Trimmed second moment going to zero by LDCT

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.

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.