0} f(x) = \infty^\#$ or $f(x_0^-) = \infty^\#$

$$
\#\lim_{x \to x_0} f(x) = \infty^\# \text{ or } f(x_0^-) = \infty^\# \qquad (1.5)
$$

if $f$ is defined on an interval $(a, x_0)$ and, for each hyperreal number $M$, there is a
$\delta \approx 0, \delta > 0$ such that $f(x) > M$ if $x_0 - \delta < x < x_0$.

Similarly we define: $\#\lim_{x \to x_0} f(x) = -\infty^\#$, $\#\lim_{x \to x_0} f(x) = \infty^\#$, $\#\lim_{x \to x_0} f(x) = -\infty^\#$, $\#\lim_{x \to x_0} f(x) = \infty^\#$.

Example 1. (i) $\#\lim_{x \to x_0} x_0^{-1} = -\infty^{\#}$, (ii) $\#\lim_{x \to x_0} x_0^{-1} = +\infty^{\#}$,
(iii) $\#\lim_{x \to x_0} x_0^{-1} = \infty^{\#}$

**Remark 1.** Throughout this paper, $\#\lim_{x \to x_0} f(x)$ exists” will mean that
$\#\lim_{x \to x_0} f(x) = L$, where $L$ is finite or hyperfinite.

To leave open the possibility that $L = \pm\infty^\#$, we will say that
$\#\lim_{x \to x_0} f(x)$ exists in the extended hyperreals.

This convention also applies to one-sided limits and limits as $x$ approaches $\pm\infty^\#$.

§ 2. Monotonic Functions $f : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$.

**Definition 6.** A function $f : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$ is nondecreasing on an interval $I \subset \tilde{\mathbb{R}}_c^\#$ if

$$
f(x_1) \leq f(x_2) \tag{1.6}
$$

whenever $x_1$ and $x_2$ are in $I$ and $x_1 < x_2$, or nonincreasing on $I$ if

$$
f(x_1) \geq f(x_2) \tag{1.7}
$$

whenever $x_1$ and $x_2$ are in $I$ and $x_1 < x_2$.

In either case, $f$ is on $I$. If $\le$ can be replaced by $<$ in (1.6), $f$ is increasing on $I$. If $\ge$
$\mu^{\#}(\emptyset) = 0$
$\mu^{\#}\left(\bigcup_{n \in \mathbb{N}^{\#}} A_n\right) = \text{Ext-}\sum_{n \in \mathbb{N}^{\#}} \mu^{\#}(A_n)$ 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 $0 < |\mu^{\#}(B)| < {}^{*}\infty$.

**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 $\mu^\#(X) < {}^*\infty$, $\{f_n\}_{n=1}^{*\infty}$ 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

$E_n(k) := \bigcup_{m=n}^{*\infty} \{x : |f_m(x)| \ge k - 1\}$. 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 $\varepsilon \approx 0$, 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
can be replaced by > in (1.7), f is decreasing on I. In either of these two cases, f is strictly monotonic on I.

**Theorem 4.** Suppose that $f(x)$ is monotonic on $(a, b)$ and define
$\alpha = \inf_{a<x<b} f(x)$ and $\beta = \sup_{<x<b} f(x)$. Suppose that $\exists \alpha$ and $\exists \beta$, then:

(a) If $f$ is nondecreasing, then $f(a^+) = \alpha$ and $f(b^-) = \beta$.

(b) If $f$ is nonincreasing; then $f(a^+) = \beta$ and $f(b^-) = \alpha$.

Here $a^+ = -\infty^+$ if $a = -\infty^+$ and $b^+ = \infty^+$ if $b = \infty^+$.

