3

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...

Mathbeginner
  • 901
  • Are you sure you don't want each $A_j$ to be measurable? Otherwise, $\mathbb{A}$ being infinite implies $A$ is infinite. Hence, there exists a sequence $(a_n){n\in\mathbb{N}}$ of distinct elements of $A$. Then $A=(\cup{n=1}^{\infty} {a_n})\cup (A\setminus (\cup_{n=1}^{\infty} {a_n})),$ the union has infinitely many "summands" and each of these sets is clearly disjoint. – WoolierThanThou Sep 17 '19 at 18:23

2 Answers2

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.

Marios Gretsas
  • 21,906
  • Example: $A = [0,1]$, $\Bbb A$ the Borel subsets of $[0,1]$, and $A_n=[0,1/n]$ for $n=1,2,\ldots$. In this case $B_1=[0,1]$, $B_k = \emptyset$ for $k=2,3,\ldots$. Your procedure does not produce a partition (of $A$, into $\Bbb A$-measurable sets, which is the (implicit) goal). – John Dawkins Sep 18 '19 at 17:49
  • 1
    @JohnDawkins the qeustion in the post is not clear and i though that the O.P wanted a partition of the sigma algebra – Marios Gretsas Sep 18 '19 at 17:56
  • First of all thanks for your answers, it helps clarify things for me greatly. I don't see how, if the space is countably infinite, the sigma algebra would be uncountable. Also, I tried the following:

    Let $C := \bigcup\limits_{i=1}^{\infty} \mathbb{A}{i}$. Then if $C$ is not already pairwise disjoint, we let $D=C{i} \setminus ( B_{i}^{c} \cup B_{j}^{c})$. Then $B_{i} \setminus D = B_{i}^{c} \cup B_{j}$, which is an elements of $\mathbb{A}$. Same applies for $B_{j}$, therefore we have now divided our set into smaller sets of pairwise disjoint sets. Do this for all sets done. Correct?

     – Mathbeginner Sep 19 '19 at 08:56
  • @Mathbeginner take a look at this for a proof of the uncountability of the sigma algebra https://math.stackexchange.com/questions/320035/if-s-is-an-infinite-sigma-algebra-on-x-then-s-is-not-countable#targetText=An%20immediate%20consequence%20is%20that%20the%20%CF%83%2Dalgebra%20is%20uncountable.&targetText=all%20lie%20in%20S%2C%20because,that%20they%20are%20pairwise%20disjoint.&targetText=So%20you%20get%20an%20injection,N%20is%20an%20uncountable%20set. – Marios Gretsas Sep 19 '19 at 14:22
  • @Mathbeginner...i meant that if the sigma algebra is infinite the it is uncountable...not the space..my bad. – Marios Gretsas Sep 19 '19 at 22:56
  • @MariosGretsas in your first answer, how do you construct that initially countable sequence ${ A_{n} : n \in \mathbb{N} } \subset \mathbb{A}$? I don't really understand your $\mathbb{A} = { A_{i} : i \in I }$... – Mathbeginner Sep 20 '19 at 07:54
  • @Mathbeginner the set ${A_i:i \in I}$ is the whole sigma algebra which is uncountable (if we assume tha has finite elements) form this family then i chose a countable sequence of sets $A_n$ – Marios Gretsas Sep 20 '19 at 08:09
  • If the sigma algebra has uncountable elements and in general is infinite ,then i can create a family of distinct countable sets by axiom of choise for example – Marios Gretsas Sep 20 '19 at 08:10
  • But you let $\mathbb{A} = { A_{i} : i \in I }$, then I would say each $A_{i}$ contains exactly one element from $A$, namely $a_{i}$, correct? But then all intersections are empty by your definition of $A_{i}$ and therefore we wouldn't need anything else... Where do I go wrong? – Mathbeginner Sep 20 '19 at 08:22
  • How do you know that singleton belong to $\Bbb{A}$? – Marios Gretsas Sep 20 '19 at 08:26
  • Since $\mathbb{A}$ has infinite elements? – Mathbeginner Sep 20 '19 at 08:48
  • So what...? it note need to have singletons for having infinite elements. – Marios Gretsas Sep 20 '19 at 08:51
  • That does not guarantee that has infinitely many singletons...If it has then your idea is correct. – Marios Gretsas Sep 20 '19 at 08:53
1

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 $|\mathcal B|\ge2^{\aleph_0}$.

Henno Brandsma
  • 242,131