Is it true that $|f\_n|\le M$ and $f\_n\to f$ in measure implies $|f|\le M$ a.e.? It appears so to me, as justified below. I'd appreciate it if someone can confirm or refute it for me.

Let $A\triangleq \{x:|f(x)|>M\}$. We want to show $\mu(A)=0$. Let $A\_k\triangleq \{x:|f(x)|>M+\frac{1}{k}\}, k\in\mathbb N.$ Then $A=\bigcup\_{k=1}^\infty A\_k$, and $A\_{k}\subset A\_{k+1}$. Therefore, $\mu(A)=\lim\_{k\to \infty}\mu(A\_k).$ 

Now consider $E\_n\triangleq\{x:|f\_n(x)-f(x)|>\frac{1}{k}\}$. Clearly $A\_k \subset E\_n$, so $\mu(A\_k)\le \mu(E\_n)$ for any $n$. But $\mu(E\_n)\to 0$, since $f\_n\to f$ in measure. So $\mu(A\_k)=0$ for any $k$, and hence $\mu(A)=0.$

Is this proof correct? Are there flaws or simpler proof? Thanks a lot!