Lower-bound for $\underset{p \le \gamma_d(A) \le q}{\inf} \gamma(A^\epsilon)$, where $\gamma_d$ is the standard gaussian distribution on $\mathbb R^d$ Let $\gamma_d = \gamma_1 \otimes \ldots \otimes \gamma_1$ be the standard Gaussian distribution on $\mathbb R^d$, where $d$ is a large positive integer. Given $\epsilon  \ge 0$ and a measurable $A \subseteq \mathbb R^d$, let $A^\epsilon := \{x \in \mathbb R^d \mid \mbox{dist}(x,A) \le \epsilon\}$ be its epsilon-neighborhood, where $\mbox{dist}(x,A) := \inf_{a \in A} \|x-a\|$ is the distance of between $x$ and the closest point in $A$. Finally, let $0 < p \le q \le 1$.

Question. What is a good lower-bound for $r_d(\epsilon,p,q) := \underset{p \le \gamma_d(A) \le q}{\inf} \gamma(A^\epsilon)$ ?

Note. Understanding the special case $p \to 0^+$ would already be interesting. And of course, I'm perfectly fine with a bound which looks at different regimes of $\epsilon$, $p$, and $q$.
 A: The answer is $\inf_{p \leq \gamma_{d}(A) \leq q} \gamma(A^{\varepsilon}) = \Phi(\Phi^{-1}(p)+\varepsilon)$ where $\Phi(x) = \int_{-\infty}^{x} \frac{e^{-s^{2}/2}}{\sqrt{2\pi}}ds$.
Indeed, all you need is the following claim:
For any measurable $A \subset \mathbb{R}^{d}$ and any $\varepsilon>0$ we have $\gamma_{d}(A^{\varepsilon})\geq \Phi(\Phi^{-1}(\gamma_{d}(A))+\varepsilon)$.  The equality is attained if $A$ is any hafspace $H$ with $\gamma_{d}(A)=\gamma_{d}(H)$.
Since both $\Phi$ and $\Phi^{-1}$ are increasing the answer to your questions follows from the claim.
The claim  follows from the  Ehrhard inequality: for any measurable sets $U, V \subset \mathbb{R}^{d}$ and any $\alpha, \beta \geq 0$ such that $\alpha+\beta \geq 0$, $|\alpha-\beta|\leq 1$,   $\alpha U+\beta V$ is measurable (here "+" is Minkowski sum) we have
$$
\gamma_{d}(\alpha U +\beta V) \geq \Phi(\alpha \Phi^{-1}(\gamma_{d}(U))+\beta\Phi^{-1}(\gamma_{d}(V))).
$$
So, let $B$ be the unit ball, and  $\lambda \in (0,1)$. Then by Ehrhard we have
$$
\gamma_{d}(A^{\varepsilon}) = \gamma_{d}(\lambda [\lambda^{-1}A]+(1-\lambda)[(1-\lambda)^{-1}\varepsilon B]) \geq \Phi(\lambda \Phi^{-1}(\gamma_{d}(\lambda^{-1}A))+(1-\lambda)\Phi^{-1}(\gamma_{d}((1-\lambda)^{-1} \varepsilon B)))
$$
Next, let $\lambda \to 1$, $\lambda<1$. Then  $\lambda \Phi^{-1}(\gamma_{d}(\lambda^{-1}A)) \to \Phi^{-1}(\gamma_{d}(A))$, and also $(1-\lambda)\Phi^{-1}(\gamma_{d}((1-\lambda)^{-1} \varepsilon B)) \to \varepsilon$. The first limit is simple. To verify the second limit we need to show that $\lim_{r \to \infty}\frac{1}{r} \Phi^{-1}(\gamma_{d}(r B))=1$. Perhaps there are many ways to show this.  The most straightforward one will be to honestly compute all asymptotic:
$$
\gamma_{d}(r B) = 1- \frac{\sigma_{d}}{(2\pi)^{d/2}}\int_{r}^{\infty}e^{-r^{2}/2}r^{d-1}dr \stackrel{r \to \infty}{=}1-\frac{\sigma_{d}}{(2\pi)^{d/2}} e^{-r^{2}/2} r^{d-2}+o(e^{-r^{2}/2} r^{d-2}),
$$
where $\sigma_{d}$ is the surface area measure of the unit sphere.  Thus
$$
\Phi^{-1}(s) \stackrel{s \to 1^{-}}{=} (-2\ln(1-s))^{1/2}+o((-\ln(1-s))^{1/2}),
$$
therefore
$$
\lim_{r \to \infty} \frac{1}{r} \Phi^{-1}(\gamma_{d}(r B)) = \lim_{r \to \infty} \frac{1}{r} [-2\ln(r^{d-2}e^{-r^{2}/2})]^{1/2}=1. 
$$
