This is Exercise 3.1.1 part (a) from Amir Dembo's Note:
Let $S_{n}=\sum_{k=1}^{n}X_{n}$ for $X_{k}$ mutually independent such that $v_{n}=Var(S_{n})<\infty$. Show that if there exists $q>2$ such that $$\lim_{n\rightarrow\infty}v_{n}^{-q/2}\sum_{k=1}^{n}\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{q}=0,$$ then $$v_{n}^{-1/2}(S_{n}-\mathbb{E}S_{n})\Rightarrow\mathcal{N}(0,1).$$
I have worked out most of the proof, but I got stuck in the end.
My attempt is to apply Lindeberg's Central Limit Theorem to $$\widehat{S}_{n}:=\sum_{k=1}^{n}X_{n,k}$$ where $X_{n,k}:=v_{n}^{-1/2}(X_{k}-\mathbb{E}X_{k}).$
I have managed to show that $Var(\widehat{S}_{n})=1$ and $\mathbb{E}X_{n,k}=0$. However, I don't know how to show the Lindeberg's Condition that for each $\epsilon>0$, we have $$g_{n}(\epsilon):=\sum_{k=1}^{n}\mathbb{E}[X_{n,k}^{2};|X_{n,k}|\geq\epsilon]\longrightarrow 0\ \text{as}\ n\longrightarrow\infty.$$
The proof in the note works in this way:
\begin{align*} \sum_{k=1}^{n}\mathbb{E}[X_{n,k}^{2};|X_{n,k}|\geq\epsilon]&=v_{n}^{-1}\sum_{k=1}^{n}\mathbb{E}[(X_{k}-\mathbb{E}X_{k})^{2};|X_{k}-\mathbb{E}X_{k}|\geq v_{n}^{-1/2}\epsilon]\\ &\leq\epsilon^{2-q}v_{n}^{-q/2}\sum_{k=1}^{n}|X_{k}-\mathbb{E}X_{k}|^{q}\longrightarrow 0. \end{align*}
I have no problem with the first equality, which is just the expansion of $X_{n,k}^{2}$ and $|X_{n,k}|$, I have no problem with the convergence in the end since it is by hypothesis from the question.
However, I don't really understand the inequality the solution claims. I tired Chebyshev (or Markov, depending on the book), since this is the only inequality concerning with the expectation with indictor function, but I don't see how I can get this inequality.
Is there any other inequality I am missing?
Thank you!
Edit 1:
Okay I figured this out following Galton's hint, as follows:
By hypothesis, there exists $q>2$ such that $$\lim_{n\rightarrow\infty}v_{n}^{-q/2}\sum_{k=1}^{n}\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{q}=0.$$
Fix such $q>2$, note that $2-q<0$, and then we can write for each $\epsilon>0$ that \begin{align*} \sum_{k=1}^{n}\mathbb{E}[X_{n,k}^{2};|X_{n,k}|\geq\epsilon]&=v_{n}^{-1}\sum_{k=1}^{n}\mathbb{E}[(X_{k}-\mathbb{E}X_{k})^{2};|X_{k}-\mathbb{E}X_{k}|\geq v_{n}^{\frac{1}{2}}\epsilon]\\ &=v_{n}^{-1}\sum_{k=1}^{n}\mathbb{E}[(X_{k}-\mathbb{E}X_{k})^{q}(X_{k}-\mathbb{E}X_{k})^{2-q};|X_{k}-\mathbb{E}X_{k}|\geq v_{n}^{\frac{1}{2}}\epsilon]\\ &=v_{n}^{-1}\sum_{k=1}^{n}\mathbb{E}[(X_{k}-\mathbb{E}X_{k})^{q}(X_{k}-\mathbb{E}X_{k})^{2-q};|X_{k}-\mathbb{E}X_{k}|^{2-q}\leq v_{n}^{\frac{2-q}{2}}\epsilon^{2-q}]\\ &\leq v_{n}^{-1}\sum_{k=1}^{n}\mathbb{E}[|X_{k}-\mathbb{E}X_{k}|^{q}|X_{k}-\mathbb{E}X_{k}|^{2-q};|X_{k}-\mathbb{E}X_{k}|^{2-q}\leq v_{n}^{\frac{2-q}{2}}\epsilon^{2-q}]\\ &\leq v_{n}^{-1}\cdot v_{n}^{\frac{2-q}{2}}\epsilon^{2-q}\sum_{k=1}^{n}\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{q}\\ &=\epsilon^{2-q}v_{n}^{-q/2}\sum_{k=1}^{n}\mathbb{E}|X_{k}-\mathbb{E}X_{k}|^{q}\longrightarrow 0. \end{align*}
Let $Y_k:=\left\lvert X_k-\mathbb E[X_k] \right\rvert$. Then for each positive $R$, $$ Y_k^2\mathbf 1\{Y_k>R\}=Y_k^qY_k^{2-q}\mathbf 1\{Y_k>R\}\leqslant Y_k^qR^{2-q}. $$ With $R=\varepsilon v_n^{1/2}$, we get the wanted inequality.