Let $(X,\mathcal{M},\mu)$ be a $\sigma$-finite measure space.
Let $f_1, f_2, ... : X \to \mathbb{R}$ be measurable functions such that $n^{2}\|f_n\|_2\le1$ for every $n\ge1$.
Does it imply that $f_n\to0$ a.e. where $n\to\infty$ ?
Since $f_n\to0$ in $L^2$ we know that there is a subsequence $\{f_{n_k}\}$ such that $f_{n_k}\to0$ a.e. Can't see how the fact that the measure is $\sigma$-finite is helpful...
So, I tried to disprove it using the "typewriter sequence":
$X=[0,1]$, $\mu=m$ Lebesgue measure. $f_n=\chi_{[j/2^{k},(j+1)/2^{k}]}$ where $n=2^{k}+ j$ with $0\le j<2^{k}$ . So $f_n\to0$ in $L^2$ and $\{f_n\}$ does not converge to $0$ for any point. But $k\approx\log_2{n}$ and $n^{2}\|f_n\|_2>1$ for every $n>1$.
I can choose a subsequence $\{f_{n_k}\}$ such that ${n_{k}}^{2}\|f_{n_k}\|_2\le1$ , but then this subsequence might converge to $0$. Maybe some another variation of this sequence would work?
Thanks!
The point here is that $\sum_{n\geq 1} \frac{1}{n^2}$ converges: Set $s_k:=\sum_{n=1}^k |f_n|$. By the triangle inequality we have $$\sum_{n=1}^\infty \| f_n\|_2 \leq \sum_{n= 1}^\infty \frac{1}{n^2}<\infty.$$ Since $L^2$ is complete, we have that $s_k$ converges in norm to some element $g\in L^2$. In particular, a subsequence $s_{k_j}$ converges pointwise to $g$, but since $s_k(x)\leq s_{k+1}(x)$ for every $x$ and $k\in \mathbb{N}$, we must have that the full sequence $s_k$ converges pointwise a.e. In particular, this implies that $f_n(x)\to 0$ at every $x$ for which $s_k(x)\to g(x)$ and $g(x)<\infty$.