Strong law of large numbers when the expectation is infinite

Question

Let $\{X_n\}$ be a sequence of i.i.d. positive random variables and $S_n = X_1 + \cdots + X_n$. If $\mathbb{E}(X_1)=\infty$, then we have that $$ \limsup_{n \to \infty} \frac{S_n}{n} = \infty ~~\text{a.e.}~. $$ Can we prove the stronger result that $$ \lim_{n \to \infty} \frac{S_n}{n} = \infty ~~\text{a.e.}~ ?$$

Answer 1

For fixed $k \in \mathbb{N}$ define $Y_n := \min\{X_n,k\}$. Then $Y_n$, $n \geq 1$, are iid random variables with finite mean and so, by the strong law of large numbers

$$\lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n Y_i = \mathbb{E}(\min\{X_1,k\}) \quad \text{a.s.}$$

From $X_n \geq Y_n$, we find that

$$\liminf_{n \to \infty} \frac{1}{n} \sum_{i=1}^n X_i \geq \liminf_{n \to \infty} \frac{1}{n} \sum_{i=1}^n Y_i = \mathbb{E}(\min\{X_1,k\})$$

with probability $1$. As $k \geq 1$ is arbitrary, an application of the monotone convergence theorem yields

$$\liminf_{n \to \infty} \frac{1}{n} \sum_{i=1}^n X_i \geq \sup_{k \geq 1} \mathbb{E}(\min\{X_1,k\}) = \mathbb{E}(X_1)=\infty.$$

Consequently, $\frac{1}{n} \sum_{i=1}^n X_i \to \infty$ almost surely.