Interchanging Limits: Finite Measure Subsets of Real Line 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.
 A: 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.
A: The answer is no in general:

*

*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 $f\in\mathcal{L}_1$ (https://math.stackexchange.com/a/3741777/121671).
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 $n$.


*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.
