Q:

Convergence in measure for bounded f and uniformly bounded $g_n$ implies $f_n \cdot g_n \to f \cdot g$ in measure

Let $f_n$ be convergent in measure to some bounded $f$ ($\exists_{M \in \mathbb{R}} \text{ }|f(x)| \leq M $) and $g_n$ uniformly bounded (there exists $E \in \mathcal{F}, \mu(E)=0$ and $\forall_{x \notin E} \exists_{N\in \mathbb{R}} \forall_{n} |g_n(x)| \leq N$) and convergent to $g$. Prove that $f_n \cdot g_n \to f\cdot g$ in measure. 
I tried bounding:
$$
\forall_{x \notin E} \mu\left(x: |f_n(x)g_n(x) - f(x)g(x)| > \epsilon\right) \leq
$$
$$
\forall_{x \notin E} \mu\left(x: |f_n(x)N - M g(x)| > \epsilon\right)
$$
But that did not lead me far. I have to show the limit of the above measure is 0 as n goes to infinity. How can I tackle this problem?

A:

Note that $$\{x \in E^c:|f_n(x)g_n(x)-f(x)g(x)| > \epsilon \}$$ $$\subseteq \{x \in E^c:|f_n(x)-f(x)|>\frac{\epsilon}{N+M}\} \cup\{x \in E^c: |g_n(x)-g(x)|>\frac{\epsilon}{N+M}\}$$
where $M>0$ such that $|f(x)| \leq M,\forall x \in E$
Thus $$\mu(\{x :|.....|>\epsilon\}) \leq \mu(\{x:\in E^c:|.....|>\epsilon\})+\mu(x \in E:|....|>\epsilon\}$$ $$=\mu(\{x:\in E^c:|.....|>\epsilon\})$$
Continue from here...

