This question comes from Bass, Ex 4.15. Given a finite measure space $(X, \mathcal{A}, \mu)$, we can define an outer measure $\mu^*$ as
$$
\mu^*(A) := \inf\{ \mu(B) : A \subset B , B \in \mathcal{A} \} \:.
$$
(One can check that this is an outer measure and that $\mu^*$ agrees with $\mu$ on $\mathcal{A}$).
I want to show now that if $\{A_n\}_{n \geq 1}$ is a sequence of increasing sets (not necessarily in $\mathcal{A}$) such that $A_n \nearrow A$, then $\mu^*(A_n) \nearrow \mu^*(A)$.
I have this so far. By monotonicity of outer measure, $\mu^*(A_n) \leq \mu^*(A)$ for all $n$, hence $\lim_{n \rightarrow \infty} \mu^*(A_n) \leq \mu^*(A)$. It remains to show that
$\mu^*(A) \leq \lim_{n \rightarrow \infty} \mu^*(A_n)$.
My rough idea is as follows. Fix any $\epsilon > 0$. We can pick a sequence of sets $\{G_i\}_{i \geq 1}$ in $\mathcal{A}$ such that $\mu^*(A_i) \geq \mu(G_i) – \epsilon/2^{i}$. If the $\{G_i\}$ were an increasing set then we'd be done by using the continuity of measure from below. However, they are not. But, we can define $H_n := \bigcup_{i=1}^{n} G_i$ which is an increasing sequence by definition. Since $A \subset \bigcup_{n \geq 1} H_n$, we have that
$$
\mu^*(A) \leq \mu( \bigcup_{n \geq 1} H_n ) = \lim_{n \rightarrow \infty} \mu(H_n) \:.
$$
Now I am not quite sure how to finish off. I want to say something to the effect of $\mu(H_n) \leq \mu^*(A_n) + \epsilon$. However, since the $A_n$'s are not measurable, I am having trouble making such a comparison.
Best Answer
As you wrote, you have already proved $\lim_{n \rightarrow \infty} \mu^*(A_n) \leq \mu^*(A)$. It remains to show that $\mu^*(A) \leq \lim_{n \rightarrow \infty} \mu^*(A_n)$.
Your idea to complete the proof is essentially correct, all it needs is a small adjustment.
Given any $\epsilon > 0$. We can pick a sequence of sets $\{G_i\}_{i \geq 1}$ in $\mathcal{A}$ such that $A_i\subseteq G_i$ and $\mu^*(A_i) \leq \mu(G_i) \leq \mu^*(A_i) + \epsilon/2^{i}$.
Define $H_n=\bigcap_{i=n}^\infty G_i$. Then $\{H_n\}_{n\geq 1}$ is an increasing sequence of sets in in $\mathcal{A}$.
For any $n \geq 1$, if $i \geq n$ then $A_n\subseteq A_i \subseteq G_i$. So, for any $n \geq 1$, $A_n \subseteq \bigcap_{i=n}^\infty G_i=H_n$. So we have $A_n \subseteq H_n\subseteq G_n$ and we get $$ \mu^*(A_n) \leq \mu(H_n) \leq \mu(G_n) \leq \mu^*(A_n) + \epsilon/2^{n}$$
So we have $\lim_{n \to +\infty} \mu^*(A_n) = \lim_{n \to +\infty} \mu(H_n)$.
Since $A_n \nearrow A$, it is easy to see that that $A\subseteq \bigcup_{n\geq 1} H_n$. But $H_n \nearrow \bigcup_{n\geq 1} H_n$, so using the the continuity of measure from below, we have
$$\mu^*(A)\leq \mu\left(\bigcup_{n\geq 1} H_n\right)=\lim_{n \to +\infty} \mu(H_n)= \lim_{n \to +\infty} \mu^*(A_n)$$