2
$\begingroup$

Let $\mathcal{B}^n$ be the borel sigma algebra generated by the rectangles in $\mathbb{R}^n$. I can write $f(\mathbf{x})=g_1(x_1)\cdots g_n(x_n)$. Let $\mu=\mu_1\times \cdots \times \mu_n$ be the Lebesgue product measure.

Is it possible to write $\int_Af(\mathbf{x})\ d\mu(\mathbf{x})$, where $A \in \mathcal{B}^n$, as an expression of $\prod^n_{i=1}\int_{A_i}g_i(x_i) \, d\mu_i(\mathbf{x})$, where the $A_i$ are not dependent on each other?

So, I was thinking of using the fact that the $\mathcal{B}^n$ can be generated by rectangles of the form $I_1\times \cdots \times I_n$, where $I_i \subset \mathbb{R}$ are intervals; and then use the Fubini theorem, which is valid for rectangles, to reach the expression I desire. However, I don't know how to do this...

Any help would be appreciated.

P.S.: What if $\mu$ is not lebesgue?

$\endgroup$
0
$\begingroup$

No, you can't. Look at the following example: If $A$ is the unit disc in the $(x_1,x_2)$-plane then Fubini's theorem says that $$\int_A g_1(x_1) g_2(x_2)\>{\rm d}(x_1,x_2)=\int_{-1}^1 \int_{-\sqrt{1-x_1^2}}^{\sqrt{1-x_1^2}} g_1(x_1)\> g_2(x_2)\>dx_2\>dx_1\ ,$$ as you have learned in Calculus 102. (You are allowed to take the factor $g_1(x_1)$ out of the inner integral.)

Things are different if the domain of integration $B\subset{\mathbb R}^n$ is itself a cartesian product $$A=\prod_{i=1}^n A_i\tag{1}$$ with $A_i$ embedded in the $i^{\rm th}$ factor of ${\mathbb R}^n$. In this case the $n$-fold integral indeed "separates" as envisaged in your question.

A general subset $A\subset{\mathbb R}^n$, even a finite union of boxes $B:=\prod_{i=1}^n [a_i,b_i]$, cannot be presented in the form $(1)$, and this definitively forbids the simplification you desire.

$\endgroup$
  • $\begingroup$ Thanks for your answer. I also thought about that. However, by being able to write the integral with the dependence are we proving that there's no possible way of writing the set A without that dependence? $\endgroup$ – An old man in the sea. Feb 4 '16 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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