If $f_n \to f$ in measure then there exists a subsequence s.t $f_{n_k} \to f$ almost uniformly.
I have proven that there is a subsequence such that the convergence is pointwise a.e. I am not sure what to modify in the proof to make it work.
My proof:
Pick a subsequence such that $\mu(\{x: |f_{n_k}-f|>\frac{1}{2^k}\}=A_k)<\frac{1}{2^k}$. Then by Borell-Cantelli $\mu (\limsup A_k)=0$. Now $f_{n_k}(x)\not \to f(x)$ $\implies$ $x \in \limsup A_k$, thus $f_{n_k}(x) \to f(x)$ a.e. Not sure how to make is almost uniformly.