Lemma 3.3.113FB36EAD186A

Proof. If $A=\varnothing $, choose the empty sequence. Otherwise choose any $F_1$. After $F_1,\ldots ,F_ j$, stop if they cover $A$. If they do not, a point outside their closed union belongs to sufficiently small members $F$ with $N(F,\operatorname {diam}F)$ disjoint from that union. Let $e_ j>0$ be the supremum of their diameters, and choose such an $F_{j+1}$ with diameter $>2e_ j/3$. All chosen sets are disjoint. Fix $x\in A\setminus \bigcup _{i\le j}F_ i$ and choose a member $F$ containing $x$ whose diameter is so small that its diameter-neighborhood avoids the first $j$ sets. It must meet a later chosen $F_ k$ in that neighborhood. Otherwise it remains eligible forever, so all subsequent chosen diameters exceed $2\operatorname {diam}F/3$; choosing a point in each yields an infinite uniformly separated subset of compact $X$, impossible. At the first such $k$, $\operatorname {diam}F_ k>2\operatorname {diam}F/3$. Some point of $F_ k$ is within $\operatorname {diam}F$ of $F$, hence $d(x,F_ k)<2\operatorname {diam}F<3\operatorname {diam}F_ k$. This proves the covering assertion. In the measure case the outer measure of the uncovered set is at most $C\sum _{i>j}m(F_ i)$ for every $j$; the tail tends to zero because the sets are disjoint and $m(X)<\infty $. Completeness makes every subset of the resulting null set measurable. $\square$