Proposition 9

a) There exists a crushing $(K_\alpha)_{\alpha \in A}$ of T for $\mu$. b) Let $(K_\alpha)_{\alpha \in A}$ be a crushing of T for $\mu$. If $\mu_\alpha$ is the measure on T defined by $\mu_{K_\alpha}$ (No. 3, Example 2), then the family $(\mu_\alpha)_{\alpha \in A}$ is summable, its sum is equal to $\mu$, and, for every function $f \in \mathscr{F}_+(T)$,

For a mapping $g$ of $T$ into a topological space $G$ (Hausdorff or not) to be $\mu$-measurable, it is necessary and sufficient that $g_{K_\alpha}$ be $\mu_{K_\alpha}$-measurable for every $\alpha \in A$.

A) Existence of a crushing:

The proof is a repetition of that of Prop. 14 of Ch. IV, §5, No. 9, with slight modifications. Let $\mathscr{K}$ be the set of compact subsets $K$ of $T$ such that $\mathrm{Supp}(\mu_K) = K$, and let $\mathscr{H}$ be the set (ordered by inclusion) of subsets $\mathscr{L}$ of $\mathscr{K}$ consisting of pairwise disjoint sets. Let us first show that every element $\mathscr{L}$ of $\mathscr{H}$ is locally countable. Let $x$ be a point of $T$, and $V$ an open neighborhood of $x$ such that $\mu^\bullet(V) < +\infty$; let $\mathscr{L}_V$ be the set of $K \in \mathscr{L}$ that intersect $V$. If $(K_i)_{1 \leq i \leq n}$ is a finite sequence of distinct elements of $\mathscr{L}_V$, we have, by the Cor. of Prop. 4,

because the $K_i$ are pairwise disjoint. Thus,

Now, $\mu^\bullet(K \cap V) = \mu_K^\bullet(K \cap V) > 0$ for every $K \in \mathscr{L}_V$, because $K \cap V$ is nonempty, open in $K$, and the support of $\mu_K$ is all of $K$; $\mathscr{L}_V$ is therefore countable, and $\mathscr{L}$ is indeed locally countable. It is immediate that $\mathscr{H}$ is inductive, and nonempty (one has $\varnothing \in \mathscr{H}$). Thus, let $\mathfrak{H}$ be a maximal element of $\mathscr{H}$. We are going to show that the set $N = T - \bigcup_{K \in \mathfrak{H}} K$ is locally negligible. By Prop. 2, it suffices to verify that $\mu^\bullet(N \cap L) = 0$ for every compact set $L$, or again that $\mu_L^\bullet(N \cap L) = 0$. We shall argue by contradiction. Thus, suppose that $\mu_L^\bullet(N \cap L) > 0$. Since the set of $K \in \mathfrak{H}$ that intersect $L$ is countable, $N \cap L$ is $\mu_L$-measurable; therefore there exists a compact set $J$ contained in $N \cap L$ such that $\mu_L^\bullet(J) > 0$. Let $S$ be the support of the nonzero measure $(\mu_L)_J = \mu_J$; it is contained in $N$, the measure $\mu_S$ is nonzero, and $\mathrm{Supp}(\mu_S) = S$ (Ch. IV, §5, No. 7, Lemma 2). The set $\mathfrak{H} \cup \{S\}$ therefore belongs to $\mathscr{H}$, in contradiction with the maximal character of $\mathfrak{H}$. This proves the existence of a crushing.

(1) We shall see later on (§2, No. 2) that $\mu_\alpha$ is the measure $\varphi_{K_\alpha} \cdot \mu$.

B) Proof of (7): For every $\alpha \in A$, $\mu_\alpha^\bullet(f) = \mu_{K_\alpha}^\bullet(f_{K_\alpha}) = \mu^\bullet(f \varphi_{K_\alpha})$ by formula (3) of No. 3 and Prop. 2 a) of No. 2; these formulas show that the encumbrance $\sum_{\alpha \in A} \mu_\alpha^\bullet$ is $\leq \mu^\bullet$, hence that the family $(\mu_\alpha)_{\alpha \in A}$ is summable (No. 7, Prop. 7). It thus suffices to show that $\mu = \sum_{\alpha \in A} \mu_\alpha$, that is, to establish the formula

for every compact subset K of T. Now, K being fixed, the set $A'$ of $\alpha \in A$ such that $K_\alpha$ intersects K is countable. Let $g \in \mathscr{F}_+(K)$; then $g^0 = g^0 \varphi_N + \sum_{\alpha \in A} g^0 \varphi_{K_\alpha}$, and $g^0 \varphi_{K_\alpha} = 0$ for $\alpha \in A - A'$; by Prop. 4 of No. 5, it follows that $\mu^\bullet(g^0) = \sum_{\alpha \in A} \mu^\bullet(g^0 \varphi_{K_\alpha})$, whence thus (8) is established.

C) Measurability: For a function $g$ defined on T to be $\mu$-measurable, it is necessary and sufficient that it be $\mu_\alpha$-measurable for every $\alpha \in A$ (No. 7, Prop. 8); but this amounts to saying that $g_{K_\alpha}$ is $\mu_{K_\alpha}$-measurable for all $\alpha \in A$ (No. 5, Example). Q.E.D.

As in Prop. 14 of Ch. IV, §5, No. 9, one can require the compact sets $K_\alpha$ to belong to a $\mu$-dense set of compact subsets of T, given in advance. We will only need the following result, which we shall establish directly: