Davon Irwin

2022-06-21

Let $X$ be a set and $\mathcal{F}$ a $\sigma$-algebra. Does there exist a topological space $U$ and a map $f:X\to U$ such that $f$ is $\mathcal{F},\mathcal{B}$-measurable and $\sigma \left(f\right)=\mathcal{F}$? Here $\mathcal{B}$ is the Borel $\sigma$-algebra of $U$.
Of course, this is trivial if every $\sigma$-algebra on a set is the Borel $\sigma$-algebra with respect to some topology on the set. But this needn't be true. This is a weaker problem.

Belen Bentley

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document If $\tau$ is a topology on a set $U$ and $f:X\to U$ is a function, then

is a topology on $X$ (since preimages, unlike images, commute with intersections and unions). Since per the comments above $\sigma \left(f\right)$ is defined as the $\sigma$-algebra on $X$ generated by $Op\left(f\right)$ this means that $\mathcal{F}=\sigma \left(f\right)$ for some $f$ only if $\mathcal{F}$ is generated by a topology on $X$. So this question does indeed reduce to the original question, which has a negative answer.

Jackson Duncan