6
$\begingroup$

Given a Borel set $A \subseteq \mathbb{R}^d, d ≥ 1$ and a measurable function $f: A \to [0, \infty)$, I want to consider the set:

$$E = \{(x, y) \in \mathbb{R}^{d+1}: x \in A, 0 ≤ y ≤ f(x)\} \subseteq \mathbb{R}^{d+1}$$

I first want to show that $E$ is a Borel set. Then, I want to prove that

$$\lambda_{d+1}(E) = \int_A f(x) d \lambda_d(x)$$

where $\lambda_d$ is the $d$-dimensional Lebesgue measure.

I unfortunately wasn't even successful showing that $E$ is a Borel set so far. I first thought that one could write $E$ as the product of two Borel sets ($E = A \times \text{another Borel set}$), but I then realized that it isn't that simple, seeing as the $y$ in a vector $(x, y) \in E$ is dependent on $x$. Maybe one could construct a clever measurable function that sends $E$ onto a measurable set in $\mathbb{R}$ or something like that? I'm not really all that sure though.

Once established that $E$ is measurable, wouldn't the second part follow more or less right from Fubini's theorem?

Also, I think the intuition behind this excercice is to acknowledge that, in case $d = 1$, the Lebesgue integral of $f$ over $A$ is nothing but the area inbetween the graph of $f$ and the $x$-axis; for $d = 2$, it's the volume, and so on. I'm not really sure how that helps me (formally) showing it.

$\endgroup$
2
$\begingroup$

(I will assume that $f$ is Borel measurable, and extend $f$ to all of $\Bbb R^d$ by setting $f(x)=-1$ for $x\in A^c$.) Think of $E$ as the inverse image $g^{-1}([0,\infty))$, where $g:\Bbb R^d\times[0,\infty)\to\Bbb R$ is defined by $g(x,y)=f(x)-y$. Then $g$ is the composition of the $\mathcