The question is as in the title.

Let $\alpha_n(x):=\int_{\lVert x-y\rVert \leq 1/n } \lVert x-y\rVert^2 d\mu(y)$ where $x,y \in \mathbb{R}^m$ and $\mu$ is sufficiently regular Borel proability measure; for concreteness, we can think of the standard normal Gaussian measure.

Also let $A_n:=\int_{\mathbb{R}^m} \alpha_n(x) d\mu(x)$.

Then, for any smooth real-valued bounded function $F(x)$ on $\mathbb{R}^m$, I suspect that \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} holds as $n \to \infty$.

However, I cannot justify this rigorously or know what kind of concept it is. Perhaps it is related to ergodicity?

Could anyone please provide any information?

  • $\begingroup$ It feels true. But does not seems easy to justify. $\endgroup$
    – Kroki
    May 18, 2023 at 5:33

1 Answer 1


I think you should consider $$ \frac{1}{A_n} \int_{\mathbb R^m} F(x) \alpha_n(x) dx $$ instead of the integral with respect to $d\mu(x)$. The intuitive reason for this is that you are integrating with respect to $\mu$ twice on the LHS of the convergence (once in $\alpha_n$, once in the actual LHS), but only once on the RHS.

I will nevertheless try to adress both the question as posted, and the question with my suggested edit. In what follows, $\mu$ is any Borel probability measure $\mu$ (no further regularity required).

The original question. Using $[A]$ for the indicator function of the proposition $A$ that is $1$ if and only if $A$ is true and $0$ otherwise, $$ \int F(x) \alpha_n(x) d\mu(x) = \int \int F(x) \|x-y\|^2 [\|x-y\| \leq n^{-1}] d\mu(x) d\mu(y) . $$ Similarly, $A_n$ is the same expression with $F \equiv 1$. Assuming that $\mu$ has a smooth density $p$ with respect to $dx$ and writing $\omega_m = \int_{\|x\|\leq 1} \|x\|^2 dx$, asymptotically $$ \int F(x) \alpha_n(x) d\mu(x) \sim n^{-2-m} \omega_m \int F(x) p(x)^2 dx $$ so that $$ \frac{1}{A_n} \int F(x) \alpha_n(x) d\mu(x) \to \int F(x) \frac{p^2(x)}{\int p^2} dx , $$ which is not the convergence you are looking for.

The modified question. We now integrate the LHS with respect to $dx$ and not $d\mu(x)$. Define $$ \rho(x) = \begin{cases} \frac{\|x\|^2}{\int_{\|x\|\leq 1} \|x\|^2 dx} &\text{ if } \|x\|\leq 1, \\ 0 &\text{ otherwise.} \end{cases} $$ Note $\int \rho = 1$. Then $$ \alpha_n(x) = \left( \int_{\|x\|\leq 1} \|x\|^2 dx\right) \int n^2 \rho(n(x-y)) d\mu(y) = \left( \int_{\|x\|\leq 1} \|x\|^2 dx\right) n^{2-m} \rho_n * \mu(x) , $$ where $\rho_n(t) = n^m \rho(nt)$ is such that $\int \rho_n = 1$. It is pretty standard that $$ (\rho_n \star \mu)(x) dx \to \mu $$ in the weak topology as $n\to\infty$, i.e. for every continuous and bounded $F$ (again, no further regularity required) $$ \int (\rho_n \star \mu)(x) F(x) dx \to \int F(x) d\mu(x) . $$ To use your notation with the modified $ A_n = \int \alpha_n(x) dx $ (instead of an integral against $d\mu(x)$): $$ \frac{1}{A_n} \int \alpha_n(x) F(x) dx \to \int F(x) d\mu(x) . $$

Proof of the "pretty standard" convergence. Assume first that $F$ is continuous with compact support. We use that $\rho(-x) = \rho(x)$: \begin{align} \int (\rho_n * \mu)(x) F(x) dx &= \int \int \rho_n(x-t) F(x) dx d\mu(t) = \int (\rho_n * F)(t) d\mu(t) . \end{align} The exchange of the integrals is okay because the function $(t,x) \mapsto \rho_n(x-t) F(x)$ is integrable under $dxd\mu(t)$, thus Fubini’s theorem applies. Then use convergence of convolutions and approximation of unity to prove that $\rho_n*F \to F$ uniformly enough for the convergence to hold. This proves the convergence $\rho_n * \mu(x) dx \to \mu$ in the vague topology, which is equivalent to the convergence in the weak topology, see Corollary 10.5 in Relation between vague convergence and weak convergence. (The results I link also hold on $\mathbb R^m$ and not only $\mathbb R$.)


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