The following document is a translation (from French to English) of:

GABRIELE H. GRECO

# Sur la mesurabilité d'une fonction numérique par rapport à une famille d'ensembles

*Rendiconti del Seminario Matematico della Università di Padova, tome 65 (1981), pp. 163-176.*

Translated by: Jonathan M. Keith, School of Mathematics, Monash University,
jonathan.keith@monash.edu

With thanks to: Prof. Andrea D'Agnolo, Editor-in-Chief of the above journal,
for permission to publish this translation.
# On the measurability of a numerical function with respect to a family of sets

Gabriele H. Greco *

## Introduction.

In this introduction and in the sequel, we will denote by the symbols $X, \mathcal{P}(X), [0, +\infty]^X, \overline{\mathbb{R}}^X, \mathbb{N}$ respectively, a non-empty set, the class of subsets of $X$, the class of functions defined on $X$ with values in $[0, +\infty]$, the class of functions defined on $X$ with values in the extended real line $\overline{\mathbb{R}}$, and the set of integers $n > 0$. Following the notations established by P. A. Meyer in [5], we will call a paving on $X$ any family of sets $\mathcal{E} \subset \mathcal{P}(X)$ that contains the empty set; we will call a paved set the pair $(X, \mathcal{E})$ comprised of a set $X$ and a paving $\mathcal{E}$ on $X$; the phrase “$\mathcal{E}$ is stable for $(\cup f, \cap d)$” (and likewise analogous sentences) signifies that $\mathcal{E}$ is stable for finite unions and countable intersections. We will also use the letters of the Greek alphabet $\alpha, \beta, \gamma, \delta$ to designate monotonic set functions, that is, functions defined on $\mathcal{P}(X)$ with values in $[0, +\infty]$ such that $\alpha(\emptyset) = 0$ and $\alpha(A) \le \alpha(B)$ if $A \subset B$. By $\int_X f \, d\alpha$ we will designate the real number, finite or not, called the integral of $f$ over $X$ with respect to the monotone set function $\alpha$, defined by $\int_X f \, d\alpha = \int_0^{+\infty} \alpha\{f > t\} \, dt$ if $f \in [0, +\infty]^X$, and by $\int_X f \, d\alpha = \int_X f^+ \, d\alpha - \int_X f^- \, d\alpha$ if $f \in \overline{\mathbb{R}}^X$ and if $\int_X f^+ \, d\alpha < \infty$ or $\int_X f^- \, d\alpha < \infty$; if $A \in \mathcal{P}(X)$, we set $\int_A f \, d\alpha = \int_X f \varphi_A \, d\alpha$, where $\varphi_A$ is the characteristic function of the set $A$. These integrals have been studied independently in [1] and in [3]; for their properties we will refer to [3]. If $\mathcal{F}$ is a filter of $\mathcal{P}(X)$ we will denote by $\alpha_{\mathcal{F}}$ and $\beta_{\mathcal{F}}: \mathcal{P}(X) \to \{0, 1\}$ the set functions defined by “$\alpha_{\mathcal{F}}(A) = 1$ if and only if $A \in \mathcal{F}$” and “$\beta_{\mathcal{F}}(A) = 1$ if and only if $A \cap F \ne \emptyset$ for any $F \in \mathcal{F}$”; for any filter $\mathcal{F}$ of $\mathcal{P}(X)$ and for any $f \in [0, +\infty]^X$ it holds that $\int_X f \, d\alpha_{\mathcal{F}} = \lim \inf_{\mathcal{F}} f^1$ and $\int_X f \, d\beta_{\mathcal{F}} = \lim \sup_{\mathcal{F}} f$ (see example 3° in [3]). We set $0 \cdot \infty = \infty \cdot 0 = 0$ and $\inf \emptyset = +\infty$. If $(X, \mathcal{E})$ is a paved set, we will denote by $\mathcal{S}^+(X, \mathcal{E})$ the class of functions $\sum_{i=1}^{\infty} a_i \varphi_{H_i}$, where $a_i \in [0, +\infty]$, $\{H_i\}_{i \in \mathbb{N}} \subset \mathcal{E}$ and $H_i \supset H_{i+1}$ for all $i \in \mathbb{N}$.

In this article we propose and study a definition of measurability of a numerical function with respect to a family of sets $\mathcal{E} \subset \mathcal{P}(X)$ that meets this requirement: “determine numerical functions $f$ such that the integral $\int_X f \, d\alpha$

<footnote>* Author's address: Department of Mathematics, University of Trento, 38050 Povo (Trento).</footnote>

