Let X be a standard Borel space and $\mathcal C, \mathcal D$ be countably generated sub sigma algebras of the Borel sigma algebra of X. Suppose that for each $x \in X$ we have $[x]_{\mathcal C} \subset [x]_{\mathcal D}$ where $[x]_{\mathcal C}$ means the intersection of all elements in $\mathcal C$ containing $x$. Does it follow that $\mathcal D \subset \mathcal C$?

Even for finite $\mathcal D$, I don't know how to proceed. But once the finite $\mathcal D$ case is done, the infinite case follows easily.

Special cases where this implication is known include:

  1. the case where $\mathcal D$ is the Borel sigma algebra of X
  2. the case where $\mathcal C, \mathcal D$ are both finite

A related question where ignoring null sets is allowed is:

Two possible senses of a random variable being a function of another random variable


I think I found a lemma that answers this positively in Classical Descriptive Set Theory by A. S. Kechris. The lemma is Blackwell Exercise 14.16 at 88p.

(Exercise 14.16) Let $X$ be a standard Borel space and $(A_n)$ a sequence of Borel sets in $X$. Consider the equivalence relation $E\subset X^2$ defined by $$xEy \iff \forall n (x \in A_n \iff y \in A_n)$$ Show that a Borel set $A \subset X$ is $E$-invariant if and only if it belongs to the $\sigma$-algebra generated by the sequence $(A_n)$.

Proof of the nontrivial direction (the only if direction): If there was some nice sigma algebra on the quotient set $X/E$ such that the obvious map $p: X \mapsto X/E$ is measurable and that $p(A)$ is measurable whenever $A$ is Borel, then we would be done because $A \subset X$ being $E$-invariant is equivalent to $A = p^{-1}(p(A))$.

We define a map $F$ with the following property: $$xEy \iff F(x) = F(y)$$ Define $F: X \to \{0,1\}^{\mathbb N}$ by $$x \mapsto (1_{A_1}(x), 1_{A_2}(x), \dots).$$ The map $F$ is $\sigma(A_n:n\ge 1)$-measurable, where $\sigma(A_n:n\ge 1)$ denotes the $\sigma$-algebra generated by the sequence $(A_n)$.

Suppose a Borel set $A\subset X$ is $E$-invariant, which is equivalent to saying $A = F^{-1}(F(A))$. The set $F(A)$ is an analytic set in $\{0,1\}^{\mathbb N}$ because $A$ and $F$ are Borel. The set $F(A^c)$ is also analytic. Since $A^c$ is also $E$-invariant, we have $A^c = F^{-1}(F(A^c))$, and so the two sets $F(A)$ and $F(A^c)$ are disjoint. By the Lusin separation theorem, there is a Borel set $B \subset \{0,1\}^{\mathbb N}$ separating the two sets, i.e., $F(A) \subset B$ and $F(A^c) \subset B^c$. So $A = F^{-1}(B)$ and since $F$ is $\sigma(A_n:n\ge 1)$-measurable, we conclude that $A$ is in $\sigma(A_n:n\ge 1)$. QED.

So this lemma (the exercise) then answers the question positively, and by using the assumption of countably generated on $\mathcal C$ only.

The assumption of countably generated on $\mathcal C$ cannot be dropped: for a demonstrating counter-example, let $\mathcal C$ be the sigma algebra of countable or co-countable sets and let $\mathcal D$ be the Borel sigma algebra.


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