(c) If $a < x_0 < b$, then $f(x_0^+) $ and $f(x_0^-)$ exist and are finite or hyperfinite;
moreover, $f(x_0^+) \le f(x_0) \le f(x_0^-)$ if $f$ is nondecreasing, and $f(x_0^+) \ge f(x_0) \ge f(x_0^-)$
if $f$ is nonincreasing:

**Proof (a)** We first show that $f(a^+) = \alpha$. If $M > \alpha$, there is an $x_0$ in $(a, b)$ such that $f(x_0) < M$. Since $f$ is nondecreasing, $f(x) < M$ if $a < x < x_0$. Therefore, if $\alpha = -\infty^{\#}$, then $f(a^+) = -\infty^{\#}$. If $\alpha > -\infty^{\#}$, let $M = \alpha + \varepsilon$, where $\varepsilon \approx 0, \varepsilon > 0$.

Then $\alpha \le f(x) < \alpha + \varepsilon$, so (i) $|f(x) - \alpha| < \varepsilon$ if $a < x < x_0$.

If $a = -\infty^{\#}$, this implies that $f(-\infty^{\#}) = \alpha$. If $a > -\infty^{\#}$, let $\delta = x_0 - a$. Then (i) is equivalent to $|f(x) - \alpha| < \varepsilon$ if $a < x < a + \delta$, which implies that $f(a^+) = \alpha$.

We now show that $f(b^+) = \beta$. If $M < \beta$, there is an $x_0$ in $(a, b)$ such that $f(x_0) > M$.

Since $f(x)$ is nondecreasing, $f(x) > M$ if $x_0 < x < b$. Therefore, if $\beta = \infty^{\#}$, then
$f(b^-) = \infty^{\#}$. If $\beta < \infty^{\#}$, let $M = \beta - \varepsilon$, where $\varepsilon \approx \varepsilon > 0$. Then $\beta - \varepsilon < f(x) \le \beta$,
so (ii) $|f(x) - \beta| < \varepsilon$ if $x_0 < x < b$.

If $b = \infty^{\#}$, this implies that $f(\infty^{\#}) = \beta$. If $b < \infty^{\#}$, let $\delta = b - x_0$. Then (ii) is
equivalent to $f(x) < f(b - \delta) < x < b$, which implies that $f(b^-) = \beta$.

(b) The proof is similar to the proof of (a).

(c) Suppose that $f(x)$ is nondecreasing. Applying (a) to $f(x)$ on $(a, x_0)$ and $(x_0, b)$
separately shows that $f(x_0^-) = \sup_{a<x<x_0} f(x)$ and $f(x_0^+) = \inf_{x_0<x<b} f(x)$.

However, if $x_1 < x_0 < x_2$, then $f(x_1) \le f(x_0) \le f(x_2)$ and hence, $f(x_0^-) \le f(x_0) \le f(x_0^+)$.

§ 3. #-Limits Inferior and Superior

**Definition 7.** We say that: (i) $f$ is bounded on a set $S \subseteq \tilde{\mathbb{R}}_c^\#$ if there is a constant $M \in \mathbb{R}$, $M < \infty$ such that $f(x) \le M$ for all $x \in S$, (ii) $f$ is hyperbounded on a set $S \subseteq \tilde{\mathbb{R}}_c^\#$ if $f$ is not bounded on a set $S$ and there is a constant $M \in \tilde{\mathbb{R}}_c^\#/\tilde{\mathbb{R}}_{c,fin}^\#$, $M < \infty^\#$ such that $f(x) \le M$ for all $x \in S$.

**Definition 8.** Suppose that $f$ is bounded or hyperbounded on $[a, x_0)$, where $x_0$
may be finite or hyperfinite or $\infty^\#$. For $a \le x < x_0$, define (i) $S_f(x; x_0) = \sup_{x \le t < x_0} f(t)$
and

(ii) $I_f(x; x_0) = \inf_{x \le t \le x_0} f(t)$.

