On the definition on almost sure convergence

Question

I am a bit confused about the definition of almost sure convergence. Of course, we say that $X_n \underset{n \to \infty}{\longrightarrow} X$ almost surely if $$ \mathbb{P} \big( \{ \lim_{n \to \infty} X_n = X \} \big) = 1 $$ The almost sure event of convergence is $$ \big\{ \lim_{n \to \infty} X_n = X \big\} = \big\{ \lim_{n \to \infty} | X_n - X | = 0 \big\} = \big\{ \forall \, \varepsilon > 0 \quad \exists \, n \geq 1 : \forall \, m \geq n \quad |X_m - X| \leq \varepsilon \big\} $$ and of course the "$\forall \, \varepsilon > 0$" is an uncountable intersection.

But the wikipedia page on the topic says that almost sure convergence can be equivalently expressed as $$ \mathbb{P} \big( \big\{ \limsup_{n \to \infty} |X_n - X| > \varepsilon \big\} \big) = 0 \quad \forall \, \varepsilon > 0 $$ which is of course the same as $$ \mathbb{P} \big( \big\{ \limsup_{n \to \infty} |X_n - X| \leq \varepsilon \big\} \big) = 1 \quad \forall \, \varepsilon > 0 $$ that is, $$ \forall \, \varepsilon > 0 \quad \mathbb{P} \big( \big\{ \exists \, n \geq 1 : \forall \, m \geq n \quad |X_m - X| \leq \varepsilon \big\} \big) = 1 $$ But (because of the intersection being uncountable) this seems different than the statement that I got from the definition of limit, which is $$ \mathbb{P} \big( \big\{ \forall \, \varepsilon > 0 \quad \exists \, n \geq 1 : \forall \, m \geq n \quad |X_m - X| \leq \varepsilon \big\} \big) = 1 $$ and I feel stuck on this point.

Moreover, I suspect that in the first statement (with "$\forall \, \varepsilon > 0$" outside) $n$ should be intended as random, while in the other statement $n$ should be taken as deterministic. However, I wouldn't see any sign that this should be the case beyond mere speculation. I would say that $n$ should be intended as random the whole time.

Any help would be much appreciated.

Answer 1

Define events ( in some $(\Omega,\mathcal F,P)$ ) $$ B_n(\varepsilon) := \{|X_n-X|>\varepsilon\} \quad (\varepsilon >0, n\in\mathbb N) $$ as well as $$ B(\varepsilon) := \limsup B_n(\varepsilon) = \{|X_n-X|>\varepsilon\ \mbox{i.o}\}. $$ Let $B := \{X_n\to X\}$. Note that $$ B^c = \bigcup _{\varepsilon>0} B(\varepsilon) \overset{*}= \bigcup _{k=1}^\infty B\left(\frac{1}{k}\right)\in\mathcal F.\tag{1}\label{eq:1} $$

Suppose $X_n\to X$ a.s. Then, $P(B)=1$ together with \eqref{eq:1} implies that $P(B(\varepsilon))=0$ for all $\varepsilon >0$.

Conversely, if $P(B(\varepsilon)) =0$ for every $\varepsilon >0$, then, again by \eqref{eq:1}, $P(B) = 1$.

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The equality * follows from $\delta \leqslant \varepsilon \Rightarrow B(\varepsilon)\subseteq B(\delta)$

Answer 2

Note that, for $Z_n = |X- X_n|,$ the statement $\forall \varepsilon, P(\limsup Z_n > \varepsilon) = 0 $ is equivalent to $\forall \varepsilon > 0, P(\limsup Z_n \le \varepsilon) = 1$, but is not equivalent to $\forall \varepsilon > 0, P(\liminf Z_n \le \varepsilon) = 1$, which is a weaker statement. As a counterexample, consider $Z_n$ that are independent, with $P(Z_n = 0) = 1 -P(Z_n = 1) = 1/n$ (I'll leave it to you to work out the difference).

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Given this, the equivalence demanded is that $P(Z_n \to 0) = 1 \iff \forall \varepsilon > 0, P(\limsup Z_n > \varepsilon) = 0$. But notice by using the nonnegativity of $Z_n$ that $\{Z_n \to 0\}^c = \{\exists \varepsilon > 0 : \limsup Z_n > \varepsilon \}$. So the question is the same as showing

$$ P(\exists \varepsilon > 0: \limsup Z_n > \varepsilon) = 0 \iff \forall \varepsilon > 0, P(\limsup Z_n > \varepsilon) = 0. $$ This is not obvious because the second term "pulls out" the $\varepsilon$ from within the probability.

Of course, the forward implication in is easy: for any $\varepsilon > 0,$ $\{\limsup Z_n > \varepsilon\} \subset \{\exists \varepsilon > 0: \limsup Z_n > \varepsilon\},$ and the latter event having probability $0$ implies the former to have the same. So the question is why the reverse implication holds.

For this, the main observation is that $$ \{\exists \varepsilon > 0 : \limsup Z_n > \varepsilon\} = \{ \exists \varepsilon > 0, \varepsilon \in \mathbb{Q}: \limsup Z_n > \varepsilon\}, $$ which holds because for any $x > 0,$ there is always a rational in the interval $(0,x]$. Let me denote $\mathbb{Q}_+ = \mathbb{Q} \cap (0,\infty)$.

Now, observe that due to the countability of the union over $\mathbb{Q}_+$, $$P(\exists \varepsilon \in \mathbb{Q}_+ :\limsup Z_n > \varepsilon) \le \sum_{\varepsilon \in \mathbb{Q}_+} P(\limsup Z_n > \varepsilon). $$ But notice that under the hypothesis, the sum is $0$ for each $\varepsilon > 0$ and hence for each $\varepsilon \in \mathbb{Q}_+$, and thus the hypothesis implies that $P(\exists \varepsilon > 0: \limsup Z_n > \varepsilon) = 0$.