Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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 $\pi_i(x_1,x_2)=x_i$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.