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 $x \in \mathbb R$?

Question

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 $x \in \mathbb R$?

Answer 1

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. $x$.