Problem: Let $(\Omega,\mathcal F,\mu)$ be a measure space and $f_n,f:\Omega\to\mathbb C$ be integrable and $E_n,E\in\mathcal F$ such that $$\lim_{n\to\infty}\mu(E_n\triangle E)=0\quad\text{and}\quad\lim_{n\to\infty}\int_\Omega\vert f_n-f\vert\,d\mu=0.$$ Prove that $$\lim_{n\to\infty}\int_{E_n}f_n\,d\mu=\int_E f\,d\mu=0.$$
My Attempt: Recall that $E_n\triangle E=\left(E_n\cap E^\complement\right)\cup\left(E_n^\complement\cap E\right)$, and clearly, $\left(E_n\cap E^\complement\right)\cap\left(E_n^\complement\cap E\right)=\varnothing.$ From the latter it follows that $\mu\left(E_n^\complement\cap E\right)\to0$ and also $\mu\left(E_n\cap E^\complement\right)\to0$. Now the properties of the Lebesgue integral yield \begin{align} \left\vert\int_{E_n}f_n\,d\mu-\int_E f\,d\mu\right\vert&= \left\vert\int_{E_n\cap E}f_n\,d\mu+\int_{E_n\cap E^\complement}f_n\,d\mu-\int_{E_n^\complement\cap E}f\,d\mu-\int_{E_n\cap E}f\,d\mu\right\vert\\ &\leq2\int_\Omega\vert f_n-f\vert\,d\mu+\int_{E_n\cap E^\complement}\vert f\vert\,d\mu+\int_{E_n^\complement\cap E}\vert f\vert\,d\mu.\\ \end{align} By hypothesis, we have that $$\lim_{n\to\infty}\int_\Omega\vert f_n-f\vert\,d\mu=0.$$ Since $f$ is integrable, $\mu\left(E_n^\complement \cap E\right)\to0$ and $\mu\left(E_n\cap E^\complement\right)\to0$, the absolute continuity of the Lebesgue integral implies that given $\varepsilon>0$ it follows that there is some $N\in\mathbb N$ such that for all $n>N$ we have $$\int_{E_n\cap E^\complement}\vert f\vert\,d\mu+\int_{E_n^\complement\cap E}\vert f\vert\,d\mu<\varepsilon.$$ The result follows.
Do you agree with my proof above? Any comments are most welcomed and appreciated.
Thank you for your time.
Note: If the title seems misleading, I apologize. If anyone has a better title, feel free to change mine.