Almost sure convergence and equivalent definition

Question

First of all this question has related answers here and here, but I am still struggling to understand the proof and the nuances that go along with it.

Show that a.s convergence of a sequence of random variables $X_k$ to a constant $\mu$ is equivalent to the condition that for all $\epsilon > 0$: $$ \lim_{n\to\infty}\mathbb{P}[\omega:\sup_{k>n}|X_k(\omega) - \mu|>\epsilon] = 0$$

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I have interpreted the proof in the second link as follows. I fixed $\epsilon>0$ as given and defined,

$${A_k}: = \left\{ {\left| {{X_k} - \mu } \right| \geq \epsilon } \right\}\qquad {\text{and }}\qquad {B_n}: = \bigcup\limits_{k \geqslant n} {{A_k}} $$

Then if $B = \bigcap\limits_{n \geq 1} {{B_n}}$ we have $P\{B\} = 0$ since, $$P\left\{ {\mathop {\lim }\limits_{n \to \infty } {X_n} = \mu } \right\} = 1\quad \Rightarrow \quad P\left\{ {\mathop {\lim }\limits_{n \to \infty } {X_n} \ne \mu} \right\} = 0{\text{ }}\quad {\text{or}}\quad {\text{ }}\mathop {\lim }\limits_{n \to \infty } {X_n}=\; \left({\text{D.N.E}}\right)$$ Then, $$\begin{aligned} P\left\{ B \right\} & = 0 \\ & = P\left\{ {\bigcap\limits_{n \geqslant 1} {\bigcup\limits_{k > n} {{A_k}} } } \right\} = P\left\{ {\mathop {\lim \sup }\limits_{k \to \infty } {A_k}} \right\} = P\left\{ {\mathop {\lim }\limits_{n \to \infty } \mathop {\sup }\limits_{k > n} {A_k}} \right\} \\ &= \mathop {\lim }\limits_{n \to \infty } P\left\{ {\mathop {\sup }\limits_{k > n} \left| {{X_k} - \mu } \right| \geqslant \varepsilon } \right\} \hfill \\ \end{aligned}$$ But I am not sure how the last step - pulling the limit out of the probability - is justified.

I understand that $B_n$ are decreasing sets, but the continuity of probability result tell us that the limit outside becomes a union or intersection when taken inside. But this seems to be a limit remaining a limit when taken outside. Is it true that we can always pull a limit outside?

As for the converse I do not understand it at all, and my attempts to define similar sets $A_k,B_n$ etc. have failed. I know that the condition gives us, $$ \mathop {\lim }\limits_{n \to \infty } P\left\{ {\mathop {\sup }\limits_{k > n} \left| {{X_k} - \mu } \right| < \varepsilon } \right\} = 1 \quad \forall \varepsilon > 0$$ But I don't know how to proceed from here. My attempts have been to take the limit inside and show somehow that $\mathop {\lim \sup }\limits_{k \to \infty } \left| {{X_k} - \mu } \right| \leq 0$. Since $\mathop {\lim \inf }\limits_{k \to \infty } \left| {{X_k} - \mu } \right| \geq 0$ if we show the previous, then the limit exists and the desired equality : $ {\mathop {\lim }\limits_{k \to \infty } \left| {{X_k}} \right| = \mu } $ is achieved.

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Any help would be much appreciated. I have been mulling over this for three days, and I loathe to post questions that are similar to ones already I asked, but I really have not been able to crack it or find a source that enumerates and explains each step and why it is justified.

Answer 1

As mentioned in the comment, the key is to show $$\lim \sup_{k \to \infty} A_k = \{\lim \sup_{k \to \infty} |X_k - \mu| \geq \epsilon \}$$ You can do that in a couple steps, showing $\cup_{k \geq n} A_k = \{\omega \, : \,\, \sup_{k \geq n} |X_k(\omega) - \mu| \geq \epsilon \}$ holds for all $n$ using a double-inclusion argument.

Then the step of pulling the limit outside the probability is OK, as you said, from monotone limit properties of probability measures.

For the other direction, take $U_N = \{\lim \sup_{k \to \infty} |X_k - \mu| \geq 1/N \}$. Then you get $$P(U_N) = \lim_n P(\sup_{k \geq n} |X_k - \mu| \geq 1/N) = 0$$ for any $N$ by assumption.

$U_N \uparrow \{\lim \sup_{k \to \infty} |X_k - \mu| > 0 \} = U$, so $P(U) = \lim_N P(U_N) = 0$. That means

$$0 \leq \lim \inf_{k \to \infty} |X_k - \mu| \leq \lim \sup_{k \to \infty} |X_k - \mu| = 0$$ with probability one, so the limit exists and is zero.

About your question in italics, it is always true that for a real-valued random variable sequence $\{f_k\}$ you have for any $a \geq 0$ $$\lim \sup_{k \to \infty} \{|f_k| \geq a \} = \{\lim \sup_{k \to \infty} |f_k| \geq a \}$$

which is what you would prove in the step mentioned in the comment. But if you switch the inequality direction it doesn't need to hold.