TITLE: Product of a sequence of functions that converges in measure and a measurable function converges in measure.
QUESTION [5 upvotes]: I'm working through the following problem in Royden (Chapter 5, exercise 7) and would like to understand the solution:

Let $E$ have finite measure, $\{f_n\}\to f$ in measure on $E$, and $g$ be a measurable function on $E$ that is finite $a.e.$ on $E$. Prove that $\{f_n\cdot g\}\to f\cdot g$ in measure.  

$\newcommand{\N}{\mathbb{N}}$
Proof. Since $g$ is finite a.e. on $E$, $m(\bigcap_{n\in\N}\{x\in E\,\,|\,\,g(x)>n\})=0\implies \forall\epsilon>0\,\,\, \exists N_1\in\N$:$n\geq N_1 \implies m\{x\in E\,\,|\,\,|g(x)|>n\}<\epsilon$. 
So, let $\epsilon>0$ be given. Then there exists $M\in\N$ so that $m\{x\in E\,\,|\,\,|g(x)|>M\}<\frac{\epsilon}{2}$. 
Let $\eta>0$. Since $\{f_n\}\to f$ in measure on $E$, $\exists N\in\N :n\geq N\implies m\{x\in E\,\,|\,\,|f_n(x)-f(x)|>\frac{\eta}{2M}\}<\frac{\epsilon}{2}$. Letting $n\geq N$, notice that:
\begin{align*}
&\{x\in E\,\,|\,\,|(f_n\cdot g)(x)-(f\cdot g)(x)|>\frac{\eta}{2}\}\\
=&\{x\in E\,\,|\,\,|g(x)\cdot f_n(x)-g(x)\cdot f(x)|>\frac{\eta}{2}\}\\
=&\{x\in E\,\,|\,\,|g(x)|\cdot|f_n(x)-f(x)|>\frac{\eta}{2}\}\\
\subset &\{x\in E\,\,|\,\,|g(x)|>M_1\}\cup \{x\in E\,\,|\,\,|f_n(x)-f(x)|>\frac{\eta}{2M}\}.\\
\end{align*}
We then infer that:
\begin{align*}
&m\{x\in E\,\,|\,\,|(f_n\cdot g)(x)-(f\cdot g)(x)|>\frac{\eta}{2}\}\\
\leq &m\{x\in E\,\,|\,\,|g(x)|>M_1\}+m\{x\in E\,\,|\,\,|f_n(x)-f(x)|>\frac{\eta}{2M}\}\\
< &\frac{\epsilon}{2}+\frac{\epsilon}{2}\\
=&\epsilon.\\
\end{align*} 
$\blacksquare$
Specifically, I'm having some trouble understanding why:
$\{x\in E\,\,|\,\,|g(x)|\cdot|f_n(x)-f(x)|>\frac{\eta}{2}\}
\subset \{x\in E\,\,|\,\,|g(x)|>M_1\}\cup \{x\in E\,\,|\,\,|f_n(x)-f(x)|>\frac{\eta}{2M}\}$. 
Sorry if I'm being a bit handwavy. Any constructive feedback will be helpful!

REPLY [2 votes]: For convenience we set
$$F=\{x\in E:|g(x)||f_n(x)-f(x)|>\frac{\eta}{2}\}$$
$$F_1=\{x\in E:|g(x)||f_n(x)-f(x)|>\frac{\eta}{2}\text{ and }|g(x)|>M\}$$
$$F_2=\{x\in E:|g(x)||f_n(x)-f(x)|>\frac{\eta}{2}\text{ and }|g(x)|\leq M\}$$
$$F_3=\{x\in E:|g(x)|>M\}$$
$$F_4=\{x\in E:|f_n(x)-f(x)|>\frac{\eta}{2M}\}$$
It is trivial that
$$F=F_1\cup F_2,$$
$$F_1\subseteq F_3.$$
On the other hand $|g(x)||f_n(x)-f(x)|>\frac{\eta}{2}$ together with $|g(x)|\leq M$ imply $|f_n(x)-f(x)|>\frac{\eta}{2M}$, therefore 
$$F_2\subseteq F_4.$$
Hence
$$F\subseteq F_3\cup F_4.$$