Proposition 11

Let $X$ be a $\mu$-measurable subset of $T$. There exists a locally countable family $(L_\alpha)_{\alpha \in A}$ of compact subsets of $X$, pairwise disjoint, such that $X - \bigcup_{\alpha \in A} L_\alpha$ is locally $\mu$-negligible. If, in addition, $X$ is the union of a sequence $(X_n)$ of measurable sets such that $\mu^\bullet(X_n) < +\infty$, then the set $B$ of $\alpha \in A$ such that $\mu^\bullet(L_\alpha) \neq 0$ is countable, and $X - \bigcup_{\beta \in B} L_\beta$ is locally $\mu$-negligible.

Let $f$ be the characteristic function of $X$, and let $(K_\alpha)_{\alpha \in A}$ be a crushing of $T$ such that the restriction of $f$ to each of the $K_\alpha$ is continuous (Prop. 10). The set $L_\alpha = K_\alpha \cap X$ is then compact for every $\alpha \in A$, and $(L_\alpha)_{\alpha \in A}$ is the desired family. Let us pass to the second assertion; the measurable sets $X_n$ may clearly be assumed to be disjoint, and it suffices to establish the assertion for each of them. In other words, changing notation if necessary, we can suppose that $\mu^\bullet(X) < +\infty$. The set $B$ of $\alpha \in A$ such that $\mu^\bullet(L_\alpha) > 0$ is then countable, and it only remains to prove that the set $N = \bigcup_{\alpha \in A - B} L_\alpha$ is locally negligible. But $K$ is a compact set; the family $(L_\alpha)_{\alpha \in A}$ being locally countable, the set $K \cap N$ is the union of a countable subfamily of the family $(K \cap L_\alpha)_{\alpha \in A - B}$, and this set is therefore locally negligible. The same is then true of $N$ (No. 2, Prop. 2) and the proposition is established.