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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!

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  • $\begingroup$ The definitions cannot be fully equal because a measureable space need not be a topological space (also, that would replace $Borel set" with "measurable set" in the general definition) $\endgroup$ – Hagen von Eitzen Nov 14 '14 at 10:14
  • $\begingroup$ Thanks, is there any simple example of this? Maybe the power set on $\mathbb R $, which is a $\sigma $ algebra but not a measure space? $\endgroup$ – Alexander Nov 14 '14 at 10:19
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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:

  1. $\forall U \subseteq Y$ open $f^{-1}(U)$ is measurable
  2. $\forall U \subseteq Y$ borel $f^{-1}(U)$ is measurable

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.

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