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I started studying measure theory by myself using a book by D. Werner. One thing (apparently easy to prove) that I can't work out is the following (${\mathcal P}(S)$ denotes the power set of S):

${\bf Definition}$: Let $S$ be a set, $E \subset S$ and ${\mathcal E} \subset {\mathcal P}(S)$. Define
\begin{equation} {\mathcal E}\cap E:=\{A \in {\mathcal P}(E): \exists F \in {\mathcal E} \text{ with } A = F\cap E \} \end{equation} to be the trace of ${\mathcal E}$ on $E$.

I could show that ${\mathcal E} \cap E$ is again a $\sigma$ algebra on $E$, if ${\mathcal E}$ is a $\sigma$-algebra on S, but for the following I am lost. The claim is: \begin{equation} \sigma({\mathcal E \cap E}) = \sigma({\mathcal E}) \cap E \end{equation} Maybe this is really trivial, but right now I have no clue how to start.

jkds
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2 Answers2

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From $\mathcal E\subseteq\sigma(\mathcal E)$ it follows immediately that: $$\mathcal E\cap E\subseteq\sigma(\mathcal E)\cap E\tag1$$

You proved that the RHS is a $\sigma$-algebra on $E$ so $(1)$ allows the conclusion:$$\sigma(\mathcal E\cap E)\subseteq\sigma(\mathcal E)\cap E\tag2$$

Now have a look at the collection: $$\mathcal A:=\{F\in\wp(S)\mid F\cap E\in\sigma(\mathcal E\cap E)\}$$

It can be proved that $\mathcal A$ is a $\sigma$-algebra on $S$ (give this a try yourself, and let me know if you get stuck), and this obviously with $\mathcal E\subseteq\mathcal A$, so that also:$$\sigma(\mathcal E)\subseteq\mathcal A$$or equivalently:$$\sigma(\mathcal E)\cap E\subseteq\sigma(\mathcal E\cap E)\tag3$$

Actually the statement $\sigma(\mathcal E\cap E)=\sigma(\mathcal E)\cap E$ can be written as:$$\sigma(i^{-1}(\mathcal E))=i^{-1}(\sigma(\mathcal E))$$where $i:E\to S$ denotes the inclusion.

This can be recognized as a special case of:$$\sigma(f^{-1}(\mathcal E))=f^{-1}(\sigma(\mathcal E))\tag4$$ where $f:T\to S$ is a function.

For a proof of the more general $(4)$ have a look at this answer.

drhab
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  • Thanks a bunch. This already helps a lot. I feel silly asking this, but for showing that $\mathcal A$ is a sigma algebra on $S$, I need to show $S\in\mathcal A$ which means $S\cap E=E \in \sigma(\mathcal E \cap E)$. But, if I understand correctly $\mathcal E$ can be any subset of $\mathcal{P}(S)$, so how can I be sure that $E\in \sigma(\mathcal E \cap E)$? – jkds Sep 27 '18 at 19:05
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    Every $\sigma$-algebra on set $E$ will contain set $E$ and $sigma(\mathcal E\cap E)$ is a $\sigma$-algebra on set $E$. – drhab Sep 28 '18 at 07:13
  • Yes, that's what I want to show. But my problem is that the trace of ${\mathcal E}$ on $E$, i.e. ${\mathcal E}\cap E$ might easily cut out some part of $E$. Just for fun, if I take some ${\mathcal E}\subseteq\mathcal{P}(S)$ with $A \cap E = \emptyset$ for every $A \in \mathcal{E}$, then surely $E$ cannot be in $\mathcal{E}\cap E$. Generating a $\sigma$ algebra from $\emptyset$ makes little sense to me. That's what I find puzzling. Do we implicitly assume that the $\sigma$-algebra should be on all of $E$, so that $E$ is included per definition? – jkds Sep 28 '18 at 09:38
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    Let me repeat with a small addition: Every $\sigma$-algebra on set $E$ will contain set $E$ and $\sigma(\mathcal E\cap E)$ is by definition the smallest $\sigma$-algebra on $E$ that contains the collection $\mathcal E\cap E$ as a sub-collection. So $\sigma(\mathcal E\cap E)$ will contain set $E$. Period! That does not depend on the question $E\in\mathcal E$ or not. In your comment you are saying that “you want to show”, but there is nothing to show! – drhab Sep 28 '18 at 13:56
  • In the extremal case where $A\cap E=\varnothing$ for every $E\in\mathcal E$ we have $E\cap\mathcal E={\varnothing}$ (or even stronger where $\mathcal E=\varnothing$ so that $E\cap\mathcal E=\varnothing$). In both cases we have $\sigma(E\cap\mathcal E)={\varnothing,E}$. It certainly makes sense to look at $\sigma$-algebra's generated by ${\varnothing}$ or even by $\varnothing$. – drhab Sep 28 '18 at 13:56
  • Thanks, then everything is clear. I had wrongly understood $\sigma(\mathcal{E} \cap E)$ as being generated only by the intersections $A\cap E$ for all $A\in \mathcal{E}$, so not necessarily including $E$. I'll upvote once I have enough credit. – jkds Sep 28 '18 at 15:05
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Let $A\in \mathcal{E}\cap E$, then $A= F\cap E$ for some $F\in \mathcal E$. Since $\mathcal{E}\subset \sigma(\mathcal E)$, it follows that $A \in \sigma(\mathcal{E})\cap E$. This proves $$ \mathcal{E}\cap E \subset \sigma(\mathcal{E})\cap E. $$ Hence for the generated $\sigma$-Algebras you have$$ \sigma(\mathcal{E}\cap E) \subset \sigma(\sigma(\mathcal{E})\cap E). $$ But the trace of $\sigma$-Algebras is a $\sigma-$Algebra, i.e. $\sigma(\sigma(\mathcal{E})\cap E) = \sigma(\mathcal{E})\cap E$ and you obtain $$ \sigma(\mathcal{E}\cap E) \subset \sigma(\mathcal{E})\cap E. $$ Can you do the other inclusion yourself?

Jan Bohr
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