Extension of Weak Law of Large Number (2)

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Let $\{X_n\}$ be a sequence of pairwisely independent random variables with common distribution functions $F$.  Define $S_n=\sum_j X_j$.  Suppose that we have
(1) $\displaystyle\int_{|x|\leq n}x\,dF(x)=o(1)$,
(2) $\displaystyle n\int_{|x|>n}\,dF(x)=o(1)$;
then $$\frac{S_n}{n}\rightarrow0\mbox{ in probability.}$$

$\bullet$ Proof.

Define for each $n\geq1$ and $1\leq j\leq n$, $$Y_{n,j}=\begin{cases}X_j,&\mbox{if }|X_j|\leq n;\\ 0,&\mbox{if }|X_j|>n,\end{cases}$$and write $T_n=\sum_{j=1}^nY_{n,j}$.  We need to show our goal through the definition of $Y_{n,j}$ and the follows:
(a) $S_n$ and $T_n$ are equivalent;
(b) $\frac{\mathscr{E}(T_n)}{n}\rightarrow0$;
(c) $\sigma^2(T_n)=o(n^2)$ such that $\frac{T_n}{n}\rightarrow0$ in probability.

For (a), by Boole's inequality, $$\begin{array}{rl}\mathscr{P}\{T_n\neq S_n\}&=\mathscr{P}\left\{\bigcup_{j=1}^n(Y_{n,j}\neq X_j)\right\}\\ & \leq\sum_{j=1}^n\mathscr{P}\{Y_{n,j}\neq X_j\}\\ & =\sum_{j=1}^n\mathscr{P}\{|X_j|>n\}=n\mathscr{P}\{|X_1|>n\}=n\int_{|x|>n}\,dF(x)=o(1)\end{array}$$due to assumption (2).  Thus, we have that $S_n$ and $T_n$ are equivalent.

For (b), $$\begin{array}{rl}\frac{\mathscr{E}(T_n)}{n}=\frac{1}{n}\sum_{j=1}^n\mathscr{E}(Y_{n,j})&=\frac{1}{n}\sum_{j=1}^n\int_{|x|\leq n}x\,dF_j(x)\\ & =\int_{|x|\leq n}x\,dF(x) =o(1)\end{array}$$due to assumption (1).  Thus, we have that $\frac{\mathscr{E}(T_n)}{n}\rightarrow0$.

For (c), $$\begin{array}{rl}\sigma^2(T_n)&=\sum_{j=1}^n\sigma^2(Y_{n,j})\leq\sum_{j=1}^n\mathscr{E}(Y_{n,j}^2)\\
&=\sum_{j=1}^n\int_{|x|\leq n}x^2\,dF_j(x)\\
&=n\int_{|x|\leq n}x^2\,dF(x),\mbox{ then for }0<a_n\leq n\mbox{ and }a_n\rightarrow\infty\\
&=n\left[\int_{|x|\leq a_n}x^2\,dF(x)+\int_{a_n<x\leq n}x^2\,dF(x)\right]\\
&\leq na_n\int_{|x|\leq a_n}|x|\,dF(x)+n^2\int_{a_n<x\leq n}|x|\,dF(x)\\
&\leq na_n\int_{|x|\leq a_n}|x|\,dF(x)+n^2\int_{x>a_n}|x|\,dF(x)\\
&=O(na_n)+n^2o(1)=o(n^2)\end{array}$$since, for the first part $$\int_{|x|\leq n}x\,dF(x)=o(1)\implies\int_{|x|\leq n}|x|\,dF(x)<\infty,$$and for the second part $$\{|x|>a_n\}\rightarrow\emptyset\mbox{ as }a_n\rightarrow\infty.$$Thus, $$\frac{T_n-\mathscr{E}(T_n)}{n}=\frac{T_n}{n}\rightarrow0\mbox{ in probability.}$$By the equivalence between $S_n$ and $T_n$, we have $$\frac{S_n}{n}\rightarrow0\mbox{ in probability.}$$

$\Box$