I'm stuck in this exercise of Robert Ash's probability book:
Let $X_1,X_2,\ldots $ iid random variables. If $\tfrac1{n}\sum_{k=1}^nX_k$ converges a.e. to a finite limit, show that $\mathrm{E}[X_1]$ is finite and the limit equals $\mathrm{E}[X_1]$ a.e.
What I did: let $\bar X_n:=\frac1{n}\sum_{k=1}^n X_k$, then as the function defined by $Y:=\limsup_{n\to\infty}\bar X_n$ is a tail function then it can be shown that it is constant a.e., and note that $Y=\lim_{n\to\infty}\bar X_n$ a.e., so if the latter is finite a.e. then necessarily it must converge to the constant that defines $Y$, that is, we had shown that $\{\bar X_n\}_{n\in \mathbb N}$ converges a.e. to a constant and finite value.
However, I'm stuck trying to prove that $\mathrm{E}[X_1]$ exists, what I tried is to use the fact that if $\{\bar X_n\}_{n\in \mathbb N}$ converges a.e. then it converges also almost uniformly and in measure. But I dont find something that help me to show that $\mathrm{E}[X_1]$ exists. Some help will be appreciated, thank you.
Strong law of large numbers in appriopate form also holds for $\mathbb E[|X|] = \infty$. That said, let's assume $\mathbb E[|X_1|]$ is infinite, but $\frac{S_n}{n}$ converges almost surely to finite limit.
Then $0 \le |\frac{X_n}{n}| = |\frac{S_n}{n} - \frac{S_{n-1}}{n}| = |\frac{S_n}{n} - \frac{n-1}{n}\frac{S_{n-1}}{n-1}| \to 0 $ almost surely.
But by assumption $\infty = \mathbb E[|X_1|] \le 1+ \sum_{n=1}^\infty \mathbb P(|X_1| > n) =1 + \sum_{n=1}^\infty \mathbb P(|X_n| > n)$, where we used the same distribution. Now using independence and borel cantelli, it follows that $|X_n| > n$ infinitelly often almost surely, so that it cannot hold that $\frac{X_n}{n} \to 0$ almost surely, a contradiction. Hence $\mathbb E[|X_1|] < \infty$.