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Let $h$ be a bounded Lebesgue measurable function on $\mathbb{R}$ such that for any finite measure Lebesgue measurable subset $E$ of $\mathbb{R}$,

$$\lim_{n \rightarrow \infty}\int_{E}h(nx)dx = 0$$

My question is can we extend this to general measurable subsets of $\mathbb{R}$ in the following way: the Lebesgue measure on $\mathbb{R}$ is $\sigma$-finite. If $E$ is an arbitrary measurable subset of $\mathbb{R}$, then $E \cap [-t, t]$ is Lebesgue measurable and of finite measure, and this sequence of sets for $t \in \mathbb{N}^+$ increases to $E$, so

$$\lim_{n \rightarrow \infty}\int_{E}h(nx)dx = \lim_{n \rightarrow \infty}\lim_{t \rightarrow \infty}\int_{E \cap [-t, t]}h(nx)dx = \lim_{t \rightarrow \infty}\lim_{n \rightarrow \infty}\int_{E \cap [-t, t]}h(nx)dx = 0$$

I am unsure if the interchange of limits is justified, and if so, why it is appropriate. Comments and explanations welcome.

Joshua Siktar
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2 Answers2

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Take $h=\sum \frac 1 n I_{(n,n+1)}$ for a counter-example. The hypothesis is satisfied by DCT (with dominating function $I_E$) but $h$ is not integrable.

Kavi Rama Murthy
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The answer is no in general:

  1. Consider $h(x)=\sin(x)$. Being a bounded $2\pi$-periodic measurable function, it satisfies $$\int_\mathbb{R}f(x)h(nx)\,dx\xrightarrow{n\rightarrow\infty}\Big(\frac{1}{2\pi}\int^{2\pi}_0h\Big)\int f =0$$ for all $f\in\mathcal{L}_1$ (see here for example). For $E=\mathbb{R}$ or $E=[0,\infty)$, $\lim_n\int_E h(nx)\,dx$ is not defined; in fact $\int_Eh(nx)\,dx$ is not defined for any $n$.

  2. Consider the case $h(x)=\frac{|\sin x|}{|x|}$ ($h(0):=1$ ). As $|h|\leq1$, and $h(x)\xrightarrow{|x|\rightarrow\infty}0$, $$ \int_{\mathbb{R}}f(x)h(nx)\,dx\xrightarrow{n\rightarrow0}0 $$ for any $f\in \mathcal{L}_1$ by dominated convergence. for $E=\mathbb{R}$ or $E=[0,\infty)$, $\int_E h(nx)\,dx=\infty$.

  • In (1) neither $h_+$ nor $h_-$ is integrable, and so $h\not\in \mathcal{L}_1$.

  • In (2) $h\geq0$, but $h\notin \mathcal{L}_1$.

If (a) $h$ is bounded, (b) $h\in L_1$ and (c) $\lim_{|x|\rightarrow0}h(x)=0$, then by dominated convergence $$\int_\mathbb{R}f(x)h(nx)\,dx=0$$ for all $f\in\mathcal{L}_1$

For any $E$ measurable set (with or without finite measure), $$ \Big|\int_E h(nx)\,dx\Big| =\frac{1}{n}\Big|\int\mathbb{1}_{E}(x/n)h(x)\,dx\Big|\leq\frac{1}{n}\int|h(x)|\,dx\xrightarrow{n\rightarrow\infty}0 $$ This shows that conditions (a)-(c) on $h$ are sufficient for your statement to hold.

Mittens
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