Why weak law of large number still alive? I know the difference between WLLN and SLLN in terms of a convergence type. Then, as revealed in any statistical textbook saying sufficient conditions to two theorems are the same, I think that we do not need WLLN anymore. I also know that there is some example for which WLLN holds but SLLN not. Then, From this examples we would have needed to establish another conditions only needed for SLLN, still I could not find any explanations for this.
Is there anyone to explain difference between WLLN and SLLN in terms of this issue?
Thanks!
 A: Feller's WLLN says $X_n$ are i.i.d and let $S_n=\sum X_i$. Then $\exists C_n$(finite) such that
$$\frac{S_n-C_n}{n}\overset{P}{\rightarrow} 0 \hspace{10pt} \iff nP({X_1}>n)\rightarrow 0$$
In such a case $\frac{C_n}{n}=E(X_1I\{|X_1|\leq n\})+o(1)$.
Kolmogorov's SLLN says $X_n$ are i.i.d and let $S_n=\sum X_i$.Then
$$\frac{S_n}{n}\overset{a.s.}{\rightarrow} c \hspace{10pt}\text{for some finite c}\hspace{10pt}\iff X_1\in L_1$$
In such a case necessarily $c=E(X_1)$.
Result: $X\in L_p \Rightarrow n^pP(|X_1|>n^{1/p})\rightarrow 0$. But the otherway IS NOT TRUE!
Thus $X_1\in L_1$ means we can apply SLLN and WLLN both. The following example depicts WLLN holds but not SLLN
Let $X_n$ i.i.d with density $f(x)=\frac{c}{x^2\log{|x|}} I\{|x|>e\}$. Show that $E(X^{+})=\infty=E(X^{-})$ implying $E(X)$ does not exist. We can not apply SLLN. But it's not hard to show $nP(|X_1|>n)=o(1)$. Hence by WLLN we've
$$\frac{S_n}{n} -E(X_1I\{|X_1|\leq n\})\overset{P}{\rightarrow} 0$$
Since $X_1$ is symmetric the above expectation is $0$, i.e. $\frac{S_n}{n}\overset{P}{\rightarrow} 0$.
In this example you can also show 
$$\limsup \frac{S_n}{n}=\infty \hspace{10pt}\text{and}\hspace{10pt}\liminf \frac{S_n}{n}=-\infty\hspace{3pt}\text{almost surely}$$
Thus it's clearly false that $\frac{S_n}{n}\overset{a.s.}{\rightarrow} 0$.
