The concentration of measure on $ [0, 1]^n $ equipped with uniform probability measure $\mu_{\infty}$, states that for any $A \subset [0, 1]^n $ with $ \mu_{\infty}(A) \geq \frac{1}{2} $, we have: $$ 1 - \mu_{\infty}(A_{\epsilon}) \leq e^{- \pi \epsilon^2 }, \epsilon > 0, $$ where $A_{\epsilon} = \{ x \in [0, 1]^n; d(x, A) < \epsilon \}$, and $d = L_2$ is the Euclidean distance.

I wonder how the assumption $ \mu_{\infty}(A) \geq \frac{1}{2} $ affects the statement. Do we have a sharper inequality for $\mu_{\infty}(A) = 0.9$? Do we have a weaker inequality for $\mu_{\infty}(A) = 0.05$?

More precisely, the concentration of measure on $L^p$ balls has the following properties. Let $B^n_{p}$ denote a $p$ ball with $\mu_p$ normalized uniform measure, where $p \geq 2$. For any $A \subset B^n_{p} $, with $ \mu_p(A) > 0 $, we have:

$$ 1 - \mu_p(A_{\epsilon}) \leq \frac{1}{\mu_p(A)} e^{-2n C\epsilon^{p} } $$ where $\mu_p(A)$ directly affects the concentratioon bound, $A_{\epsilon} = \{ x \in B^n_p; d(x, A) < \epsilon \}$, and $d = L_p$ is the $p$ norm.

The above question may be too elementary. Since I got no response from https://math.stackexchange.com/questions/2637835/on-the-1-2-assumption-on-concentration-of-measure-on-continuous-cube It may not be a bad idea to ask here: it may be obvious to experts in the field, but not necessarily trivial for outsiders.