# Uncountability in smallest sigma algebra

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?

• Your statement of 1 is incorrect. You take subsets of the power set of $\Omega$, not subsets of $\Omega$. A subset of $\Omega$ will not, in general contain a collection of subsets of $\Omega$. Commented Mar 10, 2021 at 23:28

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 $$A$$.