# Answer Check: Show that $\int_{A\times B} h = \left( \int_Af \right) \left(\int_Bg \right)$ and as a corollary that $v(A\times B) = v(A)v(B)$

here is the problem:

Let $A \subset R^m$ and $B \subset R^n$ be contented sets, and $f:R^m \rightarrow R$ and $g: R^n \rightarrow R$ integrable functions. Define $h:R^{m+n} \rightarrow R$ by $h(x,y) = f(x)g(y)$, prove that $$\int_{A\times B} h = \left( \int_Af \right) \left(\int_Bg \right)$$ and as a corollary that $v(A\times B) = v(A)v(B)$

where $v(A)$ is the volume function.

I understand that I have to use fubini's theorm in maybe this way:

$$\int_{A \times B}h = \int_{A} \left( \int_Bh(x,y) dy \right)dx = \int_{A} \left( \int_Bf(x)g(y)dy \right)dx$$ and then since f does not depend on y, we can pull it out? so we get $$= \int_{A}f(x)dx \left( \int_Bg(y)dy \right)$$

then $$=\int_{A}f \left( \int_Bg \right)$$

Then for the corollary since $$v(A) = \int_a^b A(t)dt$$ it follows that $$v(A) = \int_Af$$ $$v(B)= \int_Bg$$

thus

$$v(A\times B) = v(A)v(B)$$

Thanks for the help guys

• Looks good to me. Apr 19, 2013 at 5:20