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The following problem is proving stubborn. I humbly request assistance.

If $f$ and $g$ are integrable functions on $\mathbb R$ and $F(x,y) = f(x)g(y)$, then $F$ is measurable, integrable on $\mathbb R\times \mathbb R$ and $\int_ {\mathbb R\times \mathbb R}F~d(\mu\times \mu)=\int_{\mathbb R}f~d\mu \int_{\mathbb R}g~d\mu.$


Can I do this for the first two parts of the problem?

If I let $A$ and $B$ be measurable subsets of $\mathbb R$. Set $f = 1_A, g=1_B$. Then $f = 1_{A\times B}$, $A\times B$ is measurable, so $f$ is measurable. $1_{A\times B}$ is integrable, so $f$ is integrable on $\mathbb R \times \mathbb R$. Furthermore $\int_{\mathbb R} F~d(\mu \times \mu) = (\mu\times \mu)(A\times B) = \mu(A)\cdot \mu(B) = \int_{\mathbb R} f ~d\mu \int_{\mathbb R }g~d\mu .$

This is all I'm able to do now. How about the second part?

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    This is Fubini's theorem, a short a nice proof of which you can find on the page 35 (40 in pdf) [here](http://www.math.cornell.edu/~durrett/PTE/PTE4_Jan2010.pdf)2012-05-09

1 Answers 1

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Well it is Fubini's theorem just observe (by definition) that $H(x,y)=f(x)$ and $G(x,y)=g(y)$ are measurable so is the product $F(x,y)=H(x,y)G(x,y)=f(x)g(y)$.

In other way $H(x,y)=f(\pi_1(x,y))$ and $G(x,y)=g(\pi_2(x,y))$. For $\pi_i: \mathbb R^2 \to \mathbb R$ such that $\pi_i(x_1,x_2)=x_i$.

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    @NateEldredge$H$and$G$are functions! What else could it be ?2012-05-09