I am reading Constructive discrepancy minimization for convex sets. I am stuck at Lemma 4 which states that for any measurable set $K\subset\mathbb{R}^n$ if $\gamma_n(K) \geq \exp(-\epsilon n)$, then $\gamma_n(K_{3\sqrt{\epsilon n}}) \geq 1-\exp(-\epsilon n)$. Some definitions are in order.
For a set $K$ and $x\in\mathbb{R}^n$, $d(x,K) = \min\{\Vert x-y\Vert_2 : y\in K\}$ is the distance of $x$ to $K$. For $\delta \geq 0$, $K_\delta =\{x \in \mathbb{R}^n: d(x,K) \leq \delta\}$ is the set of points that have at most distance $\delta$ to $K$. $\gamma_n$ is the $n$-dimensional Gaussian measure with density $\frac{1}{(2\pi)^{n/2}}\exp(-\Vert x\Vert_2^2/2)$.
I am interested in the proof the above lemma from the footnote. Let $F(x) = d(x,K)$ and $\mu = \mathbb{E}[F(x)]$. It is claimed that since $\gamma_n(K) \geq \exp(-\epsilon n)$, $\mu \leq \frac{3}{2}\sqrt{\epsilon n}$. I am not understanding how to prove this simple fact.
I would be thankful if someone gives a hint as to how to compute the corresponding integral.
Thank you,