Exercise 1.2.15 (Inner regularity of Lebesgue measure)

We explore two cases:

• $E$is bounded
  On one hand, any compact set $F$ contained in $E$ will have a smaller Lebesgue measure by monotonicity; thus, we trivially have an upper bound $m(E)\ge \underset{K\subseteq E,K\text{ compact}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}m(K)$.
  On the other hand, suppose we have another upper bound. Pick an arbitrary $𝜖>0$, and denote it by $m(E)-𝜖<m(E)$. By Exercise 1.2.7 we can find a closed set $F\subseteq E$ such that $m(E∕F)\le 𝜖$. Obviously, $F$ is bounded, and by countable additivity we have $m(E)-m(F)\le 𝜖∕2$ - thus, $m(F)\ge m(E)-𝜖$. Therefore, $m(E)$ is the least upper bound, as desired.

• $E$is unbounded
  Then we can construct a sequence ${\left(E\cap {(-r,r)}^d\right)}_{r\in \mathbb{N}}$. Each of its elements is (1) bounded (2) Lebesgue measurable (3) increasing sets (4) such that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{r\in \mathbb{N}}E\cap {(-r,r)}^d=E$. By (1) and (2) we have

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->r\in \mathbb{N}:m\left(E\cap {(-r,r)}^d\right)=\underset{K\subseteq E\cap {(-r,r)}^d,K\text{ compact}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}m(K)$$

  In each case $r\in \mathbb{N}$ we can give ourselves an epsilon of a room and find an inner compact cover $K$ such that $m(K_r)\ge m\left(E\cap {(-r,r)}^d\right)-𝜖$. Since every compact inner cover of $E\cap {(-r,r)}^d$ is also a compact inner cover of $E$, we have

  $$\underset{K\subseteq E,K\text{ compact}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}m(K)\ge m(K_r)\ge m\left(E\cap {(-r,r)}^d\right)-𝜖$$

  By (3) and (4) we can apply Exercise 1.2.11 by taking limits as $r\to \infty$:

  $$\underset{K\subseteq E,K\text{ compact}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}m(K)\ge \underset{r\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}m(K_r)\ge \underset{r\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}m\left(E\cap {(-r,r)}^d\right)-𝜖=m\left(\underset{r=1}{\overset{\infty }{\bigcup }}E\cap {(-r,r)}^d\right)-𝜖=m(E)-𝜖$$

  Thus, $\underset{K\subseteq E,K\text{ compact}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}m(K)\ge m(E)-𝜖$. Trivially, by monotonicity, we also have $m(E)\ge \underset{K\subseteq E,K\text{ compact}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}$ and we are done.

Monotonicity implies that it suffices to prove that $m(A)\le \underset{A\supset K}{\sup <!--\; FUNCTION\; APPLICATION\; -->}$ compact $m(K)$.

First suppose that $A$ is bounded, so $m(A)<\infty$.

Fix $𝜀>0$. Since $A$ is Lebesgue measurable there exists a closed set $K\subset A$ such that ${\mu }^{\text{∗}}(A\setminus K)<𝜀$. Since $K\subset A$ and $A$ is bounded, we have that $K$ is bounded as well, so $K$ is compact because it is bounded and closed. So $K$ is a compact set such that $K\subset A$ and $\mu (K)=\mu (A)-\mu (A\setminus K)\ge \mu (A)-𝜀$. This holds for all $𝜀>0$ establishing the claim in the case that $A$ is bounded.

For the case where $A$ is unbounded, consider $A_n=A\cap (B(0,n)\setminus B(0,n-1))$ for all $n$. These are all bounded, so for every $𝜀>0$ and every $n$ there exists a compact set $K_n\subset A_n$ such that $m\left(K_n\right)\ge m\left(A_n\right)-𝜀2^{-n}$. Also, ${\left(A_n\right)}_n$ are disjoint so $m(A)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n}m\left(A_n\right)$. Thus, if $K_n^{\prime }={\uplus <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^n{K}_j$, then $K_n^{\prime }$ is compact, $K_n^{\prime }\subset A$ and

This implies that there exists $n$ large enough so that $m\left(K_n^{\prime }\right)\ge m(A)-2𝜀$.

Conclusion, for any $𝜀>0$ there exists a compact $K\subset A$ such that $m(K)\ge m(A)-𝜀$.

This proves the unbounded case.