Let $\mu$ be a probability distribution on $\mathbb R^n$. For $\epsilon > 0$ and a Borell set $A \subseteq R^n$ with $\mu(A) > 0$, define the $\epsilon$-neighborhood of $A$ as $A^\epsilon := \{x \in \mathbb R^n \mid dist(x,A) \le \epsilon\}$ , where $dist(x,A):= \inf_{a \in A}\|x-v\|$ is the distance of the point $x$ from the set $A$. Under certain geometric constraints (e.g $\mu$ is log-concave, etc.), one can establish an exponential bound of the form
$$ \mu(A^\epsilon) \ge 1-c_1\exp(-c_2(\epsilon-c_3)^2), \;\forall \epsilon \ge c_3, $$
where $c_1,c_2,c_3 > 0$ are constants independent of the dimension $n$. Thus, the neigborhood $A^\epsilon$ will eventually "fill up" the whole space. Such inequalities date back to the works of M. Talagrand (and perhaps before).
Now, for a unit vector $u \in \mathbb R^n$, define $A_\epsilon(u):= \{a + \epsilon u \mid a \in A\}$, the Minkowski sum of $A$ and the singleton $\{\epsilon u\}$, obtained by translated $A$ in the direction specified by $u$. Define
Question. Can one play a similar game (of course, with analogous geometric constraints on $\mu$) to get lower bounds on $\sup_u\mu(A_\epsilon(u))$ as a function of $\epsilon$ ?