Use property to show that if $f_n \rightarrow f$ in measure on finite measure set $E$ then $f_n^2 \rightarrow f^2$ in measure on $E$. Consider the following problem:

Let $E$ have finite measure, $\{f_n\} \rightarrow f$ in measure on $E$ and $g$ be a measurable function on $E$ that is finite almost everywhere. Use the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ to prove that $\{f_n^2 \} \rightarrow f^2$ in measure on $E$. Infer from this that if $\{g_n\} \rightarrow g$ in measure on $E$ then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure on $E$.

I'm struggling understand how I can apply the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ in measure to prove that $\{f_n^2 \} \rightarrow f^2$ in measure. I've been able to come up with a proof of the fact that if $\{g_n\} \rightarrow g$ in measure then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure, and this clearly implies the claim that $\{f_n^2 \} \rightarrow f^2$ in measure. Unfortunately my proof doesn't rely on the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ so it doesn't answer the problem since it explicitly asks us to prove it a certain way.
So far my best attempt to utilize the fact is as follows. Let $\eta > 0$ then by triangle inequality and monotonicity of measure
\begin{align*}
    \lim_{n\rightarrow \infty}\mu\left\{ \left|f_n^2 - f^2\right| > \eta \right\} &\leq \lim_{n\rightarrow \infty}\mu \left\{\left|f_n - f\right|\cdot \left|f_n\right| + \left|f_n - f\right| \cdot \left|f\right| > \eta \right\} \\
    &\leq \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right|\cdot \left|f_n\right| > \eta/2 \right\} + \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| > \eta/2 \right\}
\end{align*}
By the fact that $\{f_n \cdot f\} \rightarrow f \cdot f$ in measure
\begin{equation*}
\lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| > \eta/2 \right\} = \mu\left\{\left|f_n \cdot f - f\cdot f\right| > \eta/2 \right\} = 0.
\end{equation*}
I don't see how to deal with the other limit without first proving the most general statement $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure. 
Any help or advice is greatly appreciated.
 A: For fixed $M>0$ we clearly have
$$\begin{align*} \mu(|f_n-f| \cdot |f_n| > \eta/2) &\leq \mu(|f_n-f| \cdot |f_n|>\eta/2, |f_n| \leq M) \\ &\quad + \mu(|f_n-f| \cdot |f_n|>\eta/2, |f_n| > M), \end{align*}$$
and so
$$\mu(|f_n-f| \cdot |f_n| > \eta/2) \leq \mu(|f_n-f| >\eta/(2M)) + \mu(|f_n| > M). \tag{1}$$
As $f_n \to f$ in measure, the first term on the right-hand side converges to $0$ as $n \to \infty$. For the second one, we note that
$$\begin{align*} \mu(|f_n|>M) &= \mu(|f_n-f| < \eta, |f_n|>M) + \mu(|f_n-f|\geq \eta, |f_n|>M) \\ &\leq \mu(|f| \geq M-\eta) + \mu(|f_n-f| \geq \eta). \end{align*}$$
If we let $n \to \infty$, then $\mu(|f_n-f| \geq \eta) \to 0$ as $n \to \infty$. Hence,
$$\limsup_{n \to\infty} \mu(|f_n| >M) \leq \mu(|f| \geq M-\eta),$$
and therefore, by (1),
$$\limsup_{n \to \infty} \mu(|f_n-f| \cdot |f_n| > \eta/2) \leq \mu(|f| \geq M-\eta).$$
Since $\mu$ is a finite measure, we know that $\mu(|f| \geq M-\eta) \to 0$ as $M \to \infty$, and therefore we conclude that
$$\limsup_{n \to \infty} \mu(|f_n-f| \cdot |f_n| > \eta/2) = 0.$$ 
