# Induced and generated sigma-algebras

Let $$(\Omega, \mathcal F)$$ be a measurable space, $$\mathcal C \subseteq \mathcal F$$ a collection of generators, and $$B \in \mathcal F$$ a measurable set. I am wondering whether $$\sigma(\mathcal C \cap B) = \sigma(\mathcal C) \cap B ;$$ in other words, does the sigma-algebra on $$B$$ generated by the elements of $$\mathcal C$$ "restricted to" $$B$$ coincide with the sigma-algebra induced on $$B$$ by the generated sigma-algebra? Above, I am using the notation $$\mathcal C \cap B = \{ A \cap B \, | \, A \in \mathcal C \}$$, and $$\sigma(\mathcal C) \cap B$$ is defined in the same way.

This seems like it should be easy to prove, and at least one inclusion is obvious: because every element of $$\mathcal C \cap B$$ is in $$\sigma(\mathcal C) \cap B$$, we have $$\sigma(\mathcal C \cap B) \subseteq \sigma(\mathcal C) \cap B$$, since $$\sigma(\mathcal C) \cap B$$ is a sigma-algebra on $$B$$. However, I do not know whether the reverse inclusion is true. My idea is to explicitly describe the sets in $$\sigma(\mathcal C)$$. In my notation, this Wikipedia page says, "$$\sigma(\mathcal C)$$ consists of all the subsets of $$\Omega$$ that can be made from elements of $$\mathcal C$$ by a countable number of complement, union and intersection operations." If this is true, then it may be possible to take a set $$A \cap B \in \sigma(\mathcal C) \cap B$$, represent $$A$$ as described, and "commute" the intersection with $$B$$ with all the union/intersection/complement operations from which $$A$$ is derived. For example, suppose $$A = \bigcup_{i=1}^\infty A_i$$ is a countable union of elements of $$\mathcal C$$. Then $$A \cap B = \bigcup_{i=1}^\infty (A_i \cap B) ,$$ and the right-hand side is now in $$\sigma(\mathcal C \cap B)$$. However, this only seems possible when $$A$$ is derived from a finite number of such operations, as above (which I count as a single union operation). This leads me to two questions:

1. How can Wikipedia's description be formulated? I believe it could be read as $$\sigma(\mathcal C) = \bigg\{ \mathop{\color{red}{\ast}}\limits_{i_1=1}^{\infty} \mathop{\color{red}{\ast}}\limits_{i_1=2}^{\infty} \dotso \mathop{\color{red}{\ast}}\limits_{i_n=1}^{\infty} B_{i_1, \dotsc, i_n} \, \bigg| \, B_{i_1,\dotsc,i_n} \in \mathcal C \cup \{\varnothing\} \text{ or } B_{i_1,\dotsc,i_n}^c \in \mathcal C \bigg\} ,$$ where each $$\color{red}{\ast}$$ may stand for either the symbol $$\bigcap$$ or $$\bigcup$$, for severe want of better notation. Here, the sets $$B$$ are the elements of $$\mathcal C$$ and their complements, or the empty set. Unless I am mistaken, the collection written above is a sigma-algebra containing $$\mathcal C$$, and it is easily seen to be the smallest such.
2. Assuming the collection above correctly describes $$\sigma(\mathcal C)$$, this should prove the reverse inclusion, i.e. $$\sigma(\mathcal C) \cap B \subseteq \sigma(\mathcal C \cap B)$$. Is there a way to show this inclusion that does not rely on an explicit description of $$\sigma(\mathcal C)$$?

Let $${\cal D}:= \{A\in\sigma({\cal C})\ |\ A\cap B\in\sigma({\cal C}\cap B)\,\}\ .$$ Then
• $${\cal C}\subseteq{\cal D}$$.
• If $$\ A\ \in {\cal D}\$$, then $$\ (\Omega\setminus A)\cap B=B\setminus(A\cap B)\in\sigma({\cal C}\cap B)\$$ because $$\ A\cap B \in\sigma({\cal C}\cap B)\$$, and so $$\ (\Omega\setminus A)\in{\cal D}\$$.
• if $$\ A_i\in {\cal D}\$$ for $$\ \ i=1,2,\dots\$$ then $$\ A_i\cap B\in\sigma({\cal C}\cap B)\$$ for all $$\ i\$$. Therefore $$\ \displaystyle \left(\bigcup_iA_i\right)\cap B=\bigcup_i\left(A_i\cap B\right)\in \sigma({\cal C}\cap B)\$$, and so $$\ \displaystyle\bigcup_iA_i\in {\cal D}\$$.
Thus, $$\ {\cal D}\$$ is a $$\sigma$$-algebra containing $$\ {\cal C}\$$, and therefore $$\ \sigma({\cal C})\subseteq{\cal D}\$$. So if $$\ D\in\sigma({\cal C})\cap B\$$ then $$\ D=A\cap B\$$ for some $$\ A\in \sigma({\cal C})\subseteq{\cal D}\$$, which implies $$\ A\cap B\in\sigma({\cal C}\cap B)\$$. Therefore $$\sigma({\cal C})\cap B\subseteq \sigma({\cal C}\cap B)\$$.
• Thank you; this is exactly the sort of "nice" argument I was looking for. It seems obvious now: we would like $A \cap B$ to be in $\sigma(\mathcal C \cap B)$ for any $A$ in the generated sigma-algebra $\sigma(\mathcal C)$. Although this may not seem clear when $A$ is obtained from the generating sets in a complicated way (via unions/intersection), it is clearly true if $A$ is a generating set itself. This is enough for the result, because the collection of all such $A$ is a sigma-algebra.