Q:

convergence a.e.  implies convergence in measure

Let $f$ be a measurable function.  Let $(f_{n})_{n\ge1}$ be a sequence of measurable functions such that $f_{n}\to f$ a.e.  on $X$. Suppose $\mu(X)<+\infty$. Then, $f_{n}\to f$ in measure. 
I was trying to proof this as follows:
Proof:
Let $\epsilon > 0$ be given.  Let $A_{n}(\epsilon):=\{x\in X : |f_{n}(x)-f(x)|\ge \epsilon \,\}$. We want to show that $\lim_{n\to\infty}\mu(A_{n}(\epsilon))=0$.
Since $f_{n}\to f$ a.e, there is a measurable subset $N\subseteq X$ such that $\mu(N)=0$ and for any $x\in X \setminus N=N^{c}$, we have that $f_{n}(x)\to f(x)$. So:
$x\in N^{c} \Rightarrow f_{n}(x)\to f(x) \Rightarrow$ there is some $n_{0}\ge 1$ such that for any $n\ge n_{0}$, we have $|f_{n}(x)-f(x)|< \epsilon$.
Hence, $N^{c}\subseteq \bigcap_{n\ge n_{0}}(A_{n}(\epsilon))^{c} \Rightarrow \bigcup_{n\ge n_{0}}A_{n}(\epsilon)\subseteq N$.
Therefore, for $n\ge n_{0}$, $A_{n}(\epsilon)\subseteq N \Rightarrow \mu(A_{n}(\epsilon))=0$. Then, $\lim_{n\to\infty}\mu(A_{n}(\epsilon))=0$ follows. 
I have seen proofs of this where one use the continuity from above of the measure $\mu$, and therefore use the hypothesis that $\mu(X)<+\infty$. I know that are counterexamples if we remove $\mu(X)<+\infty$. But, I dont see in my proof where I used this fact.  So, either my proof is wrong or there is some stage where $\mu(X)<+\infty$ was used.  Any help? 

A:

Your mistake is the inclusion $N^c\subseteq \cap_{n\geq n_0} (A_n(\epsilon))^c$. The index $n_0$ depends on the point $x$. If you take some other point $y\in N^c$ it is not necessary true that $|f_{n_0}(y)-f(y)|<\epsilon$, maybe for this point you need a much bigger index.  So $y$ might not belong to the intersection $\cap_{n\geq n_0}(A_n(\epsilon))^c$. The inclusion is false.