# Approximation by Open/Compact Sets

Locally finite Borel Measures on $\mathbb {R}$ are completely characterized by the Lebesque-Stieltjes outer Measure associated to a monotone increasing right continuous functions $F:\mathbb {R} \to \mathbb {R}$ by defining the measure of a Borel set $E\subseteq \mathbb {R}$ to be

$\mu _{F}(E)=\inf \left\{\sum _{i=1}^{\infty }(F(a_{i})-F(b_{i}))\;:\;E\subseteq \bigcup _{i=1}^{\infty }(a_{i},b_{i}]\right\}$ where the infimum is taken over all coverings $\{(a_{i},b_{i}]\}_{i\in \mathbb {N} }$ of $E$ by half open intervals. This is a fairly unruly albeit useful definition so it is valuable to have alternatives at hand to approximate the measure of Borel sets. The Approximation by Open and Compact sets is a great way to do so, as open and compact sets are well studied and understood, and as the following theorem will show, they approximate measurable sets arbitrarily well.

## Theorem Statement

Let $\mu$ be a Lebesque-Stieltjes measure for a right continuous increasing function $F:\mathbb {R} \to \mathbb {R}$ . By Caratheodory's Theorem, we know that the $\sigma -$ algebra ${\mathcal {M}}_{\mu }$ of $\mu -$ measurable sets makes $(\mathbb {R} ,{\mathcal {M}}_{\mu },\mu )$ into a measure space. Then, if $E\in {\mathcal {M}}_{\mu }$ we have that

$\mu (E)=\inf\{\mu (U)\;:\;E\subseteq U\;{\text{open}}\}=\sup\{\mu (K)\;:\;K\subseteq E\;{\text{compact}}\}.$ ## Proof

To prove this statement, it is useful to state and prove the following lemma.

Lemma 1: In the notation of the Theorem Statement, we find that

$\mu (E)=\inf \left\{\sum _{i=1}^{\infty }\mu ((a_{i},b_{i}))\;|\;E\subseteq \bigcup _{i=1}^{\infty }(a_{i},b_{i})\right\}$ proof. Let us denote the value on the right as $m(E)$ . To show that $\mu (E)=m(E)$ , we will first show that $\mu (E)\geq m(E)$ and then that $\mu (E)\leq m(E)$ to conclude with equality.

Claim: $\mu (E)\geq m(E)$ Let $E\subseteq \bigcup _{i=1}^{\infty }(a_{i},b_{i})$ . Then, we let $(c_{i}^{k})_{k\in \mathbb {N} }$ be a sequence of real numbers so that $c_{i}^{1}=a_{i}$ and $\lim _{k\to \infty }c_{i}^{k}=b_{i}$ . Then, let $I_{i}^{k}=(c_{i}^{k},c_{i}^{k+1}]$ . We find that $(a_{i},b_{i})$ can be expressed as the disjoint union $(a_{i},b_{i})=\bigcup _{k=1}^{\infty }I_{i}^{k}$ so that by disjoint additivity,

$\sum _{i=1}^{\infty }\mu (a_{i},b_{i})=\sum _{i,k=1}^{\infty }I_{i}^{k}\geq \mu (E)$ given the definition of $\mu$ . Since this holds for all covers of $E$ by open sets, the claim follows.

Claim: $\mu (E)\leq m(E)$ Let $\epsilon >0$ Then, there exists a collection of half open intervals $(a_{i},b_{i}]$ with $E\subseteq \bigcup _{i=1}^{\infty }(a_{i},b_{i}]$ so that

$\sum _{i=1}^{\infty }\mu ((a_{i},b_{i}])\leq \mu (E)+\epsilon /2$ By the right continuity of $F$ , we know that for each $i$ , there exists some $\delta _{i}$ so that

$\mu (b_{i},b_{i}+\delta _{i})=F(b_{i}+\delta _{i})-F(b_{i})<\epsilon ^{-(j+1)}$ .

Then, $E\subseteq \bigcup _{i=1}^{\infty }(a_{i},b_{i}+\delta _{i})$ and

$\sum _{i=1}^{\infty }(a_{i},b_{i}+\delta _{i})=\sum _{i=1}^{\infty }(a_{i},b_{i}]+\epsilon /2\leq \mu (E)+\epsilon /2+\epsilon /2=\mu (E)+\epsilon$ proving by the definition of $m$ , that $\mu (E)\leq m(E)$ , completing the proof of the lemma.

$\square$ Theorem: $\mu (E)=\inf\{\mu (U)\;:\;E\subseteq U\;{\text{open}}\}=\sup\{\mu (K)\;:\;K\subseteq E\;{\text{compact}}\}.$ proof. By the aforementioned lemma, we find that if $\epsilon >0$ , then there exist a sequence of intervals $(a_{i},b_{i})$ covering $E$ so that $\sum _{i=1}^{\infty }\mu ((a_{i},b_{i}))\leq \mu (E)+\epsilon .$ Taking $U=\bigcup _{i=1}^{\infty }(a_{i},b_{i})$ then gives the following by countable subadditivity and monotonicity of $\mu$ .

$\mu (E)\leq \mu (U)\leq \sum _{i=1}^{\infty }\mu ((a_{i},b_{i}))\leq \mu (E)+\epsilon$ which proves that $\mu (E)=\inf\{\mu (U)\;:\;E\subseteq U\;{\text{open}}\}.$ Next, to show equality for compact sets. We will devolve into cases:$E$ is bounded and when it is unbounded.

Case 1: $E$ is bounded.

Let $\epsilon >0$ and let ${\overline {E}}\setminus E\subseteq U$ be open so that $\mu (U)\leq \mu ({\overline {E}}\setminus E)+\epsilon .$ . Let $K={\overline {E}}\setminus U$ . Then, $K\subseteq E$ is closed and bounded by construction. Then, since $E$ is measurable, we find that $\mu (U\cap E)=\mu (E)-\mu (U\setminus E)$ so that

$\mu (K)=\mu (E)-\mu (E\cap U)=\mu (E)-(\mu (E)-\mu (U\setminus E))\geq \mu (E)-\mu (U)+\mu ({\overline {E}}\setminus E)\geq \mu (E)-\epsilon$ which proves the result for this case.

Case 2: $E$ is unbounded

Let $\epsilon >0$ and $E_{j}=E\cap (j,j+1]$ . Then, $E_{j}$ is bounded so we can apply the previous case to find a compact set $K_{j}\subseteq E_{j}$ so that $\mu (K_{j})\geq \mu (E_{j})+\epsilon 2^{-|j|-1}$ . Then, $H_{n}=\bigcup _{i=-n}^{n}K_{n}$ is compact with $\mu (H_{n})\geq \mu \left(\bigcup _{i=-n}^{n}E_{i}\right)-\epsilon .$ Then, since $H_{n}\subseteq E$ for all $n$ , we can apply continuity of $\mu$ to find that

$\mu (E)=\lim _{n\to \infty }\mu \left(\bigcup _{i=-n}^{n}E_{i}\right)$ from which the desired result follows. This proves the theorem.

$\square$ 