# Dilations of Integrable Function Converge to Zero Almost Everywhere

Suppose $f:[0,\infty)\rightarrow [0,\infty)$ is integrable. Set $f_{n}(x):=f(nx)$. I want to show that $f_{n}(x)\rightarrow 0$ almost everywhere or equivalently, the set $$\{x : \limsup_{n}f_{n}(x)\geq\delta\}$$ has measure zero, for any $\delta>0$.

By dilation invariance, it's clear that $f_{n}\rightarrow 0$ in $L^{1}$ and therefore also in measure. Furthermore, we can pass to a subsequence to obtain a.e. convergence. If $f$ has compact support, then it's obvious that $f_{n}\rightarrow 0$ almost everywhere.

My thought was to try approximating $f$ in $L^{1}$ by $g\in C_{c}(\mathbb{R})$ and use something like

$$|\{\limsup f_{n}\geq\delta\}|\leq|\{\limsup|f_{n}-g_{n}|\geq\delta/2\}|+|\{\limsup|g_{n}|\geq\delta/2\}|$$

and go from there. But I'm not sure how to control the first term on the RHS. Any suggestions?

Fix $a,b$ with $0<a<b$. It's enough to show that $$\int_a^b\sum_nf_n<\infty.$$ But $\int_a^bf_n=\frac1n\int_{na}^{nb}f$, so $$\int_a^b\sum_nf_n=\int_a^\infty\phi f,$$where $$\phi=\sum\frac1n\chi_{(na,nb)}.$$So it's enough to show that $\phi$ is bounded on $(a,\infty)$.
And $\phi$ is certainly bounded on $(a,R)$ for any $R<\infty$, so we need only get a bound on $\phi(x)$ for large $x$. Now, $$\phi(x)=\sum_{n\in\left(x/b,x/a\right)}\frac1n.$$At least for large $x$ it seems clear that this is bounded by something like $$O(1)+\int_{x/b}^{x/a}\frac{dt}{t}=O(1)+\log(b/a)=O(1).$$