Then the left `-limit superior` of $f(x)$ at $x_0$ is defined to be

$$
\#\overline{\lim}_{x \to \# x_0^-} f(x) = \#\lim_{x \to \# x_0^-} S_f(x; x_0) \quad (1.8)
$$

and the left limit inferior of $f(x)$ at $x_0$ is defined to be

$$
\# \underline{\lim}_{x \to \# x_0^-} f(x) = \# \lim_{x \to \# x_0^-} I_f(x; x_0). \quad (1.9)
$$

If $x_0 = \infty^{\#}$, we define $x_0 = -\infty^{\#}$.

**Theorem 5.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then $\beta = \#\overline{\lim}_{x \to \# x_0^-} f(x)$ exists and is the unique hyperreal number with the following properties:

(a) If $\varepsilon > 0, \varepsilon \approx 0$, there is an $a_1$ in $[a, x_0)$ such that
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 $\mu^{\#}(Y_n\ \&\ X'_n) < n^{-1}$ 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 $Y_n\ \&\ X'_0 \subseteq Y_n\ \&\ X'_n$, so $\mu^{\#}(Y_n\ \&\ X'_0) < n^{-1}$
for all $n \in *\mathbb{N}$. Then $\mu^{\#}(Y_n\ \&\ X'_0) = 0$, and $\mu^{\#}((\bigcup_{n=1}^{*\infty} Y_n)\ \&\ X'_0) = 0$, or $\mu^{\#}(X\ \&\ X'_0) = 0$.

Let $X_n = (X'_n \cup (X\ \&\ X'_0) \cap Y_n)$, 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
$A_1 = X_1$ 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 $D = \bigcup_n D_n$ is an element of $P_A$ as well. If $D$ is a proper subset of $A$, then $\mu^{\#}(A \setminus D) > 0$. The set $A \setminus D$ 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 $A = Z_m = Y_m \setminus \bigcup_{k=1}^{m-1} Y_k$ 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 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 
$\lim \sup_{n \to *^\infty} \mu^\#(A_n) = 0$.

## §2.2. Vector-valued #-measures

In this section, we extend the notion of a measure. Then we study the basic
(i) $f(x) < \beta + \varepsilon$ if $a_1 \le x < x_0$

(b) If $\varepsilon > 0, \varepsilon \approx 0$ and $a_1$ is in $[a, x_0)$, then
$f(\bar{x}) > \beta - \varepsilon$ for some $\bar{x} \in [a, x_0)$.

**Proof.** Since $f(x)$ is bounded or hyperbounded on $[a, x_0)$, $S_f(x; x_0)$ is nonincreasing and bounded or hyperbounded on $[a, x_0)$. By applying Theorem 17.4(b) to $S_f(x; x_0)$, we conclude that $\beta$ exists finite or hyperfinite.

Therefore, if $\varepsilon > 0, \varepsilon \approx 0$, there is an $\bar{a}$ in $[a, x_0)$ such that

(ii) $\beta - \varepsilon/2 < S_f(x; x_0) < \beta + \varepsilon/2$ if $\bar{a} \le x < x_0$.

Since $S_f(x; x_0)$ is an upper bound of $\{f(t) | x \le t < x_0\}$, $f(x) < S_f(x; x_0)$. Therefore, the second inequality in (ii) implies the inequality (i) with $a_1 = \bar{a}$. This proves (a). To prove (b), let $a_1$ be given and define $x_1 = \max\{a_1, \bar{a}\}$. Then the first inequality in (ii) implies that (iii) $S_f(x; x_0) > \beta - \varepsilon/2$. Since $S_f(x; x_0)$ is the supremum of $\{f(t) | x_1 \le t < x_0\}$, there is an $\bar{x}$ in $[x_1, x_0)$ such that

$f(\bar{x}) > S_f(x; x_0) - \varepsilon/2$. This and (iii) imply that $f(\bar{x}) > \beta - \varepsilon/2$. Since $\bar{x}$ is in $[a_1, x_0)$, this proves (b).

