This is part of an old preliminary exam in Analysis I am reviewing to prepare for my own prelim. $\lambda$ is the Lebesgue measure. $f_n\to f$ with respect to the $L^1-$norm.
I know that there exists at most a subsequence fulfilling the condition, since, e.g., we could have $f=0$ and $f_n=0$ except at $q_n$, with $q_n$ cycling through the rational numbers (with each rational number appearing infinitely often). Then $f_n\to f$ but it is not the case that $f_n(x)\to f(x)$ for almost every $x\in[0,1]$. It's obvious in this example that there will exist a subsequence fulfilling the condition, and I feel confident that there will always be such a subsequence. That is, I think the answer is to "prove", not "disprove". But I'm having difficulty seeing how to prove it.
This is a case where I feel like I understand the problem but can't quite figure out how to structure a proof. Every attempt I've made has become overly convoluted and I've become confused. Any help would be appreciated.
This is a fairly standard result, here is a proof that uses the Borel Cantelli lemma.
A slightly more general result is true, and the more general result hints at a direction of proof.
It is not difficult to show that if $f_n\to f$ (in $L^1$) then $f_n$ converges to $f$ in probability, that is for all $\epsilon>0$, $\lim_n \lambda \{ x | |f_n(x)-f(x)| \ge \epsilon \} = 0$.
For each $k$, choose $n_k$ such that with $E_k = \{ x | |f_{n_k}(x)-f(x)| \ge {1 \over k} \}$, we have $\lambda E_k < {1 \over 2^k}$.
Let $\phi = \sum_k 1_{E_k}$, and note that $\phi$ is integrable and hence $\phi(x)$ is finite for ae. [$\lambda$] $x$ (Borel Cantelli lemma). Hence for ae. $x$, there is some $N$ such that $x \notin E_k$ for all $k \ge N$, or in other words, $|f_{n_k}(x)-f(x)| < {1 \over k}$ for all $k \ge N$. In particular, for ae. $x$ we have $f_{n_k}(x) \to f(x)$.