Let $\mathscr A$ be an algebra of sets. Let $\Sigma$ be the smallest sigma algebra (also the smallest monotone class) containing $\mathscr A$. Let $A_0 \in \Sigma$. Then $A_0 \cap \Sigma$ is a sigma algebra and is also the smallest sigma algebra (or monotone class) containing the algebra $\mathscr A \cap A_0$.

To prove $A_0 \cap \Sigma$ is a sigma algebra is easy, but I'm having trouble showing it is the smallest such sigma algebra.

In particular, I'm trying to use the monotone class theorem (c.f. Lieb Loss Analysis ch 1) to do so.

To that end, let $\{\Pi\}_\alpha$ be the collection of all monotone classes containing $\mathscr A \cap A_0$. Let $$\Pi = \bigcap_{\alpha} \Pi_\alpha.$$ We wish to show $\Pi = A_0 \cap \Sigma$. It is clear $\Pi \subset A_0 \cap \Sigma$, since $A_0 \cap \Sigma = \Pi_\beta$ for some $\beta$. I'm unclear how to show the reverse inclusion. A hint as to how to proceed would be helpful, as opposed to a full answer.

This seems as though it should be fairly straightforward seeing as it appears as a side in a proof in Lieb Loss, but it hasn't shown to be so...


You can show that

$$\Sigma(\mathcal{A}\rceil{A_{0}})=\Sigma(\mathcal{A})\rceil A_0\equiv\{B\subseteq A_0 : B\in\Sigma\},$$

where $\Sigma(\mathcal{E})$ is the $\sigma$-algebra generated by $\mathcal{E}$ (i.e. the smallest $\sigma$-algebra containing $\mathcal{E}$) and "$\rceil$" means a restriction.

Since $\mathcal{A}\rceil{A_{0}}\in \Sigma(\mathcal{A})\rceil A_0$ it's clear that $\Sigma(\mathcal{A}\rceil{A_{0}})\subseteq\Sigma(\mathcal{A})\rceil A_0$. For the reverse inclusion note that

$$\mathcal{M}=\{B : B\cap A_0\in \Sigma(\mathcal{A}\rceil{A_{0}})\}$$

is a $\sigma$-algebra containing $\Sigma(\mathcal{A})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.