From Rick Durrett, Probability Theory and Examples:
My Question:
Why $\lvert A_{n,\epsilon}\cap (-1,1)^n\rvert/2^n\rightarrow 1$? Mainly why this follows from WLLN?
I can do the calculation using integration over the region $A_{n,\epsilon}\cap (-1,1)^n$ using the distribution of $X_1^2+\dots+X_n^2$, but why WLLN directly implies that?
Note: This is theorem 2.2.3 :
$\def\X{{\bf X}}$Let $\X_n$ denote the random vector $(X_1,\dots,X_n)$. Since $\X_n$ is uniformly distributed over the cube $[-1,1]^n$, the probability of $P(\X_n\in A)$ is just he Lebesgue measure of $A\cap [-1,1]^{n}$ divided by the measure of the whole cube. Therefore, $$\begin{align} |A_{n,\epsilon}\cap[-1,1]^n|/2^n &=P(\X_n\in A_{n,\epsilon})\\ &=P((1-\epsilon)\sqrt{n/3}<|\X_n|<(1+\epsilon)\sqrt{n/3})\\ &=P((1-\epsilon)^2\cdot n/3<|\X_n|^2<(1+\epsilon)^2n/3)\\ &=P((-2\epsilon+\epsilon^2)/3<(X_1^2+\dots+X_n)^2/n-1/3<(2\epsilon+\epsilon^2)/3)\\ &\le P((-2\epsilon-\epsilon^2)/3<(X_1^2+\dots+X_n)^2/n-1/3<(2\epsilon+\epsilon^2)/3) \end{align}$$ Finally, the last probability goes to one as a direct consequence of the fact that $(X_1^2+\dots+X_n)^2/n\to {1}/3$ in probability. Indeed, the definition of $Y_n\to Y$ in probability is that $$ \forall\delta>0,P(-\delta<Y_n-Y<\delta)\to 1\text{ as }n\to\infty $$ so the result follows be letting $\delta=(\epsilon^2+2\epsilon)/3$.