This question is an extension of the one I posted on ME: <https://math.stackexchange.com/questions/4701500/if-alpha-nx-int-lvert-x-y-rvert-leq-1-n-lvert-x-y-rvert2-d-muy>


It might be elementary for here, but I would deeply appreciate any help.


Let for each $n \in \mathbb{N}$, let $r\_n : \mathbb{R}^m \to (0,\infty)$ be a sequence of smooth functions that converges to zero "pointwise" as $n \to \infty$.


Also let $\mu$ be a sufficiently regular Borel probability measure on $\mathbb{R}^m$. For concreteness, we can think of the stadard normal Gaussian measure.


Now, define
\begin{equation}
\alpha\_n(x):=\int\_{\lVert x-y \lVert \leq r\_n(x)} \lVert x-y \rVert^2 d\mu(y)
\end{equation}
for each $n$ and let $A\_n := \int\_{\mathbb{R}^m} \alpha\_n(x) d\mu(x)$.


Then, I suspect that for any bounded real-valued smooth function $F$ on $\mathbb{R}^m$, we have
\begin{equation}
\frac{1}{A\_n}\int\_{\mathbb{R}^m} F(x) \alpha\_n(x) d\mu(x) \to \int\_{\mathbb{R}^m} F(x) d\mu(x)
\end{equation}
as $n \to \infty$.


However, I cannot find a way to justify this guess rigorously. I tried to use convolution theorems but they do not fit in the above formula. Perhaps it is related to ergodicity?


Could anyone please help me?



A generalized form of the approximation to identity?
$\newcommand\al\alpha\newcommand\be\beta\newcommand\R{\mathbb R}$This is not true in general. E.g., suppose that $m=1$, $\mu=N(0,1)$, and $r\_n(x)=r:=1/n$ for all $n$ and $x$. Let $f$ be the standard normal pdf.


Then for each real $x$ (and $n\to\infty$) we have
$$\al\_n(x)\sim\int\_{x-r}^{x+r}(y-x)^2\,dy\,f(x)=\frac{2r^3}3\,f(x).$$
Next, letting $X$, $Y$, and $Z$ denote independent standard normal random variables, we get
$$A\_n=\int\al\_n f=E(X-Y)^2\,1(|X-Y|\le r) 
=2EZ^2\,1(|Z|\le r/\sqrt2) \\ 
\sim 2f(0)\int\_{-r/\sqrt2}^{r/\sqrt2}dz\,z^2=f(0)\frac{r^3\sqrt2}3.$$
So, for $F:=f$ and
$$L\_n:=\frac1{A\_n}\int F\al\_n \,d\mu$$
we have
$$\liminf\_n L\_n\ge\int f\frac{\sqrt2}{f(0)}\,f \,f=L:=\frac{2}{\sqrt{3 \pi }},$$
whereas
$$R:=\int F\,d\mu=\int f^2=\frac{1}{2 \sqrt{\pi }}<L.$$
So, $L\_n\not\to R$. $\quad\Box$