<footnote n="1">¹Translator's note: The original uses 'minlim' for 'liminf' and 'maxlim' for 'limsup' here and throughout.</footnote>
depends only on those values of $\alpha$ on $\mathcal{E}$, regardless of the monotone set function $\alpha: \mathcal{P}(X) \to [0, +\infty]$; that is, we say that a function $f \in [0, +\infty]^X$ is $\mathcal{E}$-measurable if one always has $\int_X fd\alpha = \int_X fd\beta$, whenever the monotone set functions $\alpha, \beta$ are such that $\alpha(H) = \beta(H)$ for all $H \in \mathcal{E}$. The function $f$ is then $\mathcal{E}$-measurable if and only if for each pair $a, b \in (0, +\infty)$ with $a > b$, there exists a set $H \in \mathcal{E}$ such that $\{f \ge a\} \subset H \subset \{f > b\}$ (see Th. 1). This definition permits each paved set $(X, \mathcal{E})$ to be associated with the class $M(X, \mathcal{E})$ of $\mathcal{E}$-measurable functions, which can be: the well known class of measurable functions with respect to a $\sigma$-algebra (if $\mathcal{E}$ is a $\sigma$-algebra), the class of functions without oscillatory discontinuities (if $X = \mathbb{R}$ and $\mathcal{E}$ is the set algebra generated by the intervals of $\mathbb{R}$; see Remark 2), the class of continuous functions (if $X$ is a compact and totally disconnected Hausdorff topological space and if $\mathcal{E}$ is the algebra of clopen sets), the class of functions which are continuous almost everywhere (if $X = [0, 1]$ and $\mathcal{E}$ is the algebra of Peano-Jordan measurable sets), the class of $\alpha$-measurable sets in the sense of Carathéodory, with respect to a set function $\alpha$ (if $\mathcal{E}$ is the algebra of $\alpha$-measurable sets in the sense of Carathéodory; see Remark 5); in the case where $\mathcal{E}$ is an algebra, it follows that a function $f \in \overline{\mathbb{R}}^X$ is $\mathcal{E}$-measurable if and only if $f$ extends by continuity to the Stone space associated with the algebra $\mathcal{E}$.

In this article, we also notice analogies between pseudo-compact topological spaces and semi-compact paved sets (see s. 2), and between normal topological spaces and sets on which there are two good pavings (see s. 3).

# 1. $\mathcal{E}$-measurable functions.

Let $(X, \mathcal{E})$ be a paved set. We say that a function $f \in [0, +\infty]^X$ is $\mathcal{E}$-measurable if we always have $\int_X fd\alpha = \int_X fd\beta$, in the case where $\alpha$ and $\beta$ are monotonic set functions such that $\alpha(H) = \beta(H)$ for all $H \in \mathcal{E}$. We also say that $f \in \overline{\mathbb{R}}^X$ is $\mathcal{E}$-measurable if $f^+$ and $f^-$ are $\mathcal{E}$-measurable.

We will denote by $M(X, \mathcal{E})$ the class of functions $f \in \mathbb{R}^X$ which are $\mathcal{E}$-measurable and we set $M^+(X, \mathcal{E}) = M(X, \mathcal{E}) \cap [0, +\infty]^X$.

The simplest $\mathcal{E}$-measurable functions are the functions that belong to $S^+(X, \mathcal{E})$; indeed if $\{a_i\}_{i \in \mathbb{N}} \subset [0, +\infty]$; $\{H_i\}_{i \in \mathbb{N}} \subset \mathcal{E}$ and $H_{i+1} \subset H_i$ for all $i \in \mathbb{N}$, we have $\int_X (\sum_{i=1}^\infty a_i \varphi_{H_i}) d\alpha = \sum_{i=1}^\infty a_i \alpha(H_i)$ for any monotonic set function $\alpha$.

LEMMA 1. Let $(X, \mathcal{E})$ be a paved set and let $f \in [0, +\infty]^X$. The following conditions are equivalent:

i) $f$ is $\mathcal{E}$-measurable,

ii) for each pair $\alpha, \beta: \mathcal{P}(X) \to \{0, 1\}$ of monotonic set functions such that $\alpha(H) = \beta(H)$ for all $H \in \mathcal{E}$, we have

$$ \int_X fd\alpha = \int_X fd\beta. $$
PROOF. i) = ⇒ ii): The condition ii) is a particular case of condition i). ii) = ⇒ i): let us first set, for any monotonic set function α : P ( X ) → [0 , + ∞ ] and for any s ∈ (0 , ∞ ) and H ∈ P ( X ), α$_{s}$ ( H ) = 1 or 0 if we have respectively α ( H ) > s or α ( H ) ≤ s . If α, β : P ( X ) → [0 , + ∞ ] are two montonic set functions such that α ( H ) = β ( H ) for all H ∈ E , we also have α$_{s}$ ( H ) = β$_{s}$ ( H ) for all s ∈ (0 , + ∞ ). According to ii), we therefore have ∫ $_{X}$fdα$_{s}$ = ∫ $_{X}$fdβ$_{s}$ for all s ∈ (0 , + ∞ ). We then deduce that

$$
\int_X f d\alpha = \lim_{n} \sum_{i=1}^{\infty} \frac{1}{2^n} \int_X f d\alpha_{i/2^n} = \lim_{n} \sum_{i=1}^{\infty} \frac{1}{2^n} \int_X f d\beta_{i/2^n} = \int_X f d\beta.
$$

LEMMA 2. Let ( X, E ) be a paved set. If the function f ∈ [0 , + ∞ ] X is E -
measurable, then the functions αf, f ∧ a, f ∨ a − a are also E -measurable for
all a ∈ [0 , + ∞ ).

PROOF. For any monotonic set function α defined on P ( X ) with values in { 0 , 1 } it follows that:

$$
\int_X (f \wedge a) d\alpha = \left( \int_X f d\alpha \right) \wedge a \text{ and } \int_X (f \vee a - a) d\alpha = \left( \int_X f d\alpha \right) \vee a - a
$$

for any $a \in [0, +\infty)$ and $f \in [0, +\infty]^X$. We therefore deduce Lemma 2 from
Lemma 1. ■

THEOREM 1. Let (X, E) be a paved set and let $f \in [0, +\infty]^X$. The following conditions are equivalent:

i) $f$ is $\mathcal{E}$-measurable,

ii) for each pair $a, b \in (0, +\infty)$ such that $a > b$, there exists a set $H \in \mathcal{E}$ such that $\{f \ge a\} \subset H \subset \{f > b\}$,

iii) there exists a sequence {$g_n$}$_{n \in N} \subset S^+(X, \mathcal{E})$ such that $g_n \le f$ for all $n \in N$
and the sequence {$g_n \wedge a$}$_{n \in N}$ converges uniformly to $f \wedge a$ for all $a \in$
$(0, +\infty)$.

