Uncountability in smallest sigma algebra

Question

There are a bunch of questions and answers about the "smallest $\sigma$-algebra" argument. My question is focused on uncountability.

Claim: Given a set $\Omega$ and a collection $A$ of subsets of $\Omega$, then there is a smallest $\sigma$-field containing $A$.

I think the argument is something like:

1. take the set $S$ of subsets $S_i$ of $\Omega$ with $A\subseteq S_i$.
2. $S$ has a $\sigma$-algebra, so you can make another $\sigma$-algebra $\bigcap_i S_i$ by intersecting its components
3. that new $\sigma$-algebra is big enough: $A \subseteq \bigcap_i S_i$
4. that new $\sigma$-algebra is small enough: all others are bigger

My questions are

• In (4), $\sigma$-algebras are only preserved by countable intersections, not uncountable ones. How does this work then?

• In (3), how do we know that?

Answer 1

You are not doing this quite right.

Let $A$ be a collection of subsets of $\Omega$. So $A\subseteq \mathbf{P}(\Omega)$, the power set of $\Omega$.

Now, given a $S$ subset of $\mathbf{P}(\Omega)$ we can ask two questions:

1. Is $S$ a $\sigma$-algebra?
2. Does $S$ contain $A$ as a subset?

For any subset $S$, the answer to each of those questions is either "yes" or "no". So now we can form a collection of subsets of $\mathbf{P}(\Omega)$, called $\mathscr{S}$, as follows: $$\mathscr{S} = \{S\subseteq\mathbf{P}(\Omega)\mid S\text{ is a }\sigma\text{-algebra and }A\subseteq S\}.$$ That is, the subsets of $\mathbf{P}(\Omega)$ for which the answer to both question is "yes".

Now, say you have two elements $S_1$ and $S_2$ of $\mathscr{S}$; these are $\sigma$-algebras that contain $A$. Question: is $S_1\cap S_2$ (that is, all subsets of $\Omega$ that are both in $S_1$ and in $S_2$) a $\sigma$-algebra that contains $A$? Well, clearly, since $A\subseteq S_1$ and $A\subseteq S_2$, then $A\subseteq S_1\cap S_2$. Proving that the intersection is a $\sigma$-algebra is more involved, but doable. In fact:

Lemma. Let $\{S_i\}_{i\in I}$ be a nonempty family of subsets of $\mathbf{P}(\Omega)$ such that $S_i$ is a $\sigma$-algebra on $\Omega$ for each $i\in I$. Then $$T=\bigcap_{i\in I}S_i = \{X\subseteq \Omega\mid X\in S_i\text{ for all }i\in I\}$$ is a $\sigma$-algebra on $\Omega$, regardless of the cardinality of $I$.

To prove this you should verify that $T$ is indeed a collection of subsets of $\Omega$; that $\Omega\in T$ (or equivalently, that $\Omega\in S_i$ for all $i$); that if $X\in T$ then $\Omega-X\in T$; and that if $\{X_j\}_{j=1}^{\infty}$ is a countable collection of subsets of $\Omega$, $X_j\subseteq \Omega$ for all $j$, and $X_j\in T$ for all $j$, then $\cap X_j\in T$. (This is the countable intersection issue).

This is regardless of the cardinality of $I$. Note that the intersection that creates $T$ is not an intersection of sets in a $\sigma$-algebra, it is an intersection of $\sigma$-algebras. That's why the fact that $\sigma$-algebras are closed under countable intersection only is irrelevant: you aren't doing an intersection of sets in a given $\sigma$-algebra.

You say:

"$\sigma$-algebras are only preserved by countable intersections"

and the problem, in part, is that this shorthand is confusing you. The correct statement is that $S$ is a single, particular $\sigma$-algebra, then an countable intersection of elements of $S$ must be in $S$, but an arbitrary, not-countable intersection of elements of $S$ need not be in $S$. That is, the property if being a set in $S$ is preserved under countable intersections, but not necessarily under intersections with uncountably many factors. Note that it is not $S$ that is "preserved", it's the property of being an element of $S$.

And, again, this doesn't apply here because we are not talking about intersecting elements of a particular $\sigma$-algebra, we are talking about intersecting different $\sigma$-algebras to get a new collection of sets.

Once you prove the Lemma, the way we proceed to construct the $\sigma$-algebra generated by $A$ is the following:

1. Consider the collection $\mathscr{S}$.
2. Note that $\mathscr{S}$ is not empty, because $\mathbf{P}(\Omega)$ itself is a $\sigma$-algebra on $\Omega$ that contains $A$.
3. Apply the Lemma to construct $T$.
4. Show that $A\subseteq T$.
5. Note that if $S$ is any $\sigma$-algebra that contains $A$, then by construction $S\in\mathscr{S}$, so $T\subseteq S$.
6. Conclude $T$ is the "smallest" $\sigma$-algebra that contains $A$.

See also this previous question for general remarks about this kind of procedure, which shows up often in mathematics.