how to show convergence in probability imply convergence a.s. in this case? Assume that 
$X_1,\cdots,X_n$ are independent r.v., not necessarily iid,
Let $S_n=X_1+\cdots+X_n$, suppose that $S_n$ converges in probability, how can we show that $S_n$ converges a.s.?
 A: We can use Ottaviani's inequality: if we put $M_k:=\max_{1\leqslant i\leqslant k}|S_i|$ and $S_{k,n}:=\sum_{i=k+1}^nX_i$, then for all $\varepsilon >0$ we have 
$$\min_{1\leqslant k\leqslant n}P(|S_{k,n}|\leqslant\varepsilon)P(|M_n|>2\varepsilon)\leqslant P(|S_n|>\varepsilon).$$
Put $A_m:=\sup_{k\in\mathbb N^*}|S_{m+k}-S_m|$ and $A:=\inf_{m\in\mathbb N^*}A_m$. We have 
$$\{\{S_n\}\mbox{ doesn't converge}\}=\{A\neq 0\}\subset\bigcup_{\varepsilon\in\mathbb Q^+}\bigcap_{m\in\mathbb N^*}\{A_m>\varepsilon\}$$
and 
$$\{A_m>\varepsilon\}=\bigcup_{r\in\mathbb N^*}\left\{\sup_{1\leqslant k\leqslant r}|S_{m+k}-S_m|>\varepsilon\right\}.$$
Ottaviani's inequality gives 
$$\min_{1\leqslant k\leqslant r}P(|S_{m+r}-S_{m+k}|\leqslant\varepsilon)P\left(\max_{1\leqslant k\leqslant r}|S_{m+k}-S_m|>2\varepsilon\right)\leqslant P(|S_{r+m}-S_m|>\varepsilon).$$
We fix $\delta>0$; we can choose $N=N(\varepsilon,\delta)$ such that for $0\leqslant k\leqslant r$ and $m\geqslant N$, $P(|S_{m+r}-S_{m+k}|>\varepsilon)\leqslant\delta$. We get 
$$P\left(\max_{1\leqslant k\leqslant r}|S_{m+k}-S_{m+r}|>\varepsilon\right)\leqslant\frac{\delta}{1-\delta},$$
therefore $P(A_m>\varepsilon)\leqslant\frac{\delta}{1-\delta}$. We have 
$$0\leqslant P\left(\bigcap_{p\in\mathbb N^*}(A_p>\varepsilon)\right)\leqslant P(A_m>\varepsilon)\leqslant\frac{\delta}{1-\delta},$$
so for all $\varepsilon\in\mathbb Q^+$, we have $P\left(\bigcap_{p\in\mathbb N^*}(A_p>\varepsilon)\right)=0$, hence $P(A\neq 0)=0$ and we have the almost sure convergence.
