Rudin defines a function from a measure space $X $ to a topological space $Y $ as measurable if the preimage of every open set in $Y $ is measurable in the measure space $X $.
Now Wikipedia, and my books on probability, defines a function between two measure spaces $X $ and $Y $ as measurable if the preimage of every Borel set in $Y $ is measurable in $X $.
Since the inverse image preserves set operations, one can easily see that the set of subsets of $Y $ such that $f^{-1 } $ is measurable in $X $ forms a $\sigma $ algebra $\Sigma $ in $Y$. This implies $\Sigma \supset \mathcal B $, since $\Sigma $ contains all the opens sets.
Can one show the opposite inclusion, so that the defintions are in fact equal?
Thanks in advance!
Best Answer
There is no need to show the opposite inclusion. In fact the opposite inclusion is not true ($\mathcal{B} \supset \Sigma$ means that every set whose preimage is measurable is borel, but this is clearly nonsense, take as an example $f=id_{\mathbb{R}}$).
What you proved is that the following are equivalent:
Clearly 2 implies 1 since every open is borel. Conversely, if 1. holds, then $\mathcal{B} \subset \Sigma$ (since $\Sigma$ is a $\sigma$-algebra), so 2 holds.