About Weak law for triangular arrays 
Definition: to truncate a random varialbe X at level M means to consider $\hat{X}=X 1_{(|X| \le M)}=\begin{cases} X, & \textbf{if} |X| \le M \\ 0,& \textbf{if} |X|>M \end{cases}$
Theorem(*): Let $X_1,...,X_n$ have $E({X_i}^2)<\infty$ and be uncorrelated. Then $\textbf{var}(X_1+...+X_n)=\textbf{var}(X_1)+...\textbf{var}(X_n)$ where $\textbf{var}(Y)$ is the variance of Y.
Theorem (Weak law for triangular arrays): For each n, let $X_{n,k}, 1 \le k \le n$ be independent. Let $b_n>0$ with $b_n \rightarrow \infty$, and let $\hat{X}_{n,k}=X_{n,k} 1_{(|X_{n,k} \le b_n )}$. Suppose that as $n \rightarrow \infty$. $$(i). \sum_{k=1}^n P(|X_{n,k}|>b_n) \rightarrow 0$$, and $$(ii). b^{-2}_n \sum_{k=1}^n E\hat{X}^2_{n,k} \rightarrow 0.$$ If we let $S_N=X_{n,1}+...+X_{n,n}$ and put $a_n=\sum_{k=1}^n E\hat{X}_{n,k}$, then $(S_n-a_n)/b_n \rightarrow 0$ in probability.

Proof: Let $\hat{S}_n=\hat{X}_{n,1}+...+\hat{X}_{n,n}$. Clearly, $P(|\frac{S_n-a_n}{b_n}|>\epsilon) \le P(S_n \neq \hat{S}_n)+P(\frac{\hat{S_n}-a_n}{b_n}|>\epsilon)$. To estimate the first term, we note that $$P(S_n \neq \hat{S}_n \le P(\cup_{k=1}^n \{\hat{X}_{n,k} \neq X_{n,k} \}) \le \sum_{k=1}^n P(|X_{n,k}>b_n) \rightarrow b_n) \rightarrow 0$$ by property (i). For the second term, we note that Chebyshev's inequality, $a_n=E\hat{S}_n$, by theorem (*), and $\textbf{var}(X) \le EX^2$ imply $$P(|\frac{\hat{S}_n-a_n}{b_n}|>\epsilon) \le \epsilon^{-2} E{|\frac{\hat{S}_n-a_n}{b_n}|}^2=\epsilon^{-2} b^{-2}_n \textbf{var}(\hat{S}_n)=(b_n \epsilon)^{-2} \sum_{k=1}^n \textbf{var}(\hat{X}_{n,k}) \le (b_n \epsilon)^{-2} \sum_{k=1}^n E(\hat{X}_{n,k})^2 \rightarrow 0$$ by property (ii).

Question: I don't see how "$(S_n-a_n)/b_n \rightarrow 0$ in probability" comes true. And all those inequalities in the proof, like $$P(S_n \neq \hat{S}_n \le P(\cup_{k=1}^n \{\hat{X}_{n,k} \neq X_{n,k} \}) \le \sum_{k=1}^n P(|X_{n,k}>b_n) \rightarrow b_n) \rightarrow 0$$ and $$P(|\frac{\hat{S}_n-a_n}{b_n}|>\epsilon) \le \epsilon^{-2} E{|\frac{\hat{S}_n-a_n}{b_n}|}^2=\epsilon^{-2} b^{-2}_n \textbf{var}(\hat{S}_n)=(b_n \epsilon)^{-2} \sum_{k=1}^n \textbf{var}(\hat{X}_{n,k}) \le (b_n \epsilon)^{-2} \sum_{k=1}^n E(\hat{X}_{n,k})^2 \rightarrow 0$$ don't make sense to me, especially $\epsilon^{-2}$. Thanks a lot in advance. (The proof is a bit technical to me, so I would like it to be simpler to understand).

 A: For a random variable $X$ with $E[X]=\mu$, Chebyshev's inequality is:
$$P(|X-\mu|>\epsilon)\leq \frac{\mbox{Var}(X)}{\epsilon^2}=\frac{E[|X-\mu|^2]}{\epsilon^2}.$$
If you're unfamiliar with this, then hopefully you've seen Markov's inequality:
$$P(|Y|\geq \epsilon)\leq \frac{E[Y]}{\epsilon},$$
from which you can derive Chebyshev's Inequality by substituting $Y:=X-\mu$, and using the fact that since $\epsilon>0$, $P(|Y|\geq \epsilon)=P(|Y|^2\geq \epsilon^2)$.
That should hopefully clear up the convergence in probability issue.
For $P(S_n\neq \hat{S}_n)$, notice that both $S_n,\hat{S}_n$ are sums of $X_{k,n},\hat{X}_{k,n}$ respectively. Thus, the event $\{S_n\neq \hat{S}_n\}$ implies that there is at least one $j$ such that $X_{j,n}\neq \hat{X}_{j,n}$. So:
$$\{S_n\neq \hat{S}_n\}\subseteq \cup_{j=1}^n\{X_{j,n}\neq \hat{X}_{j,n}\}.$$
Then applying $P$ to both sides, and using countable additivity, you get:
$$P(\{S_n\neq \hat{S}_n\})\leq \sum_{j=1}^nP(\{X_{j,n}\neq \hat{X}_{j,n}\}).$$
Now stare at the definition of $\hat{X}_{j,n}$ until you see that you need $X_{j,n}>b_n$ to enforce $X_{j,n}\neq \hat{X}_{j,n}$.
