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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 –  Ilya May 9 '12 at 12:07

1 Answer 1

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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This is pretty unclear. What are $H$ and $G$? –  Nate Eldredge May 9 '12 at 13:27
    
He probably meant the section integrals of $f(x)g(y)$ –  Alex R. May 9 '12 at 18:50
    
Actually you uses Tonelli's Corollary. –  checkmath May 9 '12 at 22:17
    
@NateEldredge H and G are functions! What else could it be ? –  checkmath May 9 '12 at 22:20

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