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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 $\sigma(\mathcal C)$?
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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)\ $.

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  • $\begingroup$ 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. $\endgroup$
    – nahp
    Commented Oct 4, 2020 at 4:57

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