Why does $A_1^\text{c}$ have an infinite number of measurable subsets?

Question

Let $\mathcal{A}$ be a $\sigma$-algebra. Show that if $|\mathcal{A}| = \infty$, then $\mathcal{A}$ is uncountable.

We want to construct an infinite sequence of nonempty disjoint measurable sets. Then we use that to show that $\mathcal{A}$ contains uncountably many sets.

Let $A_1 \in \mathcal{A}$ which is neither $\emptyset$ nor $X$. Without loss of generality $A_1^\text{c}$ has an infinite number of measurable subsets.

My question is, why can we say the above bolded text? I imagine it ultimately follows from the assumption that $\mathcal{A}$ is infinite, but I'm not seeing the precise chain of logic. Could anybody help?

Answer 1

Choose some $A_1\in\mathcal{A}$ such that $A_1\neq \emptyset,X$. Then at least one of $A_1,A_1^c$ contains infinitely many elements of $\mathcal{A}$.

Indeed, suppose that both $A_1$ and $A_1^c$ contain only finitely many elements of $\mathcal{A}$. For any $E\in \mathcal{A}$ we can write $$ E=(E\cap A_1)\cup(E\cap A_1^c) $$ with $E\cap A_1,E\cap A_1^c\in\mathcal{A}$. There are therefore only finitely many possibilities for $E\cap A_1$ and only finitely many possibilities for $E\cap A_1^c$, hence $\mathcal{A}$ is finite, contrary to the hypothesis.

Finally, since we have shown that at least one of $A_1,A_1^c$ contains infinitely many elements of $\mathcal{A}$, we can interchange the roles of $A_1$ and $A_1^c$ if necessary to conclude that $A_1^c$ contains infinitely many elements of $\mathcal{A}$.

Answer 2

The hidden assertion here is that either $A_1$ or $A_1^c$ must have an infinite number of measurable subsets, so wlog assume it's the latter. The justification of this assertion is: Any set $B\in\cal A$ can be written $B=(B\cap A_1)\cup (B\cap A_1^c)$, a union of a measurable subset of $A_1$ with a measurable subset of $A_1^c$. How many possible $B$'s are there if both $A_1$ and $A_1^c$ have a finite number of measurable subsets?

Answer 3

If $A_1$ is neither $\emptyset$ nor $X$, then so is $A_1^C$. If both have only a finitely many measurable subsets, then, since every subset of $X$ can be written as the union of a subset of $A_1$ and a subset of $A_1^C$, $\mathcal{A}$ would be finite. If $A_1^C$ has infinitely many measurable subsets, we are done, otherwise we can by relabeling replace $A_1$ by $A_1^C$.

To prove the result, it is I think easiest to show that if $\mathcal{A}$ is countable, then for each $x$, the set $P(x)=\bigcap\{B\in\mathcal{A}:x\in B\}$ would be measurable as a countable intersection of measurable sets. Show that the $P(x)$ form a partition of the underlying space and that each measurable set is the union of partition blocks. If there are countably infinitily many, so the partition can be written $\Pi=\{P_1,P_2,\ldots\}$, one can find for each subset $F$ of $\mathbb{N}$ a unique measurable set $A_n=\bigcup\{P_n:n\in F\}$, and there are uncountably many subsets of $\mathbb{N}$.