0
$\begingroup$

I am going through a proof on the cardinality of sigma algebras. Namely, that they are either finite or uncountably infinite. I am having trouble in one part that the proof skips.

Let $\mathcal{A}_1$ be the infinite sigma algebra on the set $X$, and for some non-empty set $B \in \mathcal{A}_1 : B\neq X$, we build the following set

$$\mathcal{A}_2 = \{(X-B)\cap A:A\in\mathcal{A_1}\}$$

Now, this is a sigma algebra on $X-B$ according to the proof. Nevertheless, I could just prove the properties of closure under countable unions and that the set does include the null set and $X-B$. I can't manage to find out a way for proving closure under complements.

I have for some $A \in \mathcal{A}_1$

$$([X-B]\cap A)^C=(X-B)^C \cup A^C$$

But after this I am lost. How can I recover $(X-B) \cap S$ for $S \in \mathcal{A}$?

$\endgroup$

1 Answer 1

3
$\begingroup$

The key is that when you take the complement, you are taking it not in $X$, but in $X-B$ (since you are trying to prove that $\mathcal{A}_2$ is a $\sigma$-algebra on $X-B$). Thus the complement you need to compute is $(X-B)-((X-B)\cap A)$, which is equal to $(X-B)\cap A^c$ (where $A^c$ is the complement of $A$ in $X$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .