A property of an integrable function on $\mathbb{R}$

Theorem. Let $f \in L^1(\mathbb{R}, \mathrm{d}x)$. Then $$ \lim_{n\to\infty} f(nx) = 0 \quad \text{for a.e. } x. $$

The proof below is essentially what is shown in [1], written in my own words.

Proof. Fix $\varepsilon \gt 0$ and define $E = \{x \in \mathbb{R} : |f(x)| \geq \varepsilon\}$. Then for any $[a, b] \subseteq (0, \infty)$, \begin{align*} \int_{a}^{b} \sum_{n=1}^{\infty} \mathbf{1}_{E}(nx) \, \mathrm{d}x &= \sum_{n=1}^{\infty} \frac{1}{n} \int_{na}^{nb} \mathbf{1}_{E}(t) \, \mathrm{d}t \\ &= \int_{E} \sum_{n=1}^{\infty} \frac{1}{n} \mathbf{1}_{\{na \leq t \leq nb\}} \, \mathrm{d}t \\ &= \int_{E} \sum_{n=1}^{\infty} \frac{1}{n} \mathbf{1}_{\{ t/b \leq n \leq t/a \}} \, \mathrm{d}t \\ &\leq \int_{E} (2 + \log(b/a)) \, \mathrm{d}t \\ &= (2 + \log(b/a)) |E|, \end{align*} where $|E|$ denotes the Lebesgue measure of $E$. Since $f$ is integrable, $|E| \lt \infty$. So the sum $\sum_{n=1}^{\infty} \mathbf{1}_{E}(nx)$ is finite and hence $nx \in E$ for only finitely many $n$ for a.e. $x$ in $[a, b]$. Since $[a, b]$ is arbitrary, the same is true for a.e. $x$ in $\mathbb{R}$. From this, it is now routine to conclude the desired claim. $\square$

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References

• [1] Emmanuel Lesigne. On the behavior at infinity of an integrable function. American Mathematical Monthly, Mathematical Association of America, 2010, 117 (2), pp.175-181. ⟨hal-00276738v3⟩