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

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

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

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 the part of the theorem for compact sets, Dear editors, I am saving and will be back in a bit to finish this proof and thus, the article. Give me a minute please and then you have free reign.