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I was hoping someone can point me in the right direction for the proof of this question. I have some idea of whats going on, but I need a little more. Anyway, here is the statement.

We are given that $a$ is an $R$-measurable function on $\Gamma$ and that $b$ is an $S$-measurable function on $\Lambda$. Then we are given $f(x,y)=a(x)b(y)$, and the goal is to prove that $f$ is $R \times S$ measurable.

I'm a little new to this product measure stuff, but I know that to prove that $f$ is $R \times S$ measurable I need to show that $f^{-1}(O) \in R \times S$ for some arbitrary open set $O$ (in the complex numbers). I just don't see how one is able to show that. Do I have to use sections of the sets (or of the functions)?


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I'm sorry but I don't see how that helps at all :( – Walter Dec 7 '12 at 23:18
What do you know about measurability? – Davide Giraudo Dec 7 '12 at 23:19
I have the definition of measurability of a function, which is what I gave above. Thats equivalent to showing that any interval's pre-image is open if you are in the Reals, and that the imaginary and complex parts are both measurable, for complex. I really don't know much else about it. Am I missing some key piece of knowledge? – Walter Dec 7 '12 at 23:25

First, define $F\colon \Gamma\times\Delta\to \Bbb C\times \Bbb C$ by $F(x,y)=(a(x),b(y))$, where $\Bbb C^2$ is endowed with the product $\sigma$-algebra. This a measurable map (as so are the projections with respect to the first and second coordinates).

So what remains to show is that $M\colon \Bbb C\times\Bbb C\to\Bbb C$ defined by $M(z_1,z_2):=z_1z_2$ is Borel-measurable. It's not hard once we notice that continuous functions are Borel-measurable.

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Interesting. So you break $f$ into two parts by doing this, in a way? My question is, why Borel-measurable? I should switch that for the arbitrary measure in my question, right? Thanks! :) – Walter Dec 7 '12 at 23:33
But your function take their values in the set of complex numbers, right (I agree they are defined on an arbitrary measure space)? – Davide Giraudo Dec 7 '12 at 23:35

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