PROOF. i) = ⇒ ii): we reason by contradiction. If $f$ does not satisfy condition ii), there exist two real numbers $a, b \in (0, +\infty)$ with $a > b$ such that

$$
(1) \quad \mathfrak{a}H \in \mathcal{E} \text{ such that } \{f > a\} \subset H \subset \{f > b\}^2.
$$

Let us then define the two set functions $\tau_1, \tau_2 : \mathcal{P}(X) \to [0, 1]$ by

$$
\tau_1(A) = \begin{cases} 0 & \text{if } A = X \\ \sup\{g(x) : x \notin A\} & \text{if } A \neq X \end{cases}
$$

and

$$
\tau_2(A) = \begin{cases} 1 & \text{if } A = \emptyset \\ \inf\{g(x) : x \in A\} & \text{if } A \neq \emptyset, \end{cases}
$$

<footnote n="2">²Translator’s note: The original has {$f \ge a$} $\subset H$ $\subset \{f > b\}$, but the proof seems to require this equivalent condition.</footnote>
where $g = (f \wedge a - f \wedge b)/(a-b)$. According to this definition of $\tau_1, \tau_2$, in virtue of (1) it follows that<fnref n="3" />

$$ (2) \quad \tau_1(\emptyset) = \tau_2(\emptyset) = 1 \text{ and the functions } \tau_1, \tau_2 \text{ are decreasing;} $$

$$ (3) \quad \tau_1(H) \le t \text{ and } \tau_2(H) = 0, \text{ if } H \in \mathcal{E}, H \supset \{g > t\} \text{ and } t \in (0, 1); $$

$$ (4) \quad \tau_1(H) = 1 \text{ and } \tau_2(H) \ge t, \text{ if } H \in \mathcal{E}, H \subset \{g > t\} \text{ and } t \in (0, 1). $$

Now the two monotonic set functions $\alpha, \beta: \mathcal{P}(X) \to [0, +\infty]$, thus defined

$$ \alpha(A) = \sup\{2 - \tau_1(H) - \tau_2(H) : H \in \mathcal{E} \text{ and } H \subset A\}, $$

$$ \beta(A) = \inf\{2 - \tau_1(H) - \tau_2(H) : H \in \mathcal{E} \text{ and } H \supset A\}, $$

are therefore equal on $\mathcal{E}$ and $\beta \ge \alpha$, according to (2). Moreover, we have $\beta\{g > t\} > 1 > \alpha\{g > t\}$ for all $t \in (0, 1)$, according to (3) and (4); this proves that $\int_X g d\beta > \int_X g d\alpha$, that is the function $g = (f \wedge a - f \wedge b)/(a-b)$ is not $\mathcal{E}$-measurable, according to the definition of $\mathcal{E}$-measurability. But this contradicts the hypothesis that $f$ is $\mathcal{E}$-measurable; indeed from the $\mathcal{E}$-measurability of $f$ we must deduce that $g$ must be $\mathcal{E}$-measurable, according to Lemma 2. We thus complete the proof of i) $\implies$ ii).

ii) $\implies$ iii): if $f \in [0, +\infty]^X$ satisfies condition ii), for any $i, n \in \mathbb{N}$ there exists a set $H_{i,n} \in \mathcal{E}$ such that

$$ \{f \ge \frac{i+1}{2^n}\} \subset H_{i,n} \subset \{f > \frac{i}{2^n}\} \quad \text{and} \quad H_{i+1,n} \subset H_{i,n} $$

We therefore have

$$ f \ge \frac{1}{2^n} \sum_{i=1}^{\infty} \varphi_{H_{i,n}} \ge f \wedge \frac{1}{2^{n-1}} - \frac{1}{2^{n-1}}. $$

The sequence $\{g_n\}_{n \in \mathbb{N}}$, where $g_n = 1/2^n \sum_{i=1}^{\infty} \varphi_{H_{i,n}}$, satisfies condition iii).

iii) $\implies$ i): the $\mathcal{E}$-measurability of all $g \in S^+(X, \mathcal{E})$ and the properties of the integral $\int_X -d\alpha$ (see [3], Prop. 1) entail the $\mathcal{E}$-measurability of $f$. ■

REMARK 1. Let $(X, \mathcal{E})$ be a paved set, let $\delta: \mathcal{E} \to [0, +\infty]$ be a set function such that $\delta(\emptyset) = 0$ and $\delta(A) \le \delta(B)$ for all $A, B \in \mathcal{E}$ with $A \subset B$. The preceding theorem makes it possible to define in a satisfying manner the integral $\int_X -d\delta$; it is defined by setting $\int_X f d\delta = \int_X f d\alpha$, where $\alpha: \mathcal{P}(X) \to [0, +\infty]$ is a monotonic set function such that $\alpha(H) = \delta(H)$ for all $H \in \mathcal{E}$, when $f \in \mathcal{M}(X, \mathcal{E})$ and $\int_X f^+ d\alpha < +\infty$ or $\int_X f^- d\alpha < +\infty$. According to this definition of the integral we can analyse the properties of $\delta$ by means of the properties of $\int_X -d\alpha$. For example let $(X, \mathcal{E})$ be a paved set that is stable for $(\cap f, \cup f)$; we will see in Proposition 2 that the family of $\mathcal{E}$-measurable functions $\mathcal{M}(X, \mathcal{E})$ is a convex cone of functions stable for $(\wedge f, \vee f)$. The following conditions are equivalent

$$ i) \delta(A \cap B) + \delta(A \cup B) = \delta(A) + \delta(B), \forall A, B \in \mathcal{E}, $$

