# Arithmetic average of random variable with infinite mean

I am doing a problem sheet related to the Law of large numbers and am given the following problem.

Assume now that the $X_n$ are non-negative and that their expectation is infinite. Let $R ∈ (0,∞)$. What does the strong law of large numbers say about the limiting behaviour of $S_n^R/n$ where $S_n^R = X_1\cdot1_{(X_1<R)}+...+X_n\cdot1_{(X_n<R)}$. Deduce that $$S_n/n \to\infty$$almost surely.

Now I see that the Strong Law of Large Numbers implies that $S_n^R/n \to E(X_1\cdot1_{(X_1<R)})$ which in turn $\to\infty$ as $R\to\infty$. But I cannot figure out how to deduce that $S_n/n \to\infty$. If I could simply interchange the limits of $n\to\infty$ and $R\to\infty$ it would follow immediately, but I doubt this is possible. I think I need to use Borel-Cantelli somehow, but I cannot figure out what events to use. Any tips?

• Doesn't Tonelli's thm apply since the rv's are positive? Apr 30, 2018 at 10:01
• Not sure how we would apply that here. Tonelli's theorem concerns the interchange of integrals, not of limits. Apr 30, 2018 at 10:03

Note that the $X_n 1_{\{X_n<R\}}$'s are non-decreasing in $R$, so almost surely $$\lim_{n\to\infty}\frac{S_n}{n} = \lim_{n\to\infty}\lim_{R\to\infty} \frac{S_n^R}{n} = \lim_{n\to\infty}\sup_{R} \frac{S_n^R}{n} \ge \sup_{R} \lim_{n\to\infty} \frac{S_n^R}{n} = \sup_{R} E[X_1 1_{\{X_1<R\}}] = \lim_{R\to\infty} E[X_1 1_{\{X_1<R\}}] \ge E[\lim_{R\to\infty} X_1 1_{\{X_1<R\}}] =E[X_1] = +\infty$$ where the last inequality follows by Fatou's lemma.
• Since for any $R$, $\sup_R \frac{S_n^R}{n} \ge \frac{S_n^R}{n}$, we have as well that $\lim_{n}\sup_R \frac{S_n^R}{n} \ge \lim_{n}\frac{S_n^R}{n}$ for any $R$, and therefore $\lim_{n}\sup_R \frac{S_n^R}{n} \ge \sup_R \lim_{n}\frac{S_n^R}{n}$. Apr 30, 2018 at 10:49