# A strong version of the Dominated Convergence Theorem

Let $(X, \Sigma, \mu)$ be a measure space, and let $f, f_n:X\rightarrow \mathbb{C}$ be measurable functions with $f_n\rightarrow f$ pointwise. Assume that there are integrable functions $G, g_n:X\rightarrow[0, \infty]$ with finite integrals such that $|f_n|\leq G+g_n$ for every $n$. Also assume that $\int g_n \rightarrow 0$.

We need to show that $\int |f-f_n|\rightarrow 0$.

I've tried using the dominated convergence theorem but couldn't find the dominating function. I have 2 main "problems" with the statement:

1. $\underset{n}{\sup} g_n$ doesn't necessarily have a finite intgeral;
2. $g_n$ doesn't have to converge pointwise to anything. That's what makes this statement different from all the others in this site I've checked.

Note that a sequence of real numbers converges to a limit $L$ iff every subsequence has a subsubsequence that converges to $L$. Since your hypotheses are preserved by passing to subsequences, it suffices to show that $\int|f-f_{n_k}|\to 0$ for some subsequence $(f_{n_k})$.
In particular, since $\int g_n\to 0$, there is a subsequence $(g_{n_k})$ of $(g_n)$ which converges to $0$ pointwise almost everywhere. Write $h_n(x)=\min(f_n(x),g_n(x))$ if $f_n(x)\geq 0$ and $h_n(x)=\max(f_n(x),-g_n(x))$ if $f_n(x)<0$. Define $F_n(x)=f_n(x)-h_n(x)$. Since $|f_n|\leq G+g_n$, $|F_n|\leq G$. Furthermore, on our subsequence we have that $F_{n_k}$ converges to $f$ almost everywhere. By the dominated convergence theorem, $\int |f-F_{n_k}|\to 0$. Since $$\int |f-f_{n_k}|\leq\int|f-F_{n_k}|+\int |h_{n_k}|\leq\int|f-F_{n_k}|+\int g_{n_k},$$ it follows that $\int |f-f_{n_k}|\to 0$.