<footnote n="3">³Translator's note: The original has $\tau_1(H) < t$ in (3) and $\tau_2(H) > t$ in (4).</footnote>
$$ \text{ii) } \int_X (f+g)d\delta = \int_X fd\delta + \int_X gd\delta, \forall f, g \in \mathcal{M}^+(X, \mathcal{E}), $$

$$ \text{iii) } \int_X (f \wedge g)d\delta + \int_X (f \vee g)d\delta = \int_X fd\delta + \int_X gd\delta, \forall f, g \in \mathcal{M}^+(X, \mathcal{E}). $$

If in i), ii) and iii) we change “=” to “≤” or “≥”, we again obtain the equivalence of i), ii) and iii) (see [4]).

## 2. Properties of $\mathcal{E}$-measurable functions.

From the definition of $\mathcal{E}$-measurability and Property ii) of Theorem 1, we obtain the following propositions on the properties of $\mathcal{E}$-measurable functions.

**PROPOSITION 1.** Let $(X, \mathcal{E})$ be a paved set and let $\psi: [0, +\infty] \to [0, +\infty]$ be a continuous and increasing function. We have

i) if $\psi(0) = 0$, $f \in \mathcal{M}^+(X, \mathcal{E})$, then $\psi \circ f \in \mathcal{M}^+(X, \mathcal{E})$ (note we can dispense with the equality $\psi(0) = 0$ if $X \in \mathcal{E}$);

ii) if the sequence $\{f_n\}_{n \in \mathbb{N}} \subset \mathcal{M}^+(X, \mathcal{E})$ converges uniformly to $f$, then $f \in \mathcal{M}^+(X, \mathcal{E})$.

**PROPOSITION 2.** Let $(X, \mathcal{E})$ be a paved set. Then

i) for $\mathcal{M}^+(X, \mathcal{E})$ to be stable for ($\wedge f$) (respectively, for ($\vee f$)) it is necessary and sufficient that $\mathcal{E}$ be stable for ($\cap f$) (respectively, for ($\cup f$));

ii) to have $f+g \in \mathcal{M}^+(X, \mathcal{E})$ for any $f, g \in \mathcal{M}^+(X, \mathcal{E})$, it is necessary and sufficient that $\mathcal{E}$ be stable for ($\cap f, \cup f$);

iii) for $\mathcal{M}^+(X, \mathcal{E})$ to be stable for ($\wedge d$) (respectively, for ($\vee d$)) it is necessary and sufficient that $\mathcal{E}$ be stable for ($\cap d$) (respectively, for ($\cup d$)); in that case, for $f \in [0, +\infty]$ to be $\mathcal{E}$-measurable it is necessary and sufficient that the set $\{f \ge t\} \in \mathcal{E}$ (respectively, $\{f > t\} \in \mathcal{E}$) for all $t \in (0, +\infty)$.

**PROOF.** We are only going to show property ii). If $f+g$ is $\mathcal{E}$ measurable for all pairs of functions $f, g \in \mathcal{M}^+(X, \mathcal{E})$, we have that $\varphi_A + \varphi_B$ is $\mathcal{E}$ measurable for all $A, B \in \mathcal{E}$; it follows therefore that $A \cap B$ and $A \cup B \in \mathcal{E}$, according to ii) of Theorem 1; that is $\mathcal{E}$ is stable for ($\cap f, \cup f$). On the other hand, suppose that $\mathcal{E}$ is stable for ($\cap f, \cup f$), let us verify that $\mathcal{M}^+(X, \mathcal{E})$ is stable for summation. If $g_1 = \sum_{i=1}^n a_i \varphi_{A_i}$ and $g_2 = \sum_{i=1}^n b_i \varphi_{B_i}$ belong to $\mathcal{S}^+(X, \mathcal{E})$, the function $g_1 + g_2$ belongs to $\mathcal{S}^+(X, \mathcal{E})$, because we have for all $t \in (0, +\infty)$ the equality

$$ \{g_1 + g_2 > t\} = \cup_{(a,b) \in \mathbb{Z}} (\{g_1 \ge a\} \cap \{g_2 \ge b\}), $$

where $Z$ is the finite set

$$ \{(g_1(x), g_2(x)) \in \mathbb{R}^2 : x \in X \text{ and } g_1(x) + g_2(x) > t\}. $$

If $f, g \in \mathcal{M}^+(X, \mathcal{E})$ are bounded, there exist two sequences of bounded $\mathcal{E}$-measurable functions $\{g_n\}_{n \in \mathbb{N}}$, $\{f_n\}_{n \in \mathbb{N}}$, which converge uniformly, respectively,
to $g$ and $f$ (see Th. 1, property iii)); since $\{g_n + f_n\}_{n \in \mathbb{N}} \subset S^+(X, \mathcal{E})$ converges uniformly to $g+f$, we therefore have that $f+g \in M^+(X, \mathcal{E})$, by property ii) of Proposition 1. Finally, taking into account the equality $\int_X (f+g)d\alpha = \lim_n \int_X (f \wedge n + g \wedge n)d\alpha$, true for any $f, g \in [0, +\infty]^X$ and for any monotonic set function $\alpha : P(X) \to [0, +\infty]$, it follows then that $f+g \in M^+(X, \mathcal{E})$, also if $f, g \in M^+(X, \mathcal{E})$ are not bounded. ■

