# SLLN when the expectation in infinite

In a Post I found it says:

Whenever ${\rm E}(X)$ exists (finite or infinite), the strong law of large numbers holds. That is, if $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with finite or infinite expectation, letting $S_n = X_1+\cdots + X_n$, it holds $n^{-1}S_n \to {\rm E}(X_1)$ almost surely. The infinite expectation case follows from the finite case by the monotone convergence theorem.

Can someone give a reference/answer to this question? I want to prove that:

If $EX^{+}_{k}=∞$ and $EX^{-}_{k}<∞$ then $n^{−1}S_{n}→∞$ a.s.

• Note that $S_n\geqslant S_n^x$ for every $x$, where $S_n^x=X_1^x+\cdots+X_n^x$ and $X_n^x=\inf(X_n,x)$ for every $n$, thus $\liminf S_n/n\geqslant\lim S_n^x/n=E(X_1^x)$ almost surely, then the limit $E(X_1^x)\to+\infty$ when $x\to\infty$ yields the conclusion.
– Did
Feb 7, 2016 at 8:58
• $X_1^n \le X_1^{n+1}$ for all $n \in \mathbb{N}$, so $\lim_{n \to \infty} \mathbb{E} X_1^n = \mathbb{E} \lim_{n \to \infty} X_1^n = \mathbb{E} X_1 = \infty$. Feb 7, 2016 at 9:14
• More precise, for all $k \in \mathbb{R}$: $n^{-1}S_{n} \ge n^{-1}S_{n}^{k}→E(X_{1}^{k})$. So in particular you can consider the $\limsup$, which is the limit if it exists to get $n^{-1}S_{n}=\limsup_{k→\infty}n^{-1}S_{n}^{k}→\limsup_{k→\infty}E(X_{1}^{k}) = \lim_{k \to \infty} E(X_1^k) =E(X_{1}) = \infty$. Feb 7, 2016 at 10:12
• Yes, that is monotone convergence. Feb 7, 2016 at 13:14
• @user428487 We do, but we can consider the R.V. $Y^k = \begin{cases} X^k_1 & X_1 > 0 \\ 0 & X_1 <0 \end{cases}$. Then we have $X^k_1 = Y^k - X_1^-$, thus we can use linearity of the expectation to split the expectation in $Y^k$ and $X_1^-$, where the latter is independent of $k$, so we can take the limit inside the expectation. For the former, $Y^k$, we can use MCT to switch limit and expectation. Then take the two expectations together again, and we have switched the expectation and limit. Jun 1, 2018 at 14:00