Now we show that there cannot be more than one hyperreal number with properties (a) and (b). Suppose that $\beta_1 < \beta_2$ and $\beta_2$ has property (b); thus, if $\varepsilon \approx 0, \varepsilon > 0$ and $a_1$ is in $[a, x_0)$ there is an $\bar{x}$ in $[a_1, x_0)$ such that $f(\bar{x}) > \beta_2 - \varepsilon$. Letting $\varepsilon = \beta_2 - \beta_1$, we see that there is an $\bar{x}$ in $[a_1, b)$ such that $f(\bar{x}) > \beta_2 - (\beta_2 - \beta_1) = \beta_1$ so $\beta_1$ cannot have property (a). Therefore, there cannot be more than one hyperreal number that satisfies both (a) and (b).

**Theorem 6.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then $\alpha = \varliminf_{x \to x_0} f(x)$ exists and there is the unique hyperreal number with the following properties:

(a) If $\varepsilon \approx 0, \varepsilon > 0$ there is an $a_1$ in $[a, x_0)$ such that
$f(x) > \alpha - \varepsilon$ if $a_1 \le x < x_0$.

(b) If $\varepsilon \approx 0, \varepsilon > 0$ and $a_1$ is in $[a, x_0)$, then
$f(\bar{x}) < \alpha + \varepsilon$ for some $\bar{x} \in [a, x_0)$.

**Theorem 7.** If $f(x)$ is bounded or hyperbounded on $[a, x_0)$, then

(i) $\#\varliminf_{x \to x_0} f(x) \le \#\varlimsup_{x \to x_0} f(x)$;

(ii) $\#\varliminf_{x \to x_0} f(-x) = \#\varlimsup_{x \to x_0} f(x)$;

(iii) $\#\varliminf_{x \to x_0} f(-x) = \#\varlimsup_{x \to x_0} f(x)$;

(iv) $\#\varliminf_{x \to x_0} f(x) = \#\varlimsup_{x \to x_0} f(x)$ if and only if $\#\lim_{x \to x_0} f(x)$ exists, in which case
$\#\lim_{x \to x_0} f(x) = \#\varliminf_{x \to x_0} f(x) = \#\varlimsup_{x \to x_0} f(x)$

**Theorem 8.** Suppose that $f(x)$ and $g(x)$ are bounded or hyperbounded on $[a, x_0)$.
Then: (i) $\#\varliminf_{x \to x_0} (f+g)(x) \le \#\varlimsup_{x \to x_0} f(x) + \#\varlimsup_{x \to x_0} g(x)$;

(ii) $\#\varliminf_{x \to x_0} (f+g)(x) \ge \#\varlimsup_{x \to x_0} f(x) + \#\varlimsup_{x \to x_0} g(x)$.

**Theorem 9.** The $\alpha = \lim_{x \to x_0} f(x)$ exists i.e., $\alpha$ is finite or hyperfinite if and only if for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that $|f(x_1) - f(x_2)| < \varepsilon$ if $x_0 - \delta < x_1, x_2 < x_0$.

**Theorem 10.** (i) Suppose that $f(x)$ is bounded or hyperbounded on an interval $(x_0, b]$, then $\#\varliminf_{x \to x_0} f(x) = \#\varlimsup_{x \to x_0} f(x)$ if and only if $\#\lim_{x \to x_0} f(x)$ exists, in which case $\#\lim_{x \to x_0} f(x) = \#\varliminf_{x \to x_0} f(x) = \#\varlimsup_{x \to x_0} f(x)$.

