From Wikipedia https://en.wikipedia.org/wiki/Convergence_of_random_variables
$X_n \to X$ almost surely if $\forall\epsilon>0$
\begin{equation}P\left(\lim\inf_{n\to\infty}\{|X_n-X|<\epsilon\}\right)=1 \tag{1}\end{equation} which I believe is equivalent to \begin{equation}P\left(\bigcap_{m\ge1}\bigcup_{n\ge 1}\bigcap_{k\ge n}\{|X_k-X|<\frac{1}{m}\}\right)=1\tag{2}\end{equation}
Thus the complement is
$$P\left(\bigcup_{m\ge1}\bigcap_{n\ge 1}\bigcup_{k\ge n}\{|X_k-X|\ge\frac{1}{m}\}\right)=0\tag{3}$$
Which is equivalent to $$P\left(\bigcup_{m\ge1}\lim\sup_{n\to \infty}\{|X_k-X|\}\ge\frac{1}{m}\right)=0\tag{4}$$
Which is equivalent to $\underline{\exists\epsilon>0} $ such that
$$P\left(\lim\sup_{n\to \infty}\{|X_k-X|\ge\epsilon\}\right)=0 \tag{5}$$
However according to this post Almost sure convergence and lim sup
It is actually $X_n \to X$ almost surely if $\underline{\forall\epsilon>0} $
$$P\left(\lim\sup_{n\to \infty}\{|X_k-X|\ge\epsilon\}\right)=0 \tag{6}$$
Part 2) At https://en.wikipedia.org/wiki/Set-theoretic_limit#Almost_sure_convergence $$P\left(\bigcup_{k\ge 1}\bigcap_{n\ge 1}\bigcup_{j \ge n} \{|Y_j – Y| > \frac{1}{k}\}\right)=0$$ seems to be given as a condition for almost sure convergence. Please let me know if my understanding is incorrect.
I would greatly appreciate help in sorting out this contradiction as I have been mulling over it for a long time.
Best Answer
(Note I fixed a misplaced brace in the first statement, and changed $n$ to $k$ which I think is what you mean.)
Those are not equivalent.
What is true is that the former would be equivalent to
$$P\left(\exists \epsilon > 0 : \limsup_{n\to > \infty}\{|X_k-X|\}\ge\epsilon\right)=0.$$
This says, more precisely, that the following set has measure zero: the set of all $\omega \in \Omega$ for which there exists $\epsilon > 0$ such that $$\limsup_{n\to > \infty}\{|X_k(\omega)-X(\omega)|\}\ge \epsilon.$$ Note the order of quantifiers: $\epsilon$ is allowed to depend on $\omega$.
In your second statement, you are asking that there be a single $\epsilon$ that works for almost every $\omega$ simultaneously. That is a much stronger condition, and so it is much easier for the corresponding set of $\omega$ to be null.
To see the problem more vividly, consider the sequence $X_k = 1$ for all $k$, and $X=0$. Using your first statement, we can see that we do not have $X_k \to X$ a.s, because for every $m > 1$ and every $k$, the set $\{|X_k - X| \ge \frac{1}{m} \}$ is all of $\Omega$. So the $\limsup_{k \to \infty}$ of these sets is also $\Omega$ for all $m > 1$, and so the union over all $m$ is also $\Omega$. Thus the probability is $1$.
But using the second statement, suppose I take $\epsilon = 2$. Then the set $\{|X_k - X| \ge \epsilon\}$ is empty for all $k$, so the limsup is also empty. The probability of the empty set is 0. So your second statement is satisfied in this example, even though $X_k$ does not converge to $X$ a.s.
Update
In your current numbering, (1), (2), (3), (4), (6) are all equivalent to one another, and (5) is weaker.
To see why (1) is equivalent to (2), for notational convenience let's write $$A_\epsilon = \limsup_{k \to \infty} \{|X_k - X| <\epsilon\} = \bigcup_{n \ge 1} \bigcap_{k \ge n} \{|X_k - X| < \epsilon\}.$$ Then (1) reads $\forall \epsilon > 0 \, P(A_\epsilon) = 1$, and (2) reads $P(\bigcap_{m \ge 1} A_{1/m}) = 1$.
Suppose that (1) holds. Then $P(A_{1/m}) = 1$ for all $m \ge 1$. By countable additivity, it follows that $P(\bigcap_{m \ge 1} A_{1/m}) = 1$ as well, so we have (2). (One way to see this is that $(\bigcap_{m \ge 1} A_{1/m})^c = \bigcup_{m \ge 1} A_{1/m}^c$ which is a countable union of measure-zero sets.)
Conversely, suppose that (2) holds. Given $\epsilon > 0$, choose $M$ so large that $1/M < \epsilon$, and note that $A_{1/M} \subseteq A_\epsilon$. On the other hand, $\bigcap_{m \ge 1} A_{1/m} \subset A_{1/M}$ since $A_{1/M}$ is one of the sets being intersected on the left. So by monotonicity of measure, we have $$1 = P\left(\bigcap_{m \ge 1} A_{1/m}\right) \le P(A_{1/M}) \le P(A_\epsilon)$$ and thus $P(A_\epsilon) = 1$. Since $\epsilon > 0$ was arbitrary, (1) is proved.
To look at (4) and (5), set $$B_\epsilon = A_\epsilon^c = \limsup_{k \to \infty} \{|X_k - X| \ge \epsilon\} = \bigcap_{n \ge 1} \bigcup_{k \ge n} \{|X_k - X| \ge \epsilon\}.$$ Then your (4) reads $P(\bigcup_{m \ge 1} B_{1/m}) = 0$ and (5) is $\exists \epsilon > 0 \, P(B_\epsilon) = 0$.
We do have (4) implies (5): take $\epsilon = 1$. Then $B_1 \subset \bigcup_{m \ge 1} B_{1/m}$, so $P(B_1) \le P(\bigcup_{m \ge 1} B_{1/m}) = 0$. But (5) does not imply (4); see the counterexample I gave above.
Why can't you prove (5) implies (4) in a similar way to the proof that (1) implies (2)? Just try and you'll see why it doesn't work. Maybe you know that one of the sets $B_{1/m}$ has measure zero (if you are lucky and the $\epsilon$ that works in (5) is smaller than 1). But we have $B_1 \subseteq B_{1/2} \subseteq B_{1/3} \subseteq \dots$ and so knowing that one of them has measure zero does not tell you anything about the measure of the later sets in the sequence.
You might like to compare the following statements. Suppose $C_1, C_2, \dots$ is a sequence of events.
(a) If $P(C_n) = 1$ for every $n$, then $P(\bigcap_n C_n) = 1$. (TRUE)
(b) If $P(C_n) = 0$ for every $n$, then $P(\bigcup_n C_n) = 0$. (TRUE)
(c) If $P(C_n) = 0$ for some $n$, then $P(\bigcup_n C_n) = 0$. (FALSE)
You were probably thinking of something like (c) when you thought that (5) would imply (4). You might think that "symmetry" should suggest (a) implies (c), but it clearly doesn't.