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?
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