(ii) Suppose that $f(x)$ is bounded or hyperbounded on an open interval containing $x_0$, then $\#\lim_{x \to x_0} f(x)$ exists if and only if
$\#\varliminf_{x \to x_0} f(x) = \#\varlimsup_{x \to x_0} f(x) = \#\varliminf_{x \to x_0} f(x) = \#\varlimsup_{x \to x_0} f(x)$.
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 $(A_k)_{k \in \Sigma^#}$;

(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 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
$$ \Phi(E) = \text{Ext}-\sum_{k=1}^m \mu_k(E \cap A_k) \times w_k, (E \in \Omega). $$

## §3. The 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

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

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

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

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

$$ (i) \quad 0 \le f_1(x) \le f_2(x) \quad \text{and} \quad (ii) \quad f_1(x) = \text{Ext}-\sum_{i=1}^n \lambda_{1,i} \chi_{A_i}(x), f_2(x) = \text{Ext}-\sum_{i=1}^n \lambda_{2,i} \chi_{A_i}(x). $$ 
$$ \text{Ext}-\sum_{i=1}^n \lambda_{1,i} \le \text{Ext}-\sum_{i=1}^n \lambda_{2,i}, \quad (3.3) $$

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
§ 4. The #-continuity of a function $f: \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$.

**Definition 9.** (i) We say that a function $f: \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$. is #-continuous at $x_0$ if $f$ is defined on an open interval $(a, b)$ containing $x_0$ and #-lim$_{x\to a} x_0$-$f(x_0) = x_0$.

(ii) We say that $f$ is #-continuous from the left at $x_0$ if $f$ is defined on an open interval $(a, x_0)$ and $f(x_0 -) = f(x_0)$.

(iii) We say that $f$ is #-continuous from the right at $x_0$ if $f$ is defined on an open interval $(x_0, b)$ and $f(x_0 +) = f(x_0)$.

**Theorem 11.** (i) A function $f$ is #-continuous at $x_0$ if and only if $f$ is defined on an open interval $(a, b)$ containing $x_0$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that

$$|f(x) - f(x_0)| < \varepsilon \quad (1.10)$$

whenever $|x - x_0| < \delta$.

(ii) A function $f$ is #-continuous from the right at $x_0$ if and only if $f$ is defined on an interval $[x_0, b)$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that (1.10) holds whenever $x_0 \le x < x_0 + \delta$.

(iii) A function $f$ is #-continuous from the left at $x_0$ if and only if $f$ is defined on an interval $(a, x_0]$ and for each $\varepsilon \approx 0, \varepsilon > 0$ there is a $\delta \approx 0, \delta > 0$ such that (1.10) holds whenever $x_0 - \delta < x \le x_0$.

Note that from Definition 1.9 and Theorem 1.8, $f$ is #-continuous at $x_0$ if and only if $f(x_0 +) = f(x_0 -) = f(x_0)$ or, equivalently, if and only if it is #-continuous from the right and left at $x_0$.

**Definition 1.10.** A function $f: \mathbb{R}_c^\# \to \mathbb{R}_c^\#$ is #-continuous on an open interval $(a, b)$ if it is #-continuous at every point in $(a, b)$. If, in addition,

$$f(b -) = f(b) \quad (1.11)$$

or

$$f(a +) = f(a) \quad (1.12)$$

then $f$ is #-continuous on $(a, b]$ or $[a, b)$, respectively. If $f$ is #-continuous on $(a, b)$ and (1.11) and (1.12) both hold, then $f$ is #-continuous on $[a, b]$. More generally, if $S$ is a subset of $\text{dom}(f)$ consisting of finitely or countably or hyper finitely or hyper infinitely many disjoint intervals, then $f$ is #-continuous on $S$ if $f$ is #-continuous on every interval in $S$.

**Definition 1.11.** A function $f: \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$ is piecewise #-continuous on $[a, b]$ if

(i) $f(x_0 +)$ exists for all $x_0$ in $[a, b]$;

(ii) $f(x_0 -)$ exists for all $x_0$ in $(a, b]$;

(iii) $f(x_0 +) = f(x_0 -) = f(x_0)$ for all but except finitely or hyper finitely many points $x_0$ in $(a, b)$.