In particular cases, the class $M^+(X, \mathcal{E})$ has new properties. For example, if $X$ is a topological space in which every countable open cover of $X$ contains a finite open cover of $X$, any function $f \in [0, +\infty)^X$ that is measurable with respect to the paving of closed sets of $X$ (that is, $f \in [0, +\infty)^X$ is upper semi-continuous) is bounded; or if $X$ is a pseudocompact topological space (that is, every continuous function $g: X \to \mathbb{R}$ is bounded), any function $f \in [0, \infty)^X$ that is measurable with respect to the paving comprised of the sets of the form $\{g=0\}$, where $g: X \to \mathbb{R}$ is continuous, is also bounded. This depends on the semi-compactness of the pavings in question (see [8]); recall that the paving $\mathcal{E}$ on $X$ is said to be semi-compact (see [5]), if $\{H_n\}_{n \in \mathbb{N}} \subset \mathcal{E}$ and $\cap_{n=1}^\infty H_n = \emptyset$ entails that there exists $n_0 \in \mathbb{N}$ such that $\cap_{n=1}^{n_0} H_n = \emptyset$.

**THEOREM 2.** Let $X$ be a set endowed with a paving $\mathcal{E}$ that is stable for $(\cap f)$. The following conditions are equivalent:

i) $\mathcal{E}$ is semi-compact,

ii) any function $f \in M^+(X, \mathcal{E}) \cap \mathbb{R}^X$ is bounded,

iii) any function $f \in M^+(X, \mathcal{E})$ has a maximum,

iv) any sequence $\{f_n\}_{n \in \mathbb{N}}$, decreasing to zero, converges uniformly to zero, if all $f_n$ are $\mathcal{E}$-measurable,

v) for any sequence $\{f_n\}_{n \in \mathbb{N}} \subset M^+(X, \mathcal{E})$, decreasing to zero, and for any monotonic $\alpha : P(X) \to [0, +\infty]$ such that $\int_X f_1 d\alpha < +\infty$, we have $\lim_n \int_X f_n d\alpha = 0$.

**PROOF.** It suffices to suitably adapt the proof of an analogous theorem of I. Glicksberg [2]. i) $\implies$ iii) by property ii) of Theorem 1, there exist sets $H_n \in \mathcal{E}$ such that $\{f \ge a_{n+1}\} \subset H_n \subset \{f > a_n\}$, for all $n \in \mathbb{N}$, if $f \in M^+(X, \mathcal{E})$, $\{a_n\}_{n \in \mathbb{N}} \subset (0, +\infty)$, $a_{n+1} > a_n$ for all $n \in \mathbb{N}$ and $\lim_n a_n = \sup\{f(x) : x \in X\}$ (we assume that $f \ne 0$); since $\cap_{n=1}^m H_n \supset \{f \ge a_{m+1}\} \ne \emptyset$ for all $m \in \mathbb{N}$, we deduce from the semi-compactness of $\mathcal{E}$ that $\cap_{n=1}^\infty H_n \ne \emptyset$; if $x_0 \in \cap_{n=1}^\infty H_n$, we therefore have $f(x_0) \ge a_n$ for all $n \in \mathbb{N}$, that is $f(x_0) = \sup(f)$.

iii) $\implies$ ii): is an immediate consequence of iii).

ii) $\implies$ i): let $\{H_n\}_{n \in \mathbb{N}} \subset \mathcal{E}$ and $\cap_{n=1}^\infty H_n = \emptyset$. The sets $G_n = \cap_{i=1}^n H_n$ belong to $\mathcal{E}$ and the function $\sum_{n=1}^\infty n\varphi_{G_n} \in M^+(X, \mathcal{E}) \cap \mathbb{R}^X$; by hypothesis ii) the function $\sum_{n=1}^\infty n\varphi_{G_n}$ is bounded; then there exists a number $n_0 \in \mathbb{N}$ such that $G_{n_0} = \emptyset$, that is $\cap_{n=1}^{n_0} H_n = \emptyset$.

i) $\implies$ iv): let $\{f_n\}_{n \in \mathbb{N}} \subset M^+(X, \mathcal{E})$ be a sequence decreasing to zero. Let us show that $\{f_n\}_n$ converges to zero uniformly, that is for any real number $a > 0$ there exists $n_0 \in \mathbb{N}$ such that $f_{n_0} \le a$. We choose a sequence $\{a_n\}_n \subset$
(0, +∞) such that $a_n < a$ and $\lim_n a_n = a$. By the measurability of all $f_n$ and property ii) of Theorem 1, there exists for all $n \in \mathbb{N}$ a set $H_n \in \mathcal{E}$ such that $\{f_n \ge a\} \subset H_n \subset \{f_n > a_n\}$. We therefore have $\cap_{n=1}^{\infty} H_n = \emptyset$, because $\lim_n f_n = 0$; we then deduce that there exists $n_0 \in \mathbb{N}$ such that $\cap_{n=1}^{n_0} H_n = \emptyset$, by hypothesis i). Therefore $\{f_{n_0} \ge a\} \subset \cap_{n=1}^{n_0} H_n = \emptyset$, that is the sequence $\{f_n\}_{n \in \mathbb{N}}$ converges uniformly to zero.

iv) $\implies$ v): this is a consequence of the properties of the integral $\int_X -d\alpha$ (see [3], Prop. 1, property (4)).

v) $\implies$ i): it suffices to demonstrate that there exists $n_0 \in \mathbb{N}$ such that $H_{n_0} = \emptyset$, if $\{H_n\}_{n \in \mathbb{N}} \subset \mathcal{E}$ with $\cap_{n=1}^{\infty} H_n = \emptyset$ and $H_{n+1} \subset H_n$ for all $n \in \mathbb{N}$. We reason by contradiction; if $H_n \ne \emptyset$ for all $n \in \mathbb{N}$, the monotone set function $\alpha_F: \mathcal{P}(X) \to \{0, 1\}$ that is equal to 1 only for those sets of the filter $\mathcal{F}$ generated by the $H_n$, is such that $\alpha_F = 1$ for all $n \in \mathbb{N}$; which is contrary to hypothesis v), since the sequence $\{\varphi_{H_n}\}_{n \in \mathbb{N}}$ decreases to zero. ■

When the paving $\mathcal{E}$ on $X$ is an algebra of sets we have the following characterisation of $\mathcal{E}$-measurability.

