## Why aren't there infinitely countable sigma-algebras?

This is a supplement to one of my analysis discussion sessions.

We will now prove a fascinating result in measure theory: there are no $\sigma$-algebras that are infinitely countable. This means that any $\sigma$-algebra $S$ is either finite (and is therefore just an algebra) or very 'BIG' in cardinality, in the sense that it is uncountable. The idea behind this proof is simple. We will take a $\sigma$-algebra $S$ on a set $X$, and show that the collection of `smallest' non-trivial sets belonging to $S$ are in one-to-one correspondence with a power set of a certain infinite set.

Let's assume by contradiction that $S$ is an infinitely countable $\sigma$-algebra defined on a set $X$. The set $X$ has to be infinite as well (otherwise any $\sigma$-algebra defined on it is smaller in cardinality than its power set that is finite as well.) Define a function

mapping

Namely, the function $f$ maps $x$ to the smallest set of the $\sigma$-algebra $S$ that contains $x$. Note that this function is well-defined (and maps into $S$) exactly because of the assumption that $S$ is countable (and therefore any countable intersection remains in $S$ by the definition of a $\sigma$-algebra.) Since $S$ is a collection of sets, the image of $X$ under $f$ is a subset of the power set of $X$. In fact, it turns out that $f(X)$ provides us with a partition of $X$.

Let's try to see why. Consider the images of two points $x,y$ under $f$ and assume by contradiction they have a non-trivial intersection $f(x) \cap f(y) \neq \emptyset$. If $x \notin f(y)$ then $f(x) \setminus f(y) \in S$ is a smaller set in the $\sigma$-algebra containing $x$ which is a contradiction to the definition of $f$. Therefore $x \in f(y)$ and by the same argument $y \in f(x)$.  But then since $f$ maps a point to the smallest set in the $\sigma$-algebra containing it, $x \in f(x) \subseteq f(y)$ and $y \in f(y) \subseteq f(x)$ and thus we conclude that $f(x)=f(y)$.

We may conclude that $f(X)$ is a partition of $X$. Each set $A$ in $S$ can be written as the union of such images,

Therefore, the partition $f(X)$ cannot be finite, as otherwise $S$ would be finite as well. But now if the partition $f(X)$ is infinite, one can form all the sets in $S$ by taking all the possible (disjoint) unions of sets in $f(X)$. This means that the cardinality of $S$ is equal to the cardinality of the power set of $f(X)$, namely

which is of course uncountable.

This is a contradiction, and therefore the $\sigma$-algebra $S$ cannot be infinitely countable.