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} 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 $\Phi^{-1}(1/2)=0$.

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 $\Phi^{-1}(1/2)=0$.

By the Tsirel’son--Ibragimov--Sudakov argument, reviewed on the first page in Bobkov, by 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} 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 $B(r)>e^{-\pi r^2}$.

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 $\Phi^{-1}(1/2)=0$.

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} 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 $\Phi^{-1}(1/2)=0$.