**THEOREM 3.** Let $(X, \mathcal{A})$ be a paved set, $\mathcal{A}$ an algebra of sets, $f \in \overline{\mathbb{R}}^x$. The following conditions are equivalent:

i) $f$ is $\mathcal{A}$-measurable,

ii) for any ultrafilter $\mathcal{F}$ of the algebra $\mathcal{A}$ there exists the limit $\lim_{\mathcal{F}} f \in \overline{\mathbb{R}}$,

iii) for any real number $a > 0$, there exists a finite partition $\{H_i\}_{i=1}^n \subset \mathcal{A}$ of $X$ such that for all $i$ we have

$$f(x) \le a + f(y) \quad \text{if } x, y \in H_i$$

or

$$f(x) \ge \frac{1}{a} \quad \text{if } x \in H_i$$

or

$$f(x) \le -\frac{1}{a} \quad \text{if } x \in H_i.$$

**PROOF.** i) $\implies$ ii): we can limit ourselves to verifying that for all $f \in [0, +\infty]^X \cap \mathcal{M}(X, \mathcal{E})$ there exists the limit $\lim_{\mathcal{F}} f \in \overline{\mathbb{R}}$. Let $\mathcal{F}' \subset \mathcal{P}(X)$ be the filter of $\mathcal{P}(X)$ generated by the ultrafilter $\mathcal{F}$ of $\mathcal{A}$, and let $\alpha_{\mathcal{F}'}$ and $\beta_{\mathcal{F}'}$ be the monotonic set functions defined in the introduction. For these set functions we have $\int_X f d\alpha_{\mathcal{F}'} = \liminf_{\mathcal{F}} f$ and $\int_X f d\beta_{\mathcal{F}'} = \limsup_{\mathcal{F}} f$; since $\alpha_{\mathcal{F}'}(H) = \beta_{\mathcal{F}'}(H)$ for any $H \in \mathcal{A}$, it follows that $\limsup_{\mathcal{F}} f = \liminf_{\mathcal{F}} f$, by the definition of $\mathcal{A}$-measurability; this proves that $\lim_{\mathcal{F}} f$ exists and belongs to $\overline{\mathbb{R}}$.

ii) $\implies$ iii): according to ii), for any ultrafilter $\mathcal{F}$ of the algebra $\mathcal{A}$ there exists a set $H_{\mathcal{F}} \in \mathcal{F}^4$ such that $f(x) - f(y) \le a$ or $f(x) \ge 1/a$ or else $f(x) \le -1/a$ for $x, y \in H_{\mathcal{F}}$. The family $\{X - H_{\mathcal{F}} : \mathcal{F}$ an ultrafilter of $\mathcal{A}\}$ cannot be contained

<footnote n="4">⁴Translator's note: The original has $H_{\mathcal{F}} \in \mathcal{A}$.</footnote>
in any ultrafilter of $\mathcal{A}$; there exists therefore a finite number $\mathcal{F}_1, \mathcal{F}_2, \dots, \mathcal{F}_n$ of
ultrafilters of $\mathcal{A}$ such that

$$ \cap_{i=1}^{n} (X - H_{\mathcal{F}_i}) = \emptyset; $$

the proof of the implication ii) $\Rightarrow$ iii) is thus complete.

iii) $\Rightarrow$ i): it suffices to verify this implication when $f \in [0, +\infty]^X$ is bounded. In that case the construction of a sequence $\{g_n\}_n \subset \mathcal{M}^+(X, \mathcal{A})$ such that $\{g_n\}_n$ converges uniformly to the function $f$ is immediate; thus we can deduce that $f$ is $\mathcal{A}$-measurable, by the properties of $\mathcal{A}$-measurable functions. ■

**REMARK 2.** Let $X = \mathbb{R}$ and let $\mathcal{A}_{\mathbb{R}}$ be the algebra generated by the bounded and unbounded intervals of $\mathbb{R}$. For all $x \in \mathbb{R}$, let $\mathcal{F}_x, \mathcal{F}_{x^+}, \mathcal{F}_{x^-}, \mathcal{F}_{+\infty}, \mathcal{F}_{-\infty}$ be the ultrafilters of $\mathcal{A}_{\mathbb{R}}$ generated, respectively, by the set $\{x\}$, by the intervals $(x, x+a)$, where $a \in (0, +\infty)$, by the intervals $(x-a, x)$, where $a \in (0, +\infty)$, by the intervals $(a, +\infty)$, where $a \in \mathbb{R}$, and by the intervals $(-\infty, a)$, where $a \in \mathbb{R}$; all the ultrafilters of $\mathcal{A}_{\mathbb{R}}$ are of these types. According to Theorem 3, it follows that a function $f \in \overline{\mathbb{R}}^X$ is $\mathcal{A}_{\mathbb{R}}$-measurable if and only if the following limits exist: $\lim_{y \to x^+} f(y)$, $\lim_{y \to x^-} f(y)$, $\lim_{y \to +\infty} f(y)$, and $\lim_{y \to -\infty} f(y)$; that is $f$ is devoid of oscillatory discontinuities. This example can easily be adapted when $\mathcal{A}$ is the algebra generated by the intervals that are semi-open on the right (or on the left), and when $X = \mathbb{R}^n$ and $\mathcal{A}$ is the algebra generated by the $n$-dimensional intervals.

