Measure theory; proving an infinite partition exists from a sigma-algebra.

Question

I'm really new to measure theory and have trouble with interpreting a question.

We have a space $A$ with $\mathbb{A}$ a $\sigma$-algebra with infinitely many elements. Now I need to show an infinite partition exists.

Of course we first need to define what an infinite partition is. It is a countable infinite sequence of non-empty and disjoint sets which union is $A.$

So we know there are infinite elements in the $\sigma$-algebra. I would argue we need to rearrange the elements of $\mathbb{A}$ to make them countable and then introduce a 'left-overs' subset that contains all elements that are outside of our countable division. I know this is probably very wrong but I would appreciate any steering in the right direction...

Answer 1

Your idea is correct.

If the space is finite then the sigma algebra if finite and your partition problem is easy enough.

If the sigma algebra is infinite then it is uncountable.

Let $\Bbb{A}=\{A_i:i \in I\}$

Then you can take a countable sequence $\{A_n:n\in \Bbb{N}\} \subset \Bbb{A}$

Now define $B_n=A_n \setminus \bigcup_{k=1}^{n-1}A_k$ and $B_1=A_1$

Note that these sets are disjoint and belong to $\Bbb{A}$(since $\Bbb{A}$ is closed under set differences and unions)

Finally take $\Bbb{A}_n:=\{B_n\}$ and $\Bbb{A}_0=\Bbb{A} \setminus \{B_n: n \in \Bbb{N}\}$

So you have a partition $\{ \Bbb{A}_n:n=0,1,2...\}$ of the sigma algebra.

Answer 2

Using @bof's argument from a related question:

Let $\mathcal B$ be an infinite $\sigma$-algebra on a set $\Omega$. Partition $\Omega$ into two disjoint nonempty sets $A_1,B_1\in\mathcal B$. At least one of $\mathcal B\cap\mathcal P(A_1)$ and $\mathcal B\cap\mathcal P(B_1)$ is infinite; otherwise, if $|\mathcal B\cap\mathcal P(A_1)|=m\lt\aleph_0$ and $|\mathcal B\cap\mathcal P(B_1)|=n\lt\aleph_0$, we would have $|\mathcal B|=mn\lt\aleph_0$. We may assume that $\mathcal B\cap\mathcal P(B_1)$ is infinite. Next, partition $B_1$ into two disjoint nonempty sets $A_2,B_2\in\mathcal B$ so that $\mathcal B\cap\mathcal P(B_2)$ is infinite. Continuing in this way, we get an infinite sequence $A_1,A_2,A_3,\dots$ of pairwise disjoint nonempty elements of $\mathcal B$. (Every infinite Boolean algebra contains such a sequence; we haven't used the $\sigma$ yet.) Since $\mathcal B$ is a $\sigma$-algebra, the union of each subsequence belongs to $\mathcal B$, showing that $|\mathcal B|\ge2^{\aleph_0}$.