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We have $0<\sigma<\infty$ and $\epsilon >0$ and $X_1, X_2,...$ iid. The argument involves CLT and it continues on until the line below:

That $\frac{1}{\sigma^2}E\big[(X_1-\mu)^2\mathbb{1}\big(|X_1-\mu|>\epsilon\sigma\sqrt(n)\big)\big]\rightarrow0$ by LDCT.

Can someone help me understand why this is the case? Something I can see is that if we consider $Z_n=X_n-\mu$ and by the iid property I know that $Var(Z_n)=Var(X_n-\mu)=E\big((X_n-\mu)^2\big)=\sigma$ as $E(Z_n)=0$.

And then $(X_1-\mu)^2\mathbb{1}\big(|X_1-\mu|>\epsilon\sigma\sqrt(n)\big)\leq\sigma$

So then I know that LDCT holds, and the limit of the integral is the integral of the limit.

So then why is this zero?

$\frac{1}{\sigma^2}E\big[\displaystyle\lim_{n\rightarrow\infty}(X_1-\mu)^2\mathbb{1}\big(|X_1-\mu|>\epsilon\sigma\sqrt(n)\big)\big]$

Quick edit:

Nevermind I think I see it.

It is because $\epsilon >0$ and $\mathbb{1}\big(\frac{|X_1-\mu|}{\sigma\sqrt{n}}>\epsilon\big)\rightarrow\mathbb{1}(0>\epsilon)$, which is over a null set.

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In order to avoid the use of too much letters, let $$ Y:=\frac{\left\lvert X_1-\mu\right\rvert}{\sigma\varepsilon}.$$ The problem reduces to prove that $\mathbb E\left[Y^2\mathbf 1\{Y\gt \sqrt n\}\right]\to 0$ for each square integrable random variable, or letting $Z=Y^2$, that $\mathbb E\left[Z\mathbf 1\{Z\gt n\}\right]\to 0$ for each integrable random variable $Z$. There are two ways (which are not so fundamentally different) to show that.

  1. Let $Z_n:=Z\mathbf 1\{Z\gt n\}$. Then $0\leqslant Z_n\leqslant Z$ and $Z_n\to 0$ almost surely hence by the dominated convergence theorem, the expectation of $Z_n$ goes to zero.

  2. Let $Z'_n:=Z\mathbf 1\{Z\leqslant n\}$. Then $0\leqslant Z'_n\leqslant Z'_{n+1}$ for all $n$ and $Z'_n\to Z$ almost surely. We conclude from the monotone convergence theorem.

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