$f\in L^1\implies \lim_{n\rightarrow\infty}f(n^2 x)=0$ a.e. $x\in\mathbb{R}$? Can someone provide a hint for the following:
$f\in L^1(\mathbb{R})\implies \lim_{n\rightarrow\infty}f(n^2 x)=0$ a.e. $x\in\mathbb{R}$
I can't really make heads or tails of it. Something makes me think convergence in measure; or even a duality argument, but the quantification on $x$ for which this is satisfied is throwing me. I am not sure how it arises.
This is from an old qualifying exam.
 A: Fix an $m\in\mathbb{Z}$, we have
\begin{align*}
\sum_{n\in\mathbb{N}}\int_{m}^{m+1}|f(n^{2}x)|dx&=\sum_{n\in\mathbb{N}}\int_{n^{2}m}^{n^{2}(m+1)}\dfrac{1}{n^{2}}|f(y)|dy\\
&\leq\sum_{n\in\mathbb{N}}\dfrac{1}{n^{2}}\cdot\int_{\mathbb{R}}|f(y)|dy\\
&<\infty,
\end{align*}
so
\begin{align*}
\int_{m}^{m+1}\sum_{n\in\mathbb{N}}|f(n^{2}x)|dx<\infty,
\end{align*}
this implies
\begin{align*}
\sum_{n\in\mathbb{N}}|f(n^{2}x)|<\infty
\end{align*}
for a.e. $x\in[m,m+1]$, which entails $\lim_{n\rightarrow\infty}f(n^{2}x)=0$.
Now varying $m\in\mathbb{Z}$ to obtain the result.
A: Here's an outline of the proof. First begin by showing that for $f \in L^1(\mathbb{R})$ it holds that $\int_\mathbb{R} f(tx)\;dm(x) = \frac{1}{t}\int_\mathbb{R}f(x)\;dm(x)$. There is a typical strategy to prove such results which I am going to present briefly:

*

*Show it when $f$ is the characteristic function of some measurable set.

*Using 1. show it for simple functions.

*Show it for positive functions by approximating $f$ with simple functions.

*Use the fact that $f = f^+ - f^-$ and the linearity of the integral to prove the general case.

Having proven that use Beppo-Levi theorem to see that:
$$\int_\mathbb{R} \sum_{n=1}^{\infty} |f(n^2x)|\;dm(x) = \sum_{n=1}^{\infty}\int_\mathbb{R}|f(n^2x)|\;dm(x) = \sum_{n=1}^{\infty}\frac{1}{n^2}\cdot||f||_1 < \infty. $$
This implies that $\sum_{n=1}^{\infty}|f(n^2x)|$ converges almost everywhere and this concludes the proof.
