How can I prove the product of two measurable functions in the product measure space is measurable? I tried but still do not know how to prove.
Question in product measure
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measure-theory
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0Why is it relevant that the measure space is a product measure space? – 2012-12-10
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0I tried to prove the inverse image of every open set is measureable, but I found it is difficult unless I assume $\sigma$-finiteness, etc. – 2012-12-10
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0You need to be more specific with what you're asking. What are the domains and codomains for your functions? You should edit your question to include this. For example, if $f,g \colon X \to \mathbb{R}$, where $X$ is an arbitrary measure space and $\mathbb{R}$ has the Borel $\sigma$-algebra, then $fg$ is measurable. The fact that you're looking at a product space would be irrelevant here. Presumably there's something about your question in particular which makes it relevant. – 2012-12-10
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0We assume $X,Y$ to be two arbitrary given measurable space. And $(X\times Y)$ were given the product measure. – 2012-12-10
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0When you say measurable function then you mean a real-valued function? Such that the preimage of an open set is measurable? Or are you interested in the general topological setting? – 2012-12-10
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0Yes. I mean a measurable function with real or complex value. – 2012-12-10
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0Got it. thanks. – 2012-12-10
1 Answers
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My proof:
It suffice to prove that $f_{1}(x,y)=g(x)$ and $f_{2}(x,y)=h(y)$ are measurable functions in $(\mathcal{S}\times \mathcal{T})$. Then we should have $f(x,y)=f_{1}(x,y)*f_{2}(x,y)$. Since the product of measurable functions is measurable, we concluded the proof.
By assumption $g(x),h(y)$ are both measurable functions in coordinates. Now for any open set $O$, $f_{1}^{-1}(O)$ is equal to $g^{-1}(O)\times \mathcal{T}$, similarly $f_{2}^{-1}(O)$ is equal to $h^{-1}(O)\times \mathcal{S}$. Both sets are measurable sets by definition of product measure (Tao, page 195). Therefore both $f_{1},f_{2}$ are measurable.
We conclude that $f$ must be measurable.