Lemma 2.3.2 · Pome

Lemma 2.3.2.BAEFE6D3AE3C Let $m^*$ be an outer measure on a metric space $X$ such that

\[ m^*(A\cup B)=m^*(A)+m^*(B)\quad \text{whenever }\operatorname {dist}(A,B)>0. \]

Then every Borel subset of $X$ is $m^*$-measurable, and the restriction of $m^*$ to the Borel sets is a measure.

Proof. We first show that the $m^*$-measurable sets form a $\sigma $-algebra on which $m^*$ is countably additive. The definition is symmetric in $F$ and $X\setminus F$, so complements of measurable sets are measurable. If $F$ and $G$ are measurable and $E\subset X$, splitting $E$ by $F$ and then $E\cap F$ by $G$ gives

\[ \begin{split} m^*(E)& =m^*(E\cap F\cap G)+m^*((E\cap F)\setminus G)+m^*(E\setminus F)\\ & \geq m^*(E\cap F\cap G)+m^*(E\setminus (F\cap G)), \end{split} \]

where the inequality follows from subadditivity, since $E\setminus (F\cap G)=((E\cap F)\setminus G)\cup (E\setminus F)$. The reverse inequality is again subadditivity, so $F\cap G$ is measurable. If $F_1,F_2,\ldots $ are disjoint measurable sets with union $F$, splitting $E$ successively by $F_1,\ldots ,F_ N$ gives

\[ m^*(E)=\sum _{j=1}^ Nm^*(E\cap F_ j)+m^*\Bigl (E\setminus \bigcup _{j=1}^ NF_ j\Bigr )\geq \sum _{j=1}^ Nm^*(E\cap F_ j)+m^*(E\setminus F). \]

Letting $N\to \infty $ and using subadditivity, we obtain $m^*(E)\geq m^*(E\cap F)+m^*(E\setminus F)$, so $F$ is measurable, and the choice $E=F$ gives $m^*(F)=\sum _ jm^*(F_ j)$. As every countable union of measurable sets is a disjoint union of measurable sets, this proves the claim.

It remains to show that every nonempty closed set $F\subset X$ is measurable. By subadditivity, we only need $m^*(E)\geq m^*(E\cap F)+m^*(E\setminus F)$, and only for sets $E\subset X$ with $m^*(E)<\infty $. For $N\geq 1$ put $E_ N=\{ x\in E:\operatorname {dist}(x,F)\geq N^{-1}\} $. Since $\operatorname {dist}(E\cap F,E_ N)\geq N^{-1}$, the hypothesis gives $m^*(E)\geq m^*(E\cap F)+m^*(E_ N)$, and it suffices to show that $m^*(E_ N)\to m^*(E\setminus F)$. Consider the layers

\[ A_ k=\{ x\in E:(k+1)^{-1}\leq \operatorname {dist}(x,F)<k^{-1}\} ,\qquad k\geq 1. \]

Since $\operatorname {dist}(\cdot ,F)$ is $1$-Lipschitz, $\operatorname {dist}(A_ k,A_{k'})\geq (k+1)^{-1}-(k+2)^{-1}>0$ whenever $k'\geq k+2$. Hence finitely many layers whose indices have the same parity are pairwise at positive distance, and repeated use of the hypothesis shows that the sum of their outer measures is at most $m^*(E)$. Consequently $\sum _ km^*(A_ k)\leq 2m^*(E)<\infty $. Since $F$ is closed, $E\setminus F\subset E_ N\cup \bigcup _{k\geq N}A_ k$, and therefore

\[ m^*(E_ N)\leq m^*(E\setminus F)\leq m^*(E_ N)+\sum _{k\geq N}m^*(A_ k). \]

The tails of the convergent series tend to zero, so $m^*(E_ N)\to m^*(E\setminus F)$. Thus the $\sigma $-algebra of $m^*$-measurable sets contains the closed sets, and hence all Borel sets. $\square$

Lemma 2.3.2`BAEFE6D3AE3C`

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Lemma 2.3.2BAEFE6D3AE3C

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Lemma 2.3.2`BAEFE6D3AE3C`