If $\{f_n\}$ is a sequence of measurable functions that converges in measure to f, then f is measurable.

I'm not sure how to start this, but I know the definitions needed.

$f_n$ converges in measure to $f$, so:

lim $\mu^*(\{x \in X: |f_n(x) - f(x)| \ge \epsilon\})=0$

So the outer measure of the set of $\{x \in X\}$ where $f_n(x)$ does not converge to $f(x)$ is 0.

But the definition my book gives for a measurable function is that $f^{-1}(O)$ is a measurable set $\forall O \subseteq \mathbb R$. I'm not seeing a route I can take to get from point A to point B. Can anyone point me in the right direction?

For each $k\in{\bf{N}}$, find strictly increasing $(n_{k})$ such that $A_{k}:=\{x\in X:|f_{n_{k}}(x)-f(x)|\geq 1/k\}$, $\mu^{\ast}(A_{k})<1/2^{k}$. Let $N=\displaystyle\bigcap_{n}\displaystyle\bigcup_{k\geq n}A_{k}$, one can show $\mu^{\ast}(N)=0$ and $f_{n_{k}}(x)\rightarrow f(x)$ for all $x\in X-N$. Now $(f_{n_{k}})$ is a sequence of measurable functions, so is the a.e. limit function $f$.