$L^1$ convergence implies pointwise a.e. uniform bound of a subsequence

Question

Is the following statement true ?

Let $f_n$ be a sequence of non-negative functions in $L^1(X,m)$, where $(X,\Sigma,\mu)$ is a probability space, such that $\|f_n - f\|_{L^1(X)}\to 0 $ for some $f \in L^1(X)$. Then there exists a subsequence $f_{n_k}$ and a function $g \in L^1(X,\mu)$ such that $$f_{n_k}(x) \le g(x) \quad \text{for $\mu$-a.e } x \in X .$$

I know that there exists a subsequence that converges almost everywhere, but I don't really know how to use this information. Any suggestions?

Answer 1

This is correct even if the functions are not necessarily non-negative. Choose $n_k$ such that $\|f_{n_k} - f \| \le 2^{-k}$ and set $$ g = |f| + \sum_{k \ge 0} |f_{n_k}-f| \, $$

The sum converges in $L^1$ by the choice of the subsequence, and clearly $f_{n_k} \le g$ for all $k$, since all terms are non-negative.