# $\sigma$-algebra generated by trace is trace of generated $\sigma$-algebra

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
$$$${\mathcal E}\cap E:=\{A \in {\mathcal P}(E): \exists F \in {\mathcal E} \text{ with } A = F\cap E \}$$$$ to be the trace of $${\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: $$$$\sigma({\mathcal E \cap E}) = \sigma({\mathcal E}) \cap E$$$$ Maybe this is really trivial, but right now I have no clue how to start.

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.

• 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
• 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 7:13
• 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 9:38
• 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

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?

• thanks heaps. I'm still working on the reverse inclusion. – jkds Sep 27 '18 at 19:07