Limit of gaussian measure of $B_n$ ball In one paper I read, the author claimed that the Gaussian measure of the $B_n$ ball tends to 1/2 when n tends to infinity. Here $$B_n = \{x \in \mathbb{R}^n : \|x \|^2 \le n \},$$
and the Gaussian measure is the probability measure on $\mathbb{R}^n$ with density
$$\frac{1}{(2\pi)^{n/2}} e^{\frac{-\|x\|^2}{2}}.$$
Can any one suggest me a proof of this statement? I am aware that the Gaussian measure concentrates on the "flatten set"  $\{\sqrt{n} - \epsilon \le \|x\| \le \sqrt{n} + \epsilon\}$, but could not find a proof anywhere for this case (when $x$ is only on one side of $\sqrt{n}$). 
The question made explicitly:
$$\lim_{n \to \infty} \int_{\|x \|^2 \le n} \frac{1}{(2\pi)^{n/2}} e^{\frac{-\|x\|^2}{2}} dx = \frac{1}{2} ?$$
 A: This is like integrating the density you mentioned from in the range $(0,\infty)$ since the radius of ball can't be negative. But since the function is symmetrical about 0 the integration in the aforementioned range should be $1/2$
i.e
$$\lim_{n\rightarrow \infty}\int_0^{\sqrt{n}}\frac{1}{(2\pi)^{n/2}} e^{\frac{-\|x\|^2}{2}}=\frac{1}{2}\lim_{n\rightarrow \infty}\int_{-\sqrt{n}}^{\sqrt{n}}\frac{1}{(2\pi)^{n/2}} e^{\frac{-\|x\|^2}{2}}$$
A: Let $Z_1,Z_2,\dots$ be i.i.d $\chi^2(1)$ variables (squares of independent normal variables). By the central limit theorem,
$$\mathbb P\left [\frac{Z_1+\dots+Z_n}n - \mathbf E[Z_1] \leq 0\right] \to \tfrac 1 2$$
which is a restatement of the claim.
A: A quick proof when $n$ is even: Spherical coordinates change yields
\begin{align*}
\int_{B_n} \frac{1}{(2\pi)^{n/2}} e^{-\|x\|^2/2} \, dx
&= \int_{0}^{\sqrt{n}} \frac{1}{(2\pi)^{n/2}} \underbrace{ \frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})} r^{n-1} }_{=|\partial B_r|} e^{-r^2/2} \, dr \\
(r = \sqrt{2u}) \quad &= \int_{0}^{\frac{n}{2}} \frac{u^{\frac{n}{2}-1}}{\Gamma(\frac{n}{2})} e^{-u} \, du.
\end{align*}
Now assume that $n = 2k$ is even and $T_1, T_2, \cdots$ are i.i.d. exponential random variables of unit rate. Then the integral above can be written as
$$ \int_{0}^{k} \frac{u^{k-1}}{(k-1)!} e^{-u} \, du
= \mathbb{P}(T_1+\cdots+T_k \leq k)
= \mathbb{P}\left( \frac{T_1+\cdots+T_k - k}{\sqrt{k}} \leq 0 \right)$$
By the classical CLT, we have $\frac{T_1+\cdots+T_k - k}{\sqrt{k}} \Rightarrow Z \sim \mathcal{N}(0,1)$ as $k\to\infty$. So the above probability converges to $\mathbb{P}(Z \leq 0) = \frac{1}{2}$.

For a more general proof, let $\nu = n/2$ and apply the change of variables $u = \nu - \sqrt{\nu}t$. Then
$$ \int_{0}^{\nu} \frac{u^{\nu-1}}{\Gamma(\nu)} e^{-u} \, du
= \frac{\nu^{\nu-\frac{1}{2}}e^{-\nu}}{\Gamma(\nu)} \int_{0}^{1} \left(1 - \frac{t}{\sqrt{\nu}} \right)^{\nu - 1} e^{\sqrt{\nu}t}  \mathbf{1}_{[0,\sqrt{\nu}]}(t)\, dt. $$
Now using the inequality $\log(1-x) \leq -x - \frac{x^2}{2}$ for $0 < x < 1$ and the quantitative form of the Stirling's approximation, we know that the integrand is uniformly bounded by
$$ \frac{\nu^{\nu-\frac{1}{2}}e^{-\nu}}{\Gamma(\nu)} \left(1 - \frac{t}{\sqrt{\nu}} \right)^{\nu - 1} e^{\sqrt{\nu}t}  \mathbf{1}_{[0,\sqrt{\nu}]}(t)
\leq e^{\frac{3}{2}} \sqrt{2\pi} e^{-t^2/2}, $$
which is integrable. So by the dominated convergence theorem and the Stirling's approximation, we have
$$ \lim_{\nu\to\infty} \frac{\nu^{\nu-\frac{1}{2}}e^{-\nu}}{\Gamma(\nu)} \int_{0}^{1} \left(1 - \frac{t}{\sqrt{\nu}} \right)^{\nu - 1} e^{\sqrt{\nu}t}  \mathbf{1}_{[0,\sqrt{\nu}]}(t)\, dt
= \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{-t^2/2} \, dt
= \frac{1}{2}. $$
A: Let $I_n := (2\pi)^{-n/2}\int_{\|x\| \le \sqrt{n}}e^{-\|x\|^2/2}dx$. By equation (*) of this post https://math.stackexchange.com/a/3496441/168758, we have the identity $I_n = \frac{\gamma(n/2,n/2)}{\Gamma(n/2)}$, where $\gamma(a,x) := \int_{0}^xt^{a-1}e^{-t}dt$ is the incomplete gamma function. Now, use standard asymptotics results (Stirling's formulae for $\gamma$), to get that $I_n \longrightarrow 1/2$.
A: $$
\begin{align}
\lim_{n\to\infty}\int_{\|x\|^2\le n}\frac1{(2\pi)^{n/2}}e^{-\|x\|^2/2}\,\mathrm{d}x
&=\lim_{n\to\infty}\frac1{(2\pi)^{n/2}}\int_0^{n^{1/2}}e^{-r^2/2}\omega_{n-1}r^{n-1}\,\mathrm{d}r\tag1\\
&=\lim_{n\to\infty}\frac1{2^{n/2}\Gamma(n/2)}\int_0^{n^{1/2}}e^{-r^2/2}r^{n-2}\,\mathrm{d}r^2\tag2\\
&=\lim_{n\to\infty}\frac1{\Gamma(n/2)}\int_0^{n/2}e^{-r}r^{n/2-1}\,\mathrm{d}r\tag3\\[3pt]
&=\lim_{n\to\infty}\frac1{n!}\int_0^ne^{-r}r^n\,\mathrm{d}r\tag4\\[3pt]
&=\lim_{n\to\infty}\left(\frac12-\frac{2/3}{\sqrt{2\pi n}}+O\!\left(\frac1n\right)\right)\tag5\\[3pt]
&=\frac12\tag6
\end{align}
$$
Explanation:
$(1)$: convert to polar coordinates
$(2)$: $\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$
$(3)$: substitute $r\mapsto(2r)^{1/2}$
$(4)$: substitute $n\mapsto2n+2$
$(5)$: Equation $(11)$ from this answer
$(6)$: evaluate the limit
