Let $h$ be a bounded, measurable function, such that, for any interval $I$ $$\bigg|\int_I h\bigg|\leq |I|^{1/2}.$$
I want show that, for any $A$, with $|A|<\infty$, $$\int_A h_\varepsilon(x)dx\rightarrow 0,\ \ \mathrm{as}\ \varepsilon \rightarrow 0,$$
where $h_\varepsilon(x) = h(\frac x\varepsilon)$.
My idea was to take $\eta>0$, and a open set $O$ such that $|O-A|<\eta$. How $O$ is open in $\mathbb{R}$, there exist disjoint intervals, $\{I_k\}$, such that $O = \cup I_k$. Then
$$\bigg| \int_{O-A} h_\varepsilon(x)dx \bigg|\leq \|h\|_{\infty}\eta$$ and $$\bigg| \int_{O} h_\varepsilon(x)dx \bigg| \leq \sum_k\bigg| \int_{I_k} h_\varepsilon(x)dx \bigg| \leq \varepsilon^{1/2} \sum_k|I_k|^{1/2}.$$
But, for instance, if $|I_k|=1/n^2$, then $|O| = \sum_k|I_k|=\pi/6$, however $\sum_k|I_k|^{1/2}=\infty$.
How can I solve this problem?
Thank you!
As $h$ is bounded and measurable, we can find simple functions $\{h_n\}$ such that $\sup_{x\in\Bbb R}|h(x)-h_n(x)|\leqslant 1/n$.
Write $h_n(x):=\sum_{j=1}^{N_n}a_{n,j}\chi_{B_{n,j}}$. We have $$\left|\int_Ah_\varepsilon(x)dx\right|\leqslant \frac 1n+\sum_{j=1}^{N_n}|a_{n,j}|\int_{\Bbb R}\chi_{B_{n,k}}(x\varepsilon^{-1})\chi_A(x)dx.$$ Use the substitution $t=x\varepsilon^{—1}$, take the $\limsup_{\varepsilon \to 0}$ to get what we want.
Note that there are problems when $f$ is not assumed bounded or integrable.