The Lindeberg condition (classic) states that for a random variable $X$ with finite mean and variance, $\mu$ and $\sigma^2$, and for every $\varepsilon>0$ $$E(X^2 \boldsymbol{I}_{\{ X>\varepsilon\sqrt{n}\sigma\}}) \rightarrow 0, \quad \mbox{as} \quad n\rightarrow \infty,$$ where $\boldsymbol{I}$ is the indicator function.
Now I'd like to prove this. Let's say $X$ is a continuous random variable with distribution $P_X$, then $$E(X^2 \boldsymbol{I}_{\{ X>\varepsilon\sqrt{n}\sigma\}}) = \int_{X>\varepsilon \sqrt{n}\sigma}x^2 dP_X$$ I think I get the intuition of this integral: for a fixed $\varepsilon$ when $n$ gets larger, the set $X>\varepsilon\sqrt{n}\sigma$ gets smaller. So I'm integrating on a set that "tends" to the empty set.
I thought the proof may be something like $$\sigma^2 + \mu^2 = \int_{\mathbb{R}}x^2 dP_X = \int_{X\leq\varepsilon \sqrt{n}\sigma}x^2 dP_X + \int_{X>\varepsilon \sqrt{n}\sigma}x^2 dP_X,$$ but I'm not sure. Thanks
Note that since the mean and variance are finite, then $$ \sigma^2 + (EX)^2 = EX^2 = M < \infty $$
Now, you know that $EX^2 = \int x^2 dP_x$, so as you did, you break down the integral $$ EX^2= \int_{X\leq \epsilon \sqrt n \sigma} x^2 dP_x + \int_{X>\epsilon \sqrt n \sigma} x^2 dP_x $$
Then, taking the limit
$$ \lim_{n \to \infty } EX^2 = \int_{X\leq \infty} x^2 dP_x +\lim_{n\to \infty} \int_{X>\epsilon \sqrt n \sigma} x^2 dP_x = M + \lim_{n \to \infty} \int_{X>\epsilon \sqrt n \sigma} x^2 dP_x $$
Which proves what you wanted to show.