Lebesgue measure as $\sup$ of measures of contained compact sets

I know, from Kolmogorov-Fomin's Элементы теории функций и функционального анализа, the definition of external measure of a bounded set $A\subset \mathbb{R}^n$ as $$\mu^{\ast}(A):=\inf_{A\subset \bigcup_k P_k}\sum_k m(P_k)$$where the infimum is extended to all the possible covers of $A$ by finite or countable families of $n$-paralleliped $P_k=\prod_{i=1}^n I_i$, where $I_i=[a_{i,k},b_{i,k}]$ or $I_i=(a_{i,k},b_{i,k})$ or $I_i=(a_{i,k},b_{i,k}]$ or $I_i=[a_{i,k},b_{i,k})$, with $a_{i,k}\le b_{i,k}$, whose measure is defined as $$m(P_k):=\prod_{i=1}^n(b_{i,k}-a_{i,k}).$$ A set is said to be elementary when it is the union of a finite number of such $n$-parallelipeds and a set $A$ is said to be measurable if, for any $\varepsilon>0$, there is an elementary set $B$ such that $$\mu^{\ast}(A\triangle B)<\varepsilon.$$The function $\mu^{\ast}$ defined on measurable sets only is called Lebesgue measure and the notation $\mu$ is used for it.

Let us come to my question. I read that if $A$ is Lebesgue measurable then $$\mu(A)=\sup_{K\subset A} \mu(K)$$ where the $K$ are compact measurable sets. My text, Kolomogorv-Fomin's does not examine the theory of Lebesgue measure in $\mathbb{R}^n$, $n>1$, in depth, therefore I have no tool to prove such a property to myself, and I would be very grateful to anybody explaining a proof to me.

Solution 1:

Assuming we've shown that $\mu$ is a measure, and that we're only concerned with bounded sets:

Say $B$ is a bounded closed set and $E\subset B$ is measurable. First, note that $$\mu(E)=\inf\{\mu(V):E\subset V\subset B, V\text{ relatively open}\}.$$

(Here "relatively open" means open relative to $B$; that is, $V=B\cap W$ where $W$ is an open subset of $\Bbb R^d$.)

The inequality $\le$ is clear. Let $\epsilon>0$, and choose rectangles $I_j$ with $E\subset \bigcup I_j$ and $\sum\mu(I_j)<\mu(E)+\epsilon$. Let $O_j$ be an open rectangle with $I_j\subset O_j$ and $$\mu(O_j)<\mu(I_j)+\epsilon2^{-j}.$$Let $V=B\cap\bigcup O_j$. Then $V$ is relatively open, $E\subset V$, and $$\mu(V)\le\sum\mu(O_j)\le \epsilon+\sum\mu(I_j)<\mu(E)+2\epsilon.$$ So $\mu(E)\ge\inf_V\mu(V)$, hence $\mu(E)=\inf_V\mu(V)$.

That's known as outer regularity. Inner regularity follows by taking complements: Given $E\subset B$, let $F=B\setminus E$. There exists a relatively open set $V\subset B$ with $F\subset V$ and $\mu(V)<\mu(F)+\epsilon$. Let $K=B\setminus V$. Then $K$ is compact, $F\subset V$ implies $K\subset E$, and $$\mu(K)=\mu(B)-\mu(V)>\mu(B)-\mu(F)-\epsilon=\mu(E)-\epsilon.$$