Question about 'inner measure equals outer measure $\iff$ measurable set' I'm trying to solve the Folland Real analysis p.32 problem 19 and it was easy to show that the inner measure of a measurable set equals the outer measure. However I'm stuck at the converse.
Could anyone help me how to prove that if the inner measure equals the outer meausure, then the set is measurable?
This is the definition of measurability in Folland.
And the problem 19 is the one that I'm stuck at.
 A: Suppose $\mu^*(E) = \mu_*(E) := \mu_0(X) - \mu^*(E^c)$. Then, because $\mu^*(X) = \mu_0(X) < \infty$, $$\mu^*(X) = \mu_0(X) = \mu^*(E) + \mu^*(E^c).$$
Now, we can use Exercise 18 (a) to show that for each $n\in \mathcal{N}$ there is an $A_n \in \mathcal{A}_\sigma$ such that $A_n \supset E$ and $\mu^*(E) \leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$. Setting $B = \bigcap A_n$, we have that $$\mu^*(E) \leq \mu^*(B) \leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n},
$$ for each $n$. Hence, $\mu^*(B) = \mu^*(E)$. Since each $A\in \mathcal{A}$ is measurable and the measurable sets form a $\sigma$-algebra, $B$ is also measurable.
Thus, we have that $$\mu^*(X) = \mu^*(E) + \mu^*(E^c) = \mu^*(B) + \mu^*(B^c).$$ Since $\mu^*(B) = \mu^*(E) < \infty$, we also have $\mu^*(B^c) = \mu^*(E^c)$. Using the measurability of $B$ again we have
$$
\begin{align*}
\mu^*(E) + \mu^*(E^c) &= \mu^*(E) + \mu^*(B\cap E^c) + \mu^*(B^c\cap E^c) \\
&= \mu^*(E) + \mu^*(B\cap E^c) + \mu^*(B^c)\\ &
= \mu^*(E) + \mu^*(B\cap E^c) + \mu^*(E^c).
\end{align*}
$$
Hence, $\mu^*(B\setminus E) = 0$, so we can apply Exercise 18 (b) to show that $E$ is measurable.