**REMARK 3.** Theorem 3 can be presented in a more suggestive manner. If we denote by $[\mathcal{A}]$ the Stone space (the space of ultrafilters of $\mathcal{A}$)(see [6]), associated with the algebra $\mathcal{A}$, for a function $f \in \overline{\mathbb{R}}^X$ to be $\mathcal{A}$-measurable it is necessary and sufficient that there exists a continuous function $\bar{f} : [\mathcal{A}] \to \overline{\mathbb{R}}$ such that $\bar{f}(\mathcal{F}_x) = f(x)$ for any ultrafilter $\mathcal{F}_x = \{H \in \mathcal{A} : x \in H\}$, where $x \in X$. The class of bounded and $\mathcal{A}$-measurable functions is therefore an algebra of functions, isomorphic to the algebra of continuous functions from $[\mathcal{A}]$ to $\mathbb{R}$.

**REMARK 4.** In the particular case where the algebra $\mathcal{A}$ is a $\sigma$-algebra, we have that $f+g \in \mathcal{M}(X, \mathcal{A})$ if $f+g$ is defined over $X$ and $f, g \in \mathcal{M}(X, \mathcal{E})$. This property is not true if $\mathcal{A}$ is only an algebra. Indeed if we consider the algebra $\mathcal{A} \subset \mathcal{P}(\mathbb{N})$ of finite and cofinite subsets of $\mathbb{N}$, for a function $f : \mathbb{N} \to \overline{\mathbb{R}}$ to be $\mathcal{A}$-measurable it is necessary and sufficient that there exists the limit $\lim_{n} f(n)$; in that case we can therefore easily choose two functions $f, g : \mathbb{N} \to \overline{\mathbb{R}}$ such that $f+g$ is defined, but the function $f+g$ is not $\mathcal{A}$-measurable.

**REMARK 5.** Let $\mu : \mathcal{P}(X) \to [0, +\infty]$ be a monotonic or non-monotonic set function with $\mu(\emptyset) = 0$. We know that the family of sets that are $\mu$-measurable in the sense of Carathéodory is a set algebra $\mathcal{A}_\mu \subset \mathcal{P}(X)$. Let $\delta : \mathcal{A}_\mu \to [0, +\infty]$ be the monotone set function defined by $\delta(H) = \mu(H)$ for all $H \in \mathcal{A}_\mu$; a function $f$ is said to be $\mu$-measurable if $f$ is $\mathcal{A}_\mu$-measurable. For these $\mu$-measurable functions we have linearity, that is $\int (f+g)d\delta = \int_X f d\delta + \int_X g d\delta$ (see Remark 1 for the definition of the integral $\int_X -d\delta$) for all $f, g \in \mathcal{M}^+(X, \mathcal{A}_\mu)$. Furthermore we have that the sets $\{t \in \mathbb{R} : \{f > t\} \notin \mathcal{A}_\mu\}$ and $\{t \in \mathbb{R} : \{f \ge t\} \notin \mathcal{A}_\mu\}$ are countable, if $f \in \mathcal{M}(X, \mathcal{A}_\mu)$ and $\int_X |f|d\delta < +\infty$. The class of functions $f \in \mathbb{R}^X$ that are $\mu$-measurable, such that $\int_X |f|d\delta < +\infty$, is a Riesz
space (see [4]).

### 3. Measurable functions with respect to two pavings.

In this paragraph we consider two pavings $\mathcal{K}, \mathcal{U}$ on $X$, both stable for $(\cap f, \cup f)$; we define on $\mathcal{P}(X)$ the relation “$\ll$” in this manner: “$A \ll B$ if and only if there exists a pair $(K, U) \in \mathcal{K} \times \mathcal{U}$ such that $A \subset K \subset U \subset B$”. Let $D$ be the set of dyadic numbers in the interval $[0, 1]$; let $\{H_t\}_{t \in D} \subset \mathcal{P}(X)$ be a family of sets such that $H_0 = X$ and $H_t \gg H_s$ for all $t, s \in D$ with $t < s$; then the function $f: X \to [0, 1]$, defined by $f(x) = \sup\{t \in D : x \in H_t\}$, is either $\mathcal{K}$-measurable or $\mathcal{U}$-measurable. Indeed for any $a, b \in (0, +\infty)$ with $a > b$, let $t$ and $r$ be two numbers $\in D$ such that $a > t > r > b$, it therefore follows that $\{f \ge a\} \subset H_t \ll H_r \subset \{f > b\}$; since there exists a pair $(K, U) \in \mathcal{K} \times \mathcal{U}$ such that $H_t \subset K \subset U \subset H_r$, we finally have that $\{f \ge a\} \subset K \subset U \subset \{f > b\}$, that is $f$ is $\mathcal{K}$-measurable and $\mathcal{U}$-measurable.

We say that the pair of pavings $(\mathcal{K}, \mathcal{U})$ satisfies property $(N)$ if for any pair $(K, U) \in \mathcal{K} \times \mathcal{U}$ with $K \subset U$, there exists a pair $(K', U') \in \mathcal{K} \times \mathcal{U}$ such that $K \subset U' \subset K' \subset U$.

