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.

$\square$