Example of continuity of measure and convergence in measure 
Is the following correct?
Since $f_n \to f$a.e. we have that $f_n \to f$ in measure. Using the continuity of measure we have that:
fix $k$; then $\{x :|f_m(x)-f(x)|>\frac{1}{k}\}$ is an increasing sequence. so
\begin{align}
\lim_{n\to\infty}m(E_{n,k})
&= \lim_{n\to\infty} m\bigg(\bigcup_{m=n}^{\infty}\{x :|f_m(x)-f(x)|>\frac{1}{k}\}\bigg) \\
&\leq \lim_{n\to\infty} \sum_{m=n}^{\infty}m\bigg(\{x :|f_m(x)-f(x)|>\frac{1}{k}\}\bigg)\\
&=\lim_{m\to\infty} m\bigg(\{x :|f_m(x)-f(x)|>\frac{1}{k}\}\bigg) = 0  \text{ by convergence in measure}
\end{align}
 A: 
fix $k$; then $\{x :|f_m(x)-f(x)|>\frac{1}{k}\}$ is an increasing sequence.

Here, you seem to be claiming that $\{x : |f_m(x)-f(x)|>\frac{1}{k}\}$ is an increasing sequence with respect to $m.$  There is no reason for this to be true.

$$\lim_{n\to\infty} \sum_{m=n}^{\infty}m\bigg(\{x :|f_m(x)-f(x)|>\frac{1}{k}\}\bigg) = \lim_{m\to\infty} m\bigg(\{x :|f_m(x)-f(x)|>\frac{1}{k}\}\bigg)$$

Why?  This also won't be true in general.  If true, it seems like this equality is equivalent to what you're trying to prove, so you need to at least explain why this equality should be true.
A: Note that the equality in the bottom with $\lim_m$ and $\lim_n$ is not true and the series might not converge also.
Here is a proof if you want:
Let $k \in \Bbb{N}$
The sequence $E_{n,k}$ are  decreasing on $n$ thus $\mu(E_{n,k}) \to \mu(\bigcap_nE_{n,k})$
But note that $\bigcap_nE_{n,k} \subseteq \{x: f_n(x)\text{ does not converge to f(x)}\}=:B$
where $\mu(B)=0$ 

$B=\bigcup_{k \in \Bbb{N}}\bigcap_{n \in \Bbb{N}}E_{n,k}$

