Is every sub-$\sigma$-algebra of $\mathcal{B}(\mathbb{R})$ "universally countably generated"? This is a seemingly basic question that I realise I don't know the answer to:
Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $\mathbb{R}$.


Is it the case that for every sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{B}$ there exists a countably generated sub-$\sigma$-algebra $\mathcal{G}'$ of $\mathcal{G}$ such that for every probability measure $\mathbb{P}$ on $\mathcal{B}$, $\mathcal{G} \subset \sigma(\mathcal{G}' \cup \mathcal{N}_{\mathbb{P}})$?


(Here, $\mathcal{N}_\mathbb{P}$ denotes the set of $\mathbb{P}$-null sets.)

Although less important for me, it would also be interesting if possible to know the following:


*

*If the answer to the above question is no (in ZFC): Is the answer to the above question being yes consistent with ZF+DC?

*But if the answer is yes: Does it still remain yes if we replace "every probability measure $\mathbb{P}$ on $\mathcal{B}\,$" with "every probability measure $\mathbb{P}$ on $\mathcal{G}'\,$"?

 A: I've now managed to work out that the answer is no:
Proposition. Let $\rho$ be a probability measure on $(\mathbb{R},\mathcal{B})$, and let $\mathcal{G}$ be a sub-$\sigma$-algebra of $\mathcal{B}$ such that


*

*$\rho(A) \in \{0,1\}$ for all $A \in \mathcal{G}$;

*for $\rho$-almost every $x \in \mathbb{R}$ there exists $A \in \mathcal{G}$ such that $x \in A$ and $\rho(A)=0$.


Then for every countably generated sub-$\sigma$-algebra $\mathcal{G}'$ of $\mathcal{G}$ there exists a probability measure $\mathbb{P}$ on $(\mathbb{R},\mathcal{B})$ such that $\mathcal{G} \not\subset \sigma(\mathcal{G}' \cup \mathcal{N}_\mathbb{P})$.
So, for instance, we could take $\mathcal{G}$ to be the countable-cocountable algebra (with $\rho$ being any atomless probability measure).
Proof. We can represent $\mathcal{G}'$ as $\sigma(N_i : i \in \mathbb{N})$ for some sequence of $\rho$-null sets $N_i \in \mathcal{G}$. Let $N:=\bigcup_{i=1}^\infty N_i$. It is clear that
$$ \mathcal{D} \ := \ \bigcup_{B \subset N} \{B , \mathbb{R} \!\setminus\! B\} $$
is a $\sigma$-algebra, and therefore $\mathcal{G}' \subset \mathcal{D}$. Fix $x \in \mathbb{R} \setminus N$ and $A \in \mathcal{G}$ such that $x \in A$ and $\rho(A)=0$. Let $\mathbb{P}=\frac{1}{2}\delta_x + \frac{1}{2}\rho$. Then $\mathbb{P}(A)=\frac{1}{2}$. However, for any $B \in \mathcal{D}$, if $B \subset N$ then $\mathbb{P}(B)=0$ but if $\mathbb{R} \!\setminus\! B \subset N$ then $\mathbb{P}(B)=1$. So in particular, every member of $\mathcal{G}'$ has $\mathbb{P}$-trivial measure, and so $A \not\in \sigma(\mathcal{G}' \cup \mathcal{N}_\mathbb{P})$. QED
[As for the additional question: the above proof is valid in ZF+DC.]
