Almost sure convergence difference from limit existence? I'm having a bit of trouble understanding how the statements:
$$
P(\lim_{n\rightarrow \infty}X_{n}=X) = 1 \equiv P(\{\omega \in \Omega: \lim_{n\rightarrow \infty}X_{n}(\omega) = X(\omega)\}) = 1
$$
Are not simply equivalent to:
$$
\lim_{n\rightarrow \infty}X_{n}(\omega) = X(\omega)
$$
Which is a statement about the limit of the sequence of random variables $X_{n}$. It seems reasonable that $\lim_{n\rightarrow \infty}X_{n}(\omega) = X(\omega)$ would imply almost sure convergence. I'm guessing it doesn't go the other way, does it? If it doesn't, is that solely attributable to the way we define our probability measure or am I missing something here?
 A: The problem is that in probability theory, there are multiple kinds of convergence. Almost sure convergence, convergence in probability, and others. The problem with your second formula is that it is not a complete statement, the $\omega$ in that statement, what does it refer to? Some specific, chosen $\omega$? Probably not ... but you need to make that statement precise, so you need to quantify over $\omega$ ... If you complete the statement saying that it is true for all $\omega\in A$, where $A$ is an event with probability one, or that some such set $A$ exists, then it is almost sure convergence.
For details see Convergence in probability vs. almost sure convergence
A: Have a look at the definition and its implications:
Let $(\Omega, \boldsymbol{\mathfrak A}, \mathbb P) $ be a probability space. Let $\langle X_n\rangle_{n\in\mathbb N}$ be a sequence of extended real-valued $\boldsymbol{\mathfrak A}$-measurable functions on a set $A\in\boldsymbol{\mathfrak A}.$
Now what does
$$ \mathbb P\{\lim_{n\to\infty}X_n=X\}= 1\tag 1\label 1$$ mean?
Almost surely convergence $\eqref 1$ means there exists a null set $N\subset A$ such that $\lim_{n\to\infty}X_n(\omega)$ exists for every $\omega\in A\setminus N. $
Note if $A_e\subseteq A$ denotes the subset of $A$ containing all the points $\omega\in A$ for which $\lim_{n\to\infty}X_n(\omega)$ exists, then $A\setminus A_e\subset N. $ That is, $\mathbb P\{A:X_n\not\rightarrow X\}= 0.$
If $A_e=A, $ then $A\setminus A_e =\emptyset.$
