0

In an attempt to extend this result to Banach spaces, I come across the existence of some Borel sets.

Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite complete measure space. Let $(E,|\cdot|)$ be a Banach space and $\mathcal B_E$ its Borel $\sigma$-algebra. Let $f:\Omega \to E$ be $\mu$-measurable. By this theorem, $f$ is $\mathcal F$-$\mathcal B_E$ measurable. Let $\nu:= f_\sharp \mu$ be the push-forward measure. This means $\nu$ is a Borel probability measure on $E$.

Is there a Borel set $B \in \mathcal B_E$ such that $B \subset \operatorname{im} f := f(\Omega)$ and $\nu (B) = 1$?

Thank you for your elaboration!

Update: I have found a closely related question to the one above: Is there a $\mu$-measurable function $g:\Omega \to E$ such that $f=g$ $\mu$-a.e. such that $g(\Omega) \in \mathcal B_E$?

Analyst
  • 5,637
  • You need some conditions on $f$ for the question to be interesting I guess. If $f$ maps into a single point $\nu$ will exactly attain the values $0$ and $\mu(\Omega)$. You probably also want more conditions on $\mu$ because for the statement to hold you, at the bare minimum, need a set $F\in \mathcal{F}$ with $\mu(F)=1$. – Jonathan Hole Oct 20 '22 at 12:15
  • @JonathanHole Could you more elaborate more on " If $f$ maps into a single point $\nu$ will exactly attain the values $0$ and $\mu (\Omega)$" and on what $F$ means? – Analyst Oct 20 '22 at 12:19

0 Answers0