# Equivalence of measurable functions

I want to prove the following Lemma:

Let $\mathcal{A}$ be a $\sigma$-algebra in $X$ and let $f:X\rightarrow\mathbb{R}$, then TFAE:

1. $f$ is measurable.
2. For each Borel set $B\subset\mathbb{R}$ holds $f^{-1}(B)\in\mathcal{A}$.
3. For each open set $B\subset\mathbb{R}$ holds $f^{-1}(B)\in\mathcal{A}$.

I don't know how to start. I think one implication is some more work, but the other must follow. I think you have to prove 1-->3-->2-->1.

Some help maybe? Thanks

• Maybe you can define a new $\sigma$-algebra $B=\{A\subset\mathbb{R}:f^{-1}(A)\in\mathcal{A}\}$, then the implication from 1 to 3 is not so hard right? – Jens25 Jan 30 '14 at 14:46
• Sorry i mean form 1 to 2 – Jens25 Jan 30 '14 at 14:53
• I think form 3) to 2). See my answer. @Jens25 idea is very good. – drhab Jan 30 '14 at 15:09
• How is '$f$ is measurable' defined here? Isn't it a copy of 2)? – drhab Jan 30 '14 at 15:13

Define $\mathcal V:=\{A\mid f^{-1}(A)\in \mathcal A$}. It is routine to show that $\mathcal V$ is a $\sigma$-algebra. In 3) it is stated that $\mathcal O\subset \mathcal V$, where $\mathcal O$ denotes the collection of open sets in $\mathbb R$. Consequently $\sigma(\mathcal O)\subset \mathcal V$ where $\sigma(\mathcal O)$ denotes the smallest $\sigma$-algebra that contains $\mathcal O$, which is exactly the collection of Borel sets. So 3) implies 2), which is the main job here.