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 $\sigma (f)=\mathcal{F}$? 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.

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

Beginner2022-06-22Added 28 answers

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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 $\sigma (f)$ is defined as the $\sigma $-algebra on $X$ generated by $Op(f)$ this means that $\mathcal{F}=\sigma (f)$ 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.

$$

is a topology on $X$ (since preimages, unlike images, commute with intersections and unions). Since per the comments above $\sigma (f)$ is defined as the $\sigma $-algebra on $X$ generated by $Op(f)$ this means that $\mathcal{F}=\sigma (f)$ 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

Beginner2022-06-23Added 10 answers

This works, thanks. It simply never occurred to me that we could generate a topology by taking preimages.