On the 1/2 assumption on concentration of measure for continuous cube 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. 
 A: By the Tsirel’son--Ibragimov--Sudakov argument, reviewed on the first page in Bobkov, 
pushing the measure forward from the cube to the canonical Gaussian on $\mathbb R^n$ and using the Gaussian isoperimetric inequality, 
we have 
\begin{equation}
  1 - \mu_{\infty}(A_r)\le B(r):= B_p(r):=
  1-\Phi\big(r\sqrt{2\pi}+\Phi^{-1}(p)\big), 
\end{equation}
where $r\ge0$, $\Phi$ is the standard normal distribution function on $\mathbb R$, and 
\begin{equation}
 p:=\mu_{\infty}(A). 
\end{equation}
We see that the bound $B_p(r)$ indeed decreases in $p$ (and, of course, in $r$). 
If $p\ge1/2$, then 
$B(r)\le 1-\Phi\big(r\sqrt{2\pi}\big)$, and 
the inequality $\Phi(u)\ge1-e^{-u^2/2}$ for $u\ge0$ indeed implies $B(r)\le e^{-\pi r^2}$. 
If $p\uparrow1$, then $B(r)\le(1-p)^{1-o(1)}e^{-\pi r^2}$, which is better than $e^{-\pi r^2}$. If $p<1/2$, then $B(r)>1-\Phi\big(r\sqrt{2\pi}\big)=e^{-(\pi+o(1)) r^2}$ as $r\to\infty$. 
So, the case $p\ge1/2$ is special only in the sense that then the expression for the bound $B_p(r)$ is simpler, since $\Phi^{-1}(1/2)=0$. 
