Working with the case $\Omega$ is countable and $\Omega$ is uncountable for a singleton set generator $\mathcal{E}$

measure-theoryprobabilityreal-analysis

Let $\Omega \neq \varnothing$. Let $\mathcal{E}:=\{ \{\omega\} : \omega \in \Omega\}$

Show:

i) If $\Omega$ is countable, then $\sigma(\mathcal{E})=2^{\Omega}$.

ii) If $\Omega$ is uncountable, then: if $A \in \sigma(\mathcal{E}) \iff A$ or $A^{C}$ is countable

My ideas:

i) I think (i) is rather trivial.

For $\sigma(\mathcal{E})\subseteq 2^{\Omega}$. Let $A \in \sigma(\mathcal{E})$. Then $A$ can be written as a countable union of singletons which are by definition $\subseteq \Omega$ and therefore $\in 2^{\Omega}$.

$2^{\Omega} \subseteq \sigma(\mathcal{E})$ is trivial.

ii) I do not really know where to begin with an uncountable $\Omega$. Any tips?

Best Answer

Here's a suggestion. By definition $\sigma(\mathcal{E})$ is the smallest sigma algebra containing $\mathcal{E}$. Thus if we want to show that $\newcommand\calA{\mathcal{A}}\newcommand\calB{\mathcal{B}}\newcommand\calE{\mathcal{E}}\sigma(\calE)=\calA$, a common strategy is to show that $\calA$ is a $\sigma$-algebra and that $\calE\subseteq \calA$ and if $\calB$ is any $\sigma$-algebra containing $\calE$ then $\calA\subseteq \calB$.

In your case, $\calA$ will be the set of all subsets $A$ of $\Omega$ such that either $A$ is countable or its complement is countable. Then to show $\sigma(\calE)=\calA$ you need to show:

  1. $\calA$ is a $\sigma$-algebra,
  2. $\calE\subseteq \calA$, and
  3. if $\calB$ is a $\sigma$-algebra containing $\calE$ then $\calA\subseteq \calB$.

Can you show these things?

Side note

In case its unclear why these properties show that $\sigma(\calE)=\calA$, note that the third property immediately gives us $\calA\subseteq \sigma(\calE)$ (since $\sigma(\calE)$ is a $\sigma$-algebra containing $\calE$), but we also have by definition that $\sigma(\calE)$ is the intersection of all $\sigma$-algebras containing $\calE$, which includes $\calA$ by properties 1 and 2, so $\sigma(\calE)\subseteq \calA$ as well.

Edit

Suggestions for part 3. Well I have a hint, but anything stronger than that is a solution, so here's the hint and the solution is in the spoiler tag.

Hint: How can a countable set be written in terms of elements of $\calE$?

Let $\calB$ be a sigma algebra containing $\calE$. First I claim any countable subset of $\Omega$ is in $\calB$, since if $A\subset\Omega$ is countable, then $A=\bigcup_{x\in A} \{x\}$, which is a countable union of elements of $\calE$, and therefore in $\calB$. If $A$ has a countable complement on the other hand, then its complement is in $\calB$ by the argument above, and so it is in $\calB$ as well. Thus $\calB$ contains $\calA$.

Related Question