Michelle Mendoza

2022-07-01

Let  be some measurable domain and $f:E\to \mathbb{R}$ a measurable map. Let $B\in \mathrm{Bor}\left(\mathbb{R}\right)$ be a Borel set. Show that ${f}^{-1}\left(B\right)$ is measurable.
I'm advised to define . Now $\mathcal{A}$ consists of sets whose preimage is measurable, and since $f$ is continuous these sets are open. This collection seems to form a $\sigma$-algebra on $\mathbb{R}$, but I'm confused about the construction here as it seems that $\mathcal{A}$ is the smallest $\sigma$-algebra containing open sets, but that would mean that it's equal to $\mathrm{Bor}\left(\mathbb{R}\right)$?

Freddy Doyle

Expert

Let's say "measurable function" means ${f}^{-1}\left(G\right)$ is measurable for all open sets $G\subseteq \mathbb{R}$.
Then define . Show that $\mathcal{A}$ contains all open sets and that $\mathcal{A}$ is a sigma-algebra. Conclude that $\mathcal{A}\supseteq \mathrm{Bor}\left(\mathbb{R}\right)$.

Pattab

Expert

Note: you cannot prove $\mathcal{A}=\mathrm{Bor}\left(\mathbb{R}\right)$.