Integral over a small set is small and dimension independent

Question

Let $\mu_n$ be the standard normal distribution over $\mathbb{R}^n$. Is the following true?

For all $\epsilon > 0$ and $d \in \mathbb{N}$, there exists $\eta > 0$ such that for all $n \geq 1$ and $T \in \mathcal{B}_{\mathbb{R}^n}$ with $\mu_n(T) \leq \eta$, the following holds. For all polynomials $f:\mathbb{R}^n \to \mathbb{R}$ of degree at most $d$, $\int_{T} |f|d\mu_n \leq \epsilon \int |f| d\mu_n$

If so, what is a sufficient condition on the class of distributions $\{\mu_n\}_{n \geq 1}$ for this statement to hold?

In this similar question, $\eta$ is allowed to depend on $f$.

Answer 1

Here is a solution for fixed $n$.

Fix $d$. What you want is equivalent to the following: for every $\epsilon >0$, there exists $\eta >0$ such that for every $T \in \mathcal{B}(\mathbb{R}^n)$ with $\mu_n(T)\leq \eta$, we have $\sup_{f\in \Phi} \int_T |f|d\mu_n\leq \epsilon$ where I set $$\Phi = \left\{ f \in \mathbb{R}_d[X],\,\int|f| d\mu_n\leq 1\right\}.$$ This property is sometimes called "uniform continuity" of the family $\Phi$. Since $\Phi$ is bounded in $L^1(\mu_n)$, this is equivalent to the uniform integrability of $\Phi$. By the Dunford-Pettis theorem, this is equivalent to the family $\Phi$ being relatively compact for the weak topology of $L^1(\mu_n)$. Now observe that $\Phi$ is the closed unit ball of the finite-dimensional space $\mathbb{R}_d[X]$ (equipped with the $L^1(\mu_n)$ norm). As such, $\Phi$ is in fact strongly compact for said norm and in particular it is weakly compact. This proves the result.

---

This proof is non constructive and possibly uses results which are too strong. Notice however that we did not need any particular properties related to the Gaussian measure $\mu_n$ so it should work for any probability measure $\mu$ with respect to which all polynomials are integrable.