Let $E\subset \mathcal{M}(\mathbb{R}^n)$ with $m(E)>0$, $\{f_j\}_{j\in \mathbb{N}}\subset \mathcal{L}^p(E)$ and $f\in \mathcal{L}^p(E).$ Let $1\leq p < + \infty$ and suppose that $f_j\to f$ almost everywhere on E and $||f_j||_p\to ||f||_p$. I want to prove the standard fact that $||f_j-f||_p\to 0.$
I know I have the inequality
$$|f-f_n|^p \leq (2\max(|f|,|f_n|))^p = 2^p \max(|f|^p,|f_n|^p) \leq 2^p (|f|^p + |f_n|^p)$$for any $p>0.$
So, supposing the sequence $\{f_n\}$ is dominated by some $g$, we have the bound:
$$\{h_n=|f-f_n|^p\}\leq 2^{p+1}g^p$$
Since $h_n \to 0$ almost everywhere, applying dominated convergence theorem we get:
$$\text{lim}||f-f_n||_p^p=\text{lim}\int_E|f-f_n|^p=0$$
hence $$\text{lim}||f-f_n||_p=\left(\text{lim}||f-f_n||_p^p\right)^{1/p}=0.$$
Is it correct until now?
Supposing it is correct, one is left to prove that $\{f_n\}$ is indeed bounded.
How to show that $f_j\to f$ almost everywhere on E and $||f_j||_p\to ||f||_p$ imply that ${f_n}$ is dominated so that one can apply dominated convergence?
Note that there already a few questions on this site that address in some way the statement I want to prove. But I would like to know if my own reasoning is correct and moreover none of them addresses my specific questions.