If $\mu(\{w \in A: |f_n(w) - f(w)| > \epsilon\,\, \text{for infinitely many many $n$}\}) = 0$ for each $\epsilon > 0$, then $f_n \rightarrow f$ $\mu$-a.e.
I can see how to solve this problem intuitively but I can't put it in mathematical words I mean if we have that set inside isn't bounded by certain values the measure of that is zero, then outside of that we must have that $f_n$ goes to $f$ almost everywhere but I having troubles writing that rigorously.
Let $X = \bigcup\limits_{j = 1}^\infty A_j$, where
$$A_j = \left\{w\in A : |f_n(w) - f(w)| > \frac{1}{j}\,\text{for infinitely many $n$}\right\}.$$
Since $\mu(A_j) = 0$ for all $j$, by countable subadditivity of $\mu$, $\mu(X) \le \sum\limits_{j = 1}^\infty m(A_j) = 0$. If $w\notin X$, for every $j \ge 1$, $|f_n(w) - f(w)| \le \frac{1}{j}$ for all $n$ sufficiently large. Thus $f_n(w) \to f(w)$. Since $w$ was arbitrary, $f_n(w)\to f(w)$ for all $w$ outside $X$, a set of $\mu$-measure zero. Hence $f_n \to f$ $\mu$-a.e..