Question: Suppose $\{f_n\}$ converges to $f$ in $L^p(\mathbb{R})$, $1\leq p<\infty$. Prove that there is a subsequence $\{f_{n_k}\}$ and $g\in L^p(\mathbb{R})$ so that $f_{n_k}\to f$ a.e. and $|f_{n_k}|\leq g$ a.e.
I believe this is similar to part of the proof of Riesz-Fischer theorem, with some differences like the $f$ is given in this question, where in Riesz-Fischer theorem, one can define their own $f$. My attempt is below, any critique and corrections will be welcome!
Sincere thanks for any help, as I do not know my proof is correct or not, or whether there is a simpler proof.
I do note that my question is similar to this question (Convergent sequence in Lp has a subsequence bounded by another Lp function) with the difference that my question also requires proving $f_{n_k}\to f$ a.e.
My attempt:
In probability theory, we have to deal with different types of convergence, and therefore - in order to avoid any confusion - it is always good to mention which kind of convergence you are talking about, e.g. "Since $f_n$ converges in $L^p(\mathbb{R}^n)$, $\{f_n\}$ is an $L^p$-Cauchy sequence."
It depends on the definition of a "subsequence", but usually it is assumed that the order of the elements is preserved, i.e. $n_1 \leq n_2 \leq \ldots$. Therefore, I suggest adding a remark that we may assume without loss of generality that $n_k = N(2^{-k}) \leq N(2^{-(k+1)} = n_{k+1}$ for all $k$.
What is the set $E$ in the definition of $h$...? (I suppose that $E^c=\{g=\infty\}$ is the exceptional set, but I don't find the definition of $E$ in your proof.)
There is no need to apply the dominated convergence theorem to conclude $h \in L^p$. It follows from the pointwise convergence and the estimate $|g_k| \leq |g|$ that
$$|h| = \sup_{k \in \mathbb{N}} |g_k| \leq |g| \in L^p$$
and so $h \in L^p$.