If (iii) fails to hold at some $x_0$ in $(a, b)$, $f$ has a jump #-discontinuity at $x_0$. Also, $f$ has a jump #-discontinuity at $a$ if $f(a +) \neq f(a)$ or at $b$ if $f(b -) \neq f(b)$.

**Theorem 1.12.** If $f$ and $g$ are #-continuous on a set $S$, then so are $f \pm g$, and $fg$. In addition, $fg$ is #-continuous at each $x_0$ in $S$ such that $g(x_0) \neq 0$.

By hyper infinite induction, it can be shown that if $\forall n \in \mathbb{N}^\# f_n(x)$ are #-continuous on a set $S$, then so are $\text{Ext-}\sum_{i \le n} f_n(x)$. Therefore, $\forall n, m \in \mathbb{N}^\#$ any rational function
hyper infinite sequence of simple external functions which #-converges uniformly to f. Then the Lebesgue #-integral of f is

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

**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

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

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 $A \in \mathcal{F}$:

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

**Lemma 3.1.** (1) Let $f : X \to \mathbb{R}_c^\#$ be an arbitrary nonnegative #-measurable function then

$$ \mathrm{Ext}-\int_X f(x)d^\#\mu^\# = \sup\left\{\mathrm{Ext}-\int_X \varphi(x)d^\#\mu^\# \mid \varphi \text{ is a simple function such that } 0 \le \varphi(x) \le_s f(x)\right\}. \quad (3.7) $$

(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 $0 \le f \le g$ #-a.e., then
$$r(x) = \mathrm{Ext} - \sum_{i \le n} a_i x^i \left( \mathrm{Ext} - \sum_{i \le m} b_i x^i \right), b_i \ne 0 \text{ is #-continuous for all values of } x \text{ except}$$

those for which its denominator vanishes.

## §5. Removable #-discontinuities.

**Definition 1.12.** Let $f(x)$ be defined on a deleted #-neighborhood of $x_0$ and #-discontinuous (perhaps even undefined) at $x_0$. Then we say that $f(x)$ has a removable #-discontinuity at $x_0$ if $\lim_{x \to x_0} f(x_0)$ exists. In this case, the function

$$g(x) = \begin{cases} f(x) & \text{if } x \in \mathrm{dom}(f) \text{ and } x \neq x_0 \\ \lim_{x \to x_0} f(x_0) & \text{if } x = x_0 \end{cases} \qquad (1.13)$$

is #-continuous at $x_0$.

## §6. Composite Functions $f : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$.

**Definition 15.13.** Suppose that $f : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$ and $g : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$ are functions with domains $\mathrm{dom}(f)$ and $\mathrm{dom}(g)$ correspondingly. If $\mathrm{dom}(g)$ has a nonempty subset $T$ such that $g(x) \in \mathrm{dom}(g)$ whenever $x \in T$, then the composite function $f \circ g : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$ is defined on $T$ by $(f \circ g)(x) = f(g(x))$

**Theorem 1.10.** Suppose that $g$ is #-continuous at $x_0$, $g(x_0)$ is an #-interior point of $\mathrm{dom}(f)$ and $f$ is #-continuous at $g(x_0)$. Then $f \circ g$ is #-continuous at $x_0$.

**Proof.** Suppose that $\varepsilon \approx 0, \varepsilon > 0$. Since $g(x_0)$ is an #-interior point of $\mathrm{dom}(f)$ and $f(x)$ is #-continuous at $g(x_0)$, there is a $\delta_1 \approx 0, \delta_1 > 0$ such that $f(t)$ is defined and

(i) $|f(t) - f(gx_0)| < \varepsilon$ if $|t - g(x_0)| < \delta_1$.

Since $g(x)$ is #-continuous at $x_0$, there is a $\delta \approx 0, \delta > 0$ such that $g(x)$ is defined and

(ii) $|g(x) - g(x_0)| < \delta_1$ if $|x - x_0| < \delta$.

