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Rudin RCA p.29

"Let us call a function $f$ defined on a set $E\in\mathfrak{M}$ 'measurable on $X$' if $\mu(X\setminus E)=0$ and if $f^{-1}(V)$ is measurable for every open set $V$. If we define $f(x)=0$ for $x\in X\setminus E$, we obtains a measurable function on $X$, in the old sense. If our measure happens to be "COMPLETE", we can define arbitrary manner, and we still get a measurable function. The integral of $f$ over any set $A\in\mathfrak{M}$ is independent of the definition of $f$ on $X\setminus E$"

There are two things very unclear to me.

  1. What's the image of $f$? Is it any topological space or simple $\mathbb{R}^k$?

  2. It can be shown that [If $g,h$ are measurable functions from $X$ to $\mathbb{R}^k$ and $g=h$ a.e. on $X$, then for any $E\in\mathfrak{M}$, $\int_E g d\mu = \int_E h d\mu$] for any measure $\mu$. If he meant $f$ to be a function maps to $\mathbb{R}^k$, why does $\mu$ have to be complete? It seems completely inessential to me. What am i misunderstanding here ?

  3. What is the definition of 'measurable function'? Rudin - RCA defines it to be a function whose pre-image of open sets in a 'topological space' is measurable. However, wikipedia defines it to be a function whose pre-image of mearable sets in a 'measurable space' is measurable. Which one is generally used?

Thank you in advance..!

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It's the second one that is generally used. The first one is the special case of the Borel measurable functions. –  Learner Jan 14 '13 at 11:04
    
@Learner Thank you! –  Katlus Jan 14 '13 at 11:07

1 Answer 1

up vote 1 down vote accepted
  1. Every topological space works for the definition, but for some things Rudin uses specific properties of Euclidean space. Not in this argument though.

  2. By assumption $\mu(E^C)=0$. If $O$ is open, we have $f^{-1}(O)=(f^{-1}(O)\cap E)\cup (f^{-1}(O)\cap E^C)$. In general, it may be the case that $f^{-1}(O)\cap E^C$ is not a Borel set, if $f$ has an arbitrary form on $E^C$. But it is always a subset of $E^C$ and has therefore outer measure $0$. It is therefore in the completion of the measure space.

  3. The definition in Wikipedia is strictly more general. One can show that the preimage of an open set is measurable if and only if the preimage of a Borel measurable set is measurable. The more general definition allows us to consider $\sigma$-algebras different from the Borel $\sigma$-algebra on the codomain.

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I didn't know you wrote an answer so i just added the question 3. Sorry for that.. –  Katlus Jan 14 '13 at 11:01
    
@Katlus I added an answer for the third point. –  Michael Greinecker Jan 14 '13 at 11:06

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