Let $p\in [1,\infty)$. A simple consequence of the dominated convergence theorem is that
If $f_n\to f$ a.e, and there exists a function $g\in L^p$ such that $|f_n|\leq g$ a.e. for all $n$, then $f_n\to f$ in $L^p$.
It is also well known that if $f_n\to f$ in $L^p$ then up to a subsequence, $f_n\to f$ a.e. This may be considered as a partial inverse of the above theorem.
My question is if the dominated convergence theorem is fully invertible. That is, does the following hold (and if so, why?):
We have $f_n\to f$ in $L^p$ if and only if any subsequence of $\left\{f_n\right\}$ (which I will not relabel) has a subsequence $\left\{f_{n_k}\right\}$ such that $f_{n_k} \to f$ a.e. and there exists a function $g\in L^p$ such that $|f_{n_k}|\leq g$ a.e. for all $k$.
More specifically, the missing part is showing that if $f_n\to f$ in $L^p$ then up to a subsequence there exists a dominating function.
I found a counterexample when we do not allow taking a subsequence, namely $$f_n(x)=n\chi_{[\frac{1}{n},\frac{1}{n+1}]}(x),\qquad x\in (0,1)$$ Here $f_n\to 0$ in $L^1$ and a.e., but if $|f_n|\leq g$ a.e., necessarily $g\notin L^1$. However, if we allow taking subsequences, this is no longer a counterexample.
If you are allowed to take subsequences, then you can indeed find a dominating function.
First, we make the assumption $f=0$. Then we consider a subsequence of $f_n$ such that $$ \| f_{n_k} \|_{L^p} \leq 2^{-k}. $$
Then you can define $$ g := \sum_{k=1}^\infty | f_{n_k} | $$ and it can be seen that $g\in L^p$ and $g$ is a dominating function of $f_{n_k}$ for all $k$.
If $f\neq 0$, you can (with the method above) find a dominating function $g$ of $ f_{n_k}-f$ by the methods above. Then, $g':= |f|+g$ is a dominating function for all $f_{n_k}$.