# Does $f_n \to 0$ in $L^1(\mathbb R^2)$ imply that $f_{n_k}(x,\cdot)\to 0$ in $L^1(\mathbb R)$ for almost every $x \in \mathbb R$?

I would like to know what you think about this question. It is a "self-posed" question: I formulated it while I was doing an exercise.

Suppose you have $(f_n)_{n\ \in \mathbb N}\subset L^1(\mathbb R^2)$ such that $f_n \to 0$ in $L^1(\mathbb R^2)$.

Is it true that there exists a subsequence $f_{n_k}$ such that $f_{n_k}(x,\cdot)\to 0$ in $L^1(\mathbb R)$ for almost every $x \in \mathbb R$?

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I like it very much... find it interesting and appealing, moreover it reflects the fact that you understand the subject.. +1 – uforoboa Aug 17 '12 at 13:47
@uforoboa Glad you liked it. Thanks a lot! – Romeo Aug 20 '12 at 9:27

The answer is yes. Let $g_n(x) = \int_{\mathbb R}|f_n(x,y)|\,dy$. Then your condition is equivalent to saying that $g_n \to 0$ in $L^1(\mathbb R)$. It is a standard fact from measure theory that if a sequence of functions on ${\mathbb R}$ converges in $L^p$, then it has a subsequence that converges pointwise a.e. to the same limit. So we can extract a subsequence $g_{n_k}(x)$ that converges pointwise a.e. to 0, which is the same as saying $f_{n_k}(x,\cdot) \to 0$ in $L^1({\mathbb R})$ for a.e. $x$.

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The current answer should be correct. – Zarrax Aug 17 '12 at 17:31
I will read it in a minute; On another note, I think that if you post an answer, get upvotes, and then realize that it was mistaken and instead come up with a correct one, you should delete it and post the correct answer separately (so that the new answer does not get upvotes from previous, incorrect one). – tomasz Aug 17 '12 at 17:34