Proposition 12

— Let $\mu$ be a measure on a locally compact space $X$, $A$ a $\mu$-measurable subset of $X$, and $\mathscr{K}$ a set of compact subsets of $A$ satisfying the following conditions: (PLI) Every closed (hence compact) subset of a set of $\mathscr{K}$ belongs to $\mathscr{K}$. (PLII) Every finite union of sets of $\mathscr{K}$ belongs to $\mathscr{K}$. The following four properties are then equivalent: a) For a subset $B$ of $A$ to be locally $\mu$-negligible, it is necessary and sufficient that $|\mu|^*(B \cap K) = 0$ for all $K \in \mathscr{K}$. b) For every compact subset $K_0$ of $A$ and every $\varepsilon > 0$, there exists a set $K \in \mathscr{K}$, contained in $K_0$ and such that $|\mu|(K_0 - K) \leq \varepsilon$. c) For every compact subset $B$ of $A$, there exists a partition of $B$ formed by a $\mu$-negligible set $N$ and a sequence $(H_n)$ of compact sets belonging to $\mathscr{K}$. d) For every compact subset $B$ of $A$, there exists an increasing sequence $(K_n)$ of compact sets belonging to $\mathscr{K}$, contained in $B$ and such that the set $N = B - \bigcup_n K_n$ is $\mu$-negligible.

It is immediate (No. 2, Prop. 5) that d) implies $a); c)$ implies d) on taking $K_n$ to be the union of the $H_p$ for $p \leq n$ and citing (PLII). To prove that b) implies c), one defines recursively a sequence $(H_p)$ of sets of $\mathscr{K}$ such that $H_{n+1} \subset B - \bigcup_{p \leq n} H_p$ and $|\mu|(B - \bigcup_{p \leq n} H_p) \leq 1/n$ (§ 4, No. 6, Th. 4).

It remains to prove that a) implies b). Let us argue by contradiction, and suppose that the supremum $\alpha$ of the numbers $|\mu|(K)$, where $K$ runs over the set of subsets of $K_0$ belonging to $\mathscr{K}$, is $< |\mu|(K_0)$. By (PLII), there exists an increasing sequence $(L_n)$ of compact subsets of $K_0$, belonging to $\mathscr{K}$ and such that $\sup_n |\mu|(L_n) = \alpha$. Set $B = \bigcup_n L_n$; $B$ is integrable and $|\mu|(B) = \alpha$, therefore $|\mu|(K_0 - B) = |\mu|(K_0) - \alpha > 0$. On the other hand, we shall see that for every set $K \in \mathscr{K}$, $|\mu|(K \cap (K_0 - B)) = 0$, which, by virtue of a), will imply a contradiction. Indeed, if there existed a set $K \in \mathscr{K}$ such that $|\mu|(K \cap (K_0 - B)) > 0$, then there would exist a compact subset $H$ of $K \cap (K_0 - B)$ such that $|\mu|(H) > 0$. By (PLI), we would have $H \in \mathscr{K}$, and, for $n$ sufficiently large,

But $L_n \cup H$ belongs to $\mathscr{K}$ by (PLII), and this contradicts the definition of $\alpha$.