**LEMMA 3.** Let $k, u \in [0, +\infty]^X$, and let $(\mathcal{K}, \mathcal{U})$ be a pair of pavings on $X$ that satisfy property $(N)$. If $k \le u$ and if $\{k \ge t\} \in \mathcal{K}$ and $\{u > t\} \in \mathcal{U}$ for any $t \in (0, +\infty)$, then there exists a function $f \in \mathcal{M}^+(X, \mathcal{K}) \cap \mathcal{M}^+(X, \mathcal{U})$ such that $k \le f \le u$.

**PROOF.** By property i) of Proposition 1 we can suppose that $u \le 1$. For any $t \in D$, where $D$ is the set of dyadic numbers in the interval $[0, 1]$, we set $K_t = \{k \ge t\}$ and $U_t = \{u > t\}$. We construct a family of sets $\{F_t\}_{t \in D} \subset \mathcal{P}(X)$, enjoying the following properties$^5$:

$$ (1) \qquad F_0 = X, \quad F_1 = \emptyset, $$

$$ (2) \qquad F_r \ll F_t, \quad F_r \ll U_t, \quad K_r \ll F_t, \quad \text{if } t, r \in D \text{ and } t < r; $$

in fact if we suppose sets $F_t$ satisfying (1) and (2) have been defined for all $t = i/2^n$, where $0 \le i \le 2^n$, then we can choose a set $F_{t_0}$ for $t_0 = (2i+1)/2^{n+1}$, where $0 \le i \le 2^n - 1$, satisfying the relations: $F_{i/2^n} \gg F_{t_0}$, $F_{t_0} \gg F_{(i+1)/2^n}$, $U_{i/2^n} \gg F_{t_0}$, $F_{t_0} \gg K_{(i+1)/2^n}$, according to the relations (true by hypothesis or by construction):

$$ F_{i/2^n} \gg F_{(i+1)/2^n}, \quad F_{i/2^n} \gg K_{(i+1)/2^n}, \quad U_{i/2^n} \gg F_{(i-1)/2^n}, \quad U_{i/2^n} \gg K_{(i-1)/2^n}. $$

The function $f: X \to [0, +\infty]$, defined by $f(x) = \sup\{t \in D : x \in F_t\}$, is thus $\mathcal{K}$-measurable and $\mathcal{U}$-measurable and $k \le f \le u$. ■

By Lemma 3 and property iii) of Proposition 2 we deduce the following theorems:

**THEOREM 4.** Let $\mathcal{K}$ and $\mathcal{U}$ be two pavings on $X$, stable for $(\cap f, \cup f)$. The following properties are equivalent:

<footnote>$^5$Translator's note: In (2), the original has $G_t$ in place of $U_t$, and $H_r$ in place of $K_r$.</footnote>
i) the pair $(\mathcal{K}, \mathcal{U})$ satisfies property (N),

ii) for any pair $(K, U) \in \mathcal{K} \times \mathcal{U}$ there exists a function $f \in \mathcal{M}^+(X, \mathcal{K}) \cap \mathcal{M}^+(X, \mathcal{U})$ such that $\varphi_K \le f \le \varphi_U$, if $K \subset U$. ■

**THEOREM 5.** Let $\mathcal{K}$ be a paving stable for $(\cap d, \cup f)$ and let $\mathcal{U}$ be a paving stable for $(\cap f, \cup d)$. The following properties are equivalent:

i) the pair $(\mathcal{K}, \mathcal{U})$ satisfies property (N),

ii) for any $(k, u) \in \mathcal{M}^+(X, \mathcal{K}) \times \mathcal{M}^+(X, \mathcal{U})$ there exists a function $f \in \mathcal{M}^+(X, \mathcal{K}) \cap \mathcal{M}^+(X, \mathcal{U})$ such that $k \le f \le u$, if $k \le u$. ■

Let $X$ be a topological space and let $\mathcal{K}$ (respectively, $\mathcal{U}$) be a paving on $X$ comprised of closed sets (respectively, open sets); the pair of pavings $(\mathcal{K}, \mathcal{U})$ satisfies property (N) if and only if $X$ is normal. In that case Theorems 4 and 5 are well known theorems on normal spaces (see [7]).

## BIBLIOGRAPHY

[1] E. DE GIORGI, G. LETTA, *Une notion général de convergence faible pour des fonctions croissantes d'ensemble*, Ann. Sc. Nor. Pisa, **4** (IV) (1977), pp. 61-99.

[2] I. GLICKSBERG, *The representation of functionals by integrals*, Duke Math. J., **19** (1952), pp. 253-261.

[3] G. H. GRECO, *Integrale monotono*, Rend. Sem. Mat. Padova, **57** (1977), pp. 149-166.

[4] G. H. GRECO, R. C. BASSANEZI, *Sull'additività di alcuni funzionali definiti su coni di funzioni* (preprint)<fnref n="6" />.

[5] P. A. MEYER, *Probabilités et potentiel*, Hermann, 1966.

[6] R. SIKORSKI, *Boolean algebras*, Springer-Verlag, 1969.

[7] H. TONG, *Some characterizations of normal and perfectly normal spaces*, Duke Math. J., **19** (1952), pp. 289-292.

[8] V. S. VARADARAYAN, *Measures on topological spaces*, Amer. Math. Soc. Transl., ser. II, **48** (1965), pp. 161-228.

<footnote n="6">⁶Translator's note: The intended reference is clearly R. C. Bassanezi, G. H. Greco, *Sull'additività dell'integrale*, Rend. Sem. Mat. Padova, **72** (1984), pp. 249-275.</footnote>