Weak Law of Large Number

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Let $\{X_n\}$ be a sequence of pairwisely independent and identically distributed random variables with finite mean $m$.  Define $S_n=\sum_j X_j$.  Then $$\frac{S_n}{n}\rightarrow m\mbox{ in probability.}$$



$\bullet$ Proof.
Truncate $X_n$ at $n$, define $$Y_n=\begin{cases}X_n, &\mbox{if }X_n\leq n\\ 0,&\mbox{o.w.}\end{cases}$$ and $T_n=\sum_j Y_j$.

First we discuss the variance of $T_n$ and then connect $S_n$ and $T_n$ with the equivalence property.
$$\sigma^2(T_n)=\sum_{j=1}^n\sigma^2(Y_j)=\sum_{j=1}^n\int_{|x|\leq j}x^2\,d\mathcal{F}.$$ Note that, it seems that we can only get $\sigma^2(T_n)=O(n^2)$ by replacing $x^2$ with $j^2$ in the integral.

Here, the key point is to find a increasing sequence which grows slower than $\{n\}$ and then split the summation apart.  Let $0<a_n< n$, $a_n=o(n)$ and $a_n\rightarrow\infty$.  $$\begin{array}{ccl}\sigma^2(T_n)&=&\sum_{j=1}^n\int_{|x|\leq j}x^2\,d\mathcal{F}\\
&=&\sum_{j\leq a_n}\int_{|x|\leq j}x^2\,d\mathcal{F}+\sum_{a_n<j\leq n}\int_{|x|\leq j}x^2\,d\mathcal{F}\\
&\leq&\sum_{j\leq a_n}a_n\int_{|x|\leq a_n}|x|\,d\mathcal{F}+\sum_{a_n<j\leq n}\left[\int_{|x|\leq a_n}x^2\,d\mathcal{F}+\int_{a_n<|x|\leq j}x^2\,d\mathcal{F}\right] \\
&\leq&\sum_{j=1}^n a_n\int_{|x|\leq a_n}|x|\,d\mathcal{F}+ \sum_{a_n<j\leq n}n\int_{a_n<|x|\leq n}|x|\,d\mathcal{F}\\
&\leq&n a_n\int |x|\,d\mathcal{F}+ n^2\int_{|x|>a_n}|x|\,d\mathcal{F}\;\mbox{(free sum. and int. space)}\\
&=&O(n a_n)+n^2o(1) \mbox{ since }a_n\rightarrow\infty\mbox{ for the last integral.}\\
&=&o(n^2) \end{array}$$  Thus, we have $$\sigma^2(T_n)=o(n^2)\implies\frac{T_n-\mathscr{E}(T_n)}{n}\overset{P}{\rightarrow}0.$$
Since $\mathscr{E}(Y_n)\rightarrow \mathscr{E}(X_n)=m$ as $n\rightarrow\infty$, then $$\frac{\mathscr{E}(T_n)}{n}\rightarrow m\implies\frac{T_n}{n}\rightarrow m\mbox{ in probability.}$$
Finally, since $\mathscr{E}|X_n|<\infty$, $\sum_n\mathscr{P}\{X_n\neq Y_n\}=\sum_n\mathscr{P}\{|X_n|>n\}<\infty$.  That is, $Y_n$ and $X_n$ are equivalent, and so are $T_n$ and $S_n$.  Thus, $$\frac{S_n}{n}\rightarrow m\mbox{ in probability.}$$

$\Box$