Now (i) and (ii) imply that $|f(g(x)) - f(g(x_0))| < \delta$ if $|x - x_0| < \varepsilon$. Therefore, $f \circ g$ is #-continuous at $x_0$.

## §7. Bounded and Hyperbounded Functions $f : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$.

**Definition 14.** (i) A function $f : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$ is bounded below on a set $S \subset \tilde{\mathbb{R}}_c^\#$ if there is a finite or hyperfinite hyperreal number $m \in \tilde{\mathbb{R}}_{c, \mathrm{fin}}^\#$ such that $f(x) \ge m$ for all $x \in S$. If in this case the set $V = \{f(x) | x \in S\}$ has infimum $\alpha$, we write $\alpha = \inf_{x \in S} f(x)$. If there is a point $x_1 \in S$ such that $f(x_1) = \alpha_>$, we say that $\alpha$ is the minimum of $f(x)$ on $S$, and write $\alpha = \min_{x \in S} f(x)$

(ii) A function $f : \tilde{\mathbb{R}}_c^\# \to \tilde{\mathbb{R}}_c^\#$ is bounded above on $S \subset \tilde{\mathbb{R}}_c^\#$ if there is a finite or hyperfinite hyperreal number $M \in \tilde{\mathbb{R}}_{c, \mathrm{fin}}^\#$ such that $f(x) \le M$ for all $x \in S$. If in this case, $V$ has a supremum $\beta$, we write $\beta = \sup_{x \in S} f(x)$. If there is a point $x_1 \in S$ such that $f(x_2) = \beta_>$, we say that $\beta$ is the maximum of $f(x)$ on $S$, and write $\alpha = \max_{x \in S} f(x)$.

(iii) If $f$ is bounded above and below on a set $S$, we say that $f$ is bounded on $S$.

**Theorem 11.** If $f$ is #-continuous on a finite or hyperfinite #-closed interval $[a, b]$,
$$ \mathrm{Ext}-\int_X f(x)d^\#\mu^\# \le \mathrm{Ext}-\int_X g(x)d^\#\mu^\#. $$

#-Convergence theorem

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

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

**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 $f_j \le s_j f_{j+1}$ for all $j$ and $f(x) = \sup_{n \in \mathbb{N}} f_n(x)$ then

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

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

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

($\ast$-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

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

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 $(E_n)_n^{\infty}$ is an increasing hyper infinite sequence of #-measurable sets whose union is $X$, and we have

$$ \mathrm{Ext}-\int_X f_n(x)d^\#\mu^\# \ge \mathrm{Ext}-\int_{E_n} f_n(x)d^\#\mu^\# \ge \alpha \left(\mathrm{Ext}-\int_{E_n} \varphi(x)d^\#\mu^\#\right) \qquad (3.13) $$

By (3.10) and by Proposition 2.5,

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

and hence

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

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

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

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

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

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

**Corollary 3.1** If $(f_n)_n^{\infty}$ is a hyper infinite sequence in $L_1^+(X)$ and $f = \mathrm{Ext}-\sum_{n=1}^{\infty} f_n$ pointwise then

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

**Corollary 3.2** If $(f_n)_n^{\infty}$ is a hyper infinite sequence in $L_1^+(X)$, $f \in L_1^+(X)$, and $f_n \to_\# f$ $\mu^\#$-a.e., then
then $f$ is bounded or hyperbounded on $[a, b]$.

