Probability that inequality holds infinitely often implies it holds almost surely I have been working on some probability problems out of Durrett and in searching for answers I have come upon this sort of explanation a lot.

Let $X_1, X_2, \dots$ be i.i.d. random variables. Suppose that $P(X_n>a \text{ i.o. })=1$ then we have that $\limsup_{n\rightarrow\infty} X_n > a$ almost surely. Note: here i.o. means infinitely often.

I'm not exactly sure why this has to be true. Could anyone explain this to me or provide me with a source that does?
Thanks!
 A: If infinitely many terms of a sequence $y_1,y_2,y_3,\ldots$ are $>a$ then $\limsup\limits_{n\to\infty} y_n \ge a$.  That statement can be made without knowing anything about probability.  Here we have to say $\text{“ }\ge\text{ ''}$ in the conclusion even though we say $\text{“ }>\text{''}$ in the hypothesis, since, for example, we could have a decreasing sequence whose limit is $a$.
The proposed conclusion is a bit stronger: it says $\text{“ }>a\text{ ''}$. That's where $\text{“ i.i.d. ''}$ comes in.  If the limsup is equal to $a$, then for every $b>a$, only finitely many terms are $\ge b$.  That can't happen with an i.i.d. sequence unless the probability of being $\ge b$ is $0$.  Thus for every $b\ge a$, we would have $\Pr(X_1\ge b)=0$.  And then
$$
\Pr(X_1>a) = \sum_{n=1}^\infty \Pr\left( a + \frac 1 {n+1} < X_1 \le a + \frac 1 n \right) = 0.
$$
A: Let me assume that you mean that the probability that $X_n \sim X$ is more than $a$ for infinitely many natural $n$ is one. Let $p$ be the probability that $X > a$. Then $p > 0$, otherwise the probability that $X_n > a$ infinitely often would be $0$. Thus the limit supremum of $(X_n)_{n\in\mathbb{N}}$ is more than $a$, otherwise for any positive $ε$ there is a point after which the sequence is bounded above by $a+ε$, which gives $X > a+ε$ with probability $0$, which implies $a < X \le a+ε$ with probability $p$, which is clearly impossible when you consider $ε \to 0$.
