Let $X_n$ be a sequence of independent real valued random variables on the same event space, with the same (finite) mean $\mu$.
Suppose that for almost every couple of points $(\omega,\omega')$ in the squared space of events, we have
$$
X_n(\omega) – X_n(\omega') \to 0.
$$
Is it true that $X_n$ converges to $\mu$ almost surely?
And if it is false, is it true at least that there exists a random variable $X$ that is an almost surely limit for the sequence?
Another equivalent way we can formulate the question is as follows:
consider two sequences of independent random variables $X_n$ and $Y_n$ where for every $n$, $X_n\equiv Y_n$ (meaning they have the same distribution), and all the variables have the same finite mean $\mu$.
If $X_n-Y_n\to 0$ almost surely, is it true that $X_n\to\mu$ almost surely? or that there exists a random variable $X$ that is an almost sure limit for $X_n$?
The idea here should be that every $X_n$ tends to be a constant function as $n$ goes to infinity, since the couples $(\omega,\omega')$ where $X_n$ differs more than $\varepsilon$ are few.
Since the mean is the same for every $X_n$, then the constant it converges to should be the mean.
The weak point is that the mean can actually be manipulated by big deviation on very small events, but it should be possible to handle from the fact that the variables are independent.
Best Answer
What we can say is that $X_n-y_n$ converges almost surely for some sequence $(y_n)$ of real numbers. (To say that $(X_n)$ itself converges you need some addition hypothesis).
Sketch of proof: Let $X=(X_n)$ and $Y=(Y_n)$. These are independent $\mathbb R^{\infty}$ valued random vectors. Let $E$ be the set of all Cauchy sequences in $\mathbb R^{\infty}$. The $E$ is a measurable set and we have $P(X-Y\in E)=1$. Fubini's Theorem shows that there exists some sequence $y=(y_n)$ such that $P(X-y\in E)=1$. Hence $X_n-y_n$ converges almost surely.