If a set has finite outer measure, and its inner measure is equal to its outer measure, then it is measurable

Question

I'm trying to prove this statement in $\mathbb{R} ^n$ with the Lebesgue measure:

$$|E|_e < +\infty \land |E|_i = |E|_e \Rightarrow E \text{ is measurable}$$

My guess is that the easy way of going about it is by trying to find an $F_{\sigma}$ set $K$ such that $E =K \cup N$ with |N| = 0

Constructing $K$ is easy: for each $n$ in $\mathbb N $ we have a closed set $F_n \subset E$ such that $|F| > |E|_i - \frac 1 n = |E|_e - \frac 1 n$. Then set $K := \bigcup_{n \in \mathbb N} F_n$

It's easy to check that $|K| = |E|_e$.

Now let $N = E - K$. I'm stuck trying to prove that $|N| = 0$. The only result I have is that $$|E-K|_e = |E - \bigcup_{n \in \mathbb N} K_n|_e = |\bigcap_{n \in \mathbb N} (E - K_n)|_e \le \lim_{n \to \infty} |E - K_n|_e$$

Where $K_n = \bigcup_{k=1}^n F_n$ is an increasing sequence of closed sets where $K = \bigcup_{n \in \mathbb N} K_n$ and so $E - K_n$ is a decreasing sequence of sets so the inequality above holds

I'm struggling to find a bound for $|E - K_n|_e$

Answer 1

Let $\varepsilon > 0$. Then there exists an open set $G \supseteq E$ such that $|G| < |E|_e + \varepsilon$. Now,

$$|E - K_n|_e \le |G - K_n| = |G| - |K_n| < |E|_e + \varepsilon- |K_n|$$

First inequality by monotonocity of outer measure, and then the equality in the middle because both $G$ and $K_n$ are measurable and $K_n \subseteq E \subseteq G$

Then taking limits on each side, since $\lim_{n\to \infty}|K_n|=|E|_e$, we get $\lim_{n \to \infty} |E-K_n|_e < \varepsilon$

But $\varepsilon$ was arbitrary so it must be 0