Let $(X, \mathcal{A}, \mu)$ be a measurable space, $(f_n)$ a sequence of integrable functions that converges in measure to some integrable function $f$. Assume further that there is some $g : X \rightarrow [0,\infty]$ integrable such that $|f_n|\leq g$ and $|f|\leq g$. Then $(f_n) \rightarrow f$ in mean.
I am not sure how to prove this. Since $(f_n)\rightarrow f$ in measure, every subsequence of $(f_n)$ admits some subsequence that converges to $f$ almost everywhere. I can prove that such a subsequence converges in measure, since convergence a.e plus being dominated by an integrable function as above implies convergence in mean.
If $(f_n)\rightarrow f$ a.e, then $(f_n-f)\rightarrow f$ a.e. Also, $$|f_n - f|\leq |f_n| + |f| \leq 2g$$ and therefore appealing to the Dominated Convergence Theorem we obtain that $\int_X |f_n -f|d\mu \rightarrow 0$ so $(f_n)\rightarrow f$ in mean.
The problem is then in showing that if a subsequence of a sequence $(f_n)$ converges in mean, then $(f_n) \rightarrow f$ in mean, and this is the part I need some help with. Thank you!