Measurability and almost sure convergence, going from epsilon to 1/j

Question

I am trying to understand the definition of almost sure convergence. For this one needs to check if there are no measurability issues. By transforming the definition of convergence to set theory language one arrives at:

$$ \bigcap_{\epsilon>0} \bigcup_{N=1}^{\infty} \bigcap_{n=N}^{\infty} \{\omega: | X_{n} - X| \leq \epsilon \} $$

The definition stated like this seems to have a huge issue. It isn't necessarily measurable since it contains an uncountable intersection (Why is this a problem? Would like more insights on this since from my understanding the infite (still countable) intersections of sets that are in the sigma-algebra are also in the sigma-algebra. This is not true for uncountable). In turn, this equivalent definition (Why is this equivalent?!).

$$ \bigcap_{j=1}^{\infty} \bigcup_{N=1}^{\infty} \bigcap_{n=N}^{\infty} \{ \omega: | X_{n} - X| \leq \frac{1}{j} \} $$

is a measurable set! Now it supposedly makes sense to take the probability of this set. This exchange does look intuitive and clear. But the passing from a uncountable to a countable intersection is not.

Answer 1

Why is this a problem? [...] this is not true for uncountable [intersections]

You've answered your own question. In a general sigma algebra we cannot be sure that an uncountable intersection belongs to the sigma algebra, and if not handled carefully this could lead to paradoxes e.g. sets for which translating them does not change their volume. We must exclude non-measurable sets so we must try to re-state the convergence definition in a countable way.

So why does it matter if generic $\epsilon>0$ are replaced with $(1/j)_{j\in\Bbb N}$? Well, you should be quite used to this from analysis. Let's express the two sets back into words:

$$\begin{align}&\tag{1}\bigcap_{\epsilon>0}\bigcup_{N\ge1}\bigcap_{n\ge N}\{\omega:|X_n-X|\le\epsilon\}\\&\tag{2}\bigcap_{j\ge1}\bigcup_{N\ge1}\bigcap_{n\ge N}\{\omega:|X_n-X|\le1/j\}\end{align}$$

Set $(1)$ contains all $\omega$ satisfying:

For all $\epsilon>0$ there exists some natural number $N$ such that for all natural numbers $n\ge N$, we have $|X_n-X|\le\epsilon$

Set $(2)$ contains all $\omega$ satisfying:

For naturals $j$ there exists some natural number $N$ such that for all natural numbers $n\ge N$, we have $|X_n-X|\le1/j$

Clearly set $(1)$ is contained in set $(2)$, since we may take $\epsilon=1/j$ in the definition. Suppose $\omega$ is in set $(2)$. Take any $\epsilon>0$. By the 'Archimedean property', there is some $j\in\Bbb N$ with $0<1/j<\epsilon$. But then because $\omega$ is in set $(2)$, there is an $N$ such that if $n\ge N$ we have $|X_n-X|\le1/j<\epsilon$. So, if $n\ge N$ we have $|X_n-X|\le\epsilon$. Great! Because $\epsilon$ could be any positive real, we find $\omega$ is also in set $(1)$.

Therefore $(1)=(2)$. As commented, you could replace the $(1/j)_{j\in\Bbb N}$ with any strictly positive sequence that converges to zero.

This kind of thing is very useful in probability. I've seen similar things where the definition of continuity is expressed set-theoretically and then we pass from $\Bbb R$ (uncountable) to $\Bbb Q$ (countable) using a density argument, so we can express the event "is continuous" in the sigma algebra.

Answer 2

The relevant intuition behind the transformation is the same for $$ \bigcap_{\epsilon>0} \{\omega: | X_{n} - X| \leq \epsilon \} = \bigcap_{j=1}^\infty \{\omega: | X_{n} - X| \leq \frac1j \} $$

Let $M_\epsilon:= \{\omega: | X_{n} - X| \leq \epsilon \}$. The key aspect is that now that $$ \forall a,A>0:\quad a<A\Rightarrow M_a\subseteq M_A $$

Let $S\subseteq \mathbb R$. Let's look at the cut $\bigcap_{\epsilon\in S} M_\epsilon$. Then if $a,A\in S$ with $a<A$, we have $\bigcap_{\epsilon\in S} M_\epsilon = \bigcap_{\epsilon\in S\setminus\{A\}} M_\epsilon$, that is, if $a\in S$, then we can remove all elements $A\in S$ with $A>a$.

So, if we could directly set $\epsilon$ infinitely small but not 0, we'd only have to look at this single set to find the result of the cut $\bigcap_{\epsilon\in S} M_\epsilon$ - but such a convenient number doesn't exist, so we instead take some sequence $(a_n)_{n\in\mathbb N}$ with $a_n\to 0$, and build the cut over all elements of the sequence.

That is $$ \bigcap_{\epsilon>0} M_\epsilon = \bigcap_{n\in\mathbb N} M_{a_n} $$ , because for every $\epsilon>0$ we can find an $n\in\mathbb N$ such that $a_n<\epsilon$.