**Proof.** Suppose that $t \in [a, b]$. Since $f$ is #-continuous at $t$, there is an open interval $I_t$ containing $t$ such that

$$|f(x) - f(t)| < 1 \text{ if } x \in I_t \cap [a, b] \qquad (1.14)$$

To see this, set $\varepsilon = 1$ in (15.10), Theorem 11. The collection $H = \{I_t | a \le t \le b\}$ 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 $t_1, t_2, \dots, t_n, n \in \mathbb{N}^\#$ such that the intervals $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$. According to (1.14) with $t = t_i$, $|f(x) - f(t_i)| < 1$ if $x \in I_{t_i} \cap [a, b]$. Therefore,

$$f(x) = |(f(x) - f(t_i)) + f(t_i)| \le |f(x) - f(t_i)| + |f(t_i)| \le 1 + |f(t_i)| \qquad (1.15)$$

if $x \in I_{t_i} \cap [a, b]$. Let $M = 1 + \max_{1 \le j \le n} |f(t_i)|$. Since $[a, b] \subset \bigcup_{i=1}^n (I_{t_i} \cap [a, b])$, (1.15) implies that $|f(x)| \le M$ if $x \in [a, b]$.

**Theorem 12.** Suppose that $f$ is #-continuous on a finite or hyperfinite #-closed interval $[a, b]$. Let $V_{a,b} = \{f(x) | x \in [a, b]\}$. Assume that the set $V_{a,b}$ is admissible above and below. Let

$$\alpha = \inf V_{a,b} = \inf_{a \le x \le b} f(x) \text{ and } \beta = \sup V_{a,b} = \sup_{a \le x \le b} f(x). \qquad (1.16)$$

Then $\alpha$ and $\beta$ are respectively the minimum and maximum of $f$ on $[a, b]$; that is there are points $x_1$ and $x_2$ in $[a, b]$ such that $\alpha = f(x_1)$ and $\beta = f(x_2)$.

**Proof.** We show that $x_1$ exists. Note that $\alpha = \inf V_{a,b}$ and $\beta = \sup V_{a,b}$ exist since the set $V_{a,b}$ is admissible below and above. Suppose that there is no $x_1$ in $[a, b]$ such that $f(x_1) = \alpha$. Then $f(x) > \alpha$ for all $x \in [a, b]$. We will show that this leads to a contradiction. Suppose that $t \in [a, b]$. Then $f(t) > \alpha$, so $f(t) > [f(t) + \alpha]/2 > \alpha$. Since $f$ is #-continuous at $t$, there is an open interval $I_t$ about $t$ such that

$$f(x) > \frac{f(t) + \alpha}{2} \qquad (1.17)$$

if $x \in I_t \cap [a, b]$. The collection $H = \{I_t | a \le t \le b\}$ 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 $t_1, t_2, \dots, t_n$ such that the intervals $I_{t_1}, I_{t_2}, \dots, I_{t_n}$ cover $[a, b]$. Define $\alpha_1 = \min_{1 \le j \le n} [f(t_j) + \alpha]/2$. Then, since $[a, b] \subset \bigcup_{i=1}^n (I_{t_i} \cap [a, b])$, (1.17) implies that $f(t) > \alpha_1, a \le t \le b$. But $\alpha_1 > \alpha$, so this contradicts the definition of $\alpha$. Therefore, $f(x_1) = \alpha$ for some $x_1 \in [a, b]$.

## §8. Generalized Intermediate Value Theorem.

**Theorem 13. (Generalized Intermediate Value Theorem)** Suppose that:

(i) $f$ is #-continuous on $[a, b]$, (ii) $f(a) \ne f(b)$ and $f(a) < \mu < f(b)$,

(iii) the set $S = \{x | (a \le x \le b) \land (f(x) \le \mu)\}$ is admissible above. Then $f(c) = \mu$ for some $c \in (a, b)$.

**Proof.** Suppose that $f(a) < \mu < f(b)$. Note that $\sup S$ exists, since the set $S$ is admissible above. Let $c = \sup S$. We will show that $f(c) = \mu$. If $f(c) > \mu$, then $c > a$ and, since $f$ is #-continuous at $c$, there is an $\varepsilon > 0, \varepsilon \approx 0$ such that $f(x) > \mu$ if $c - \varepsilon < x \le c$. Therefore, $c$ is an upper bound for $S$, which contradicts the definition of $c$ as the supremum of $S$. If $f(c) < \mu$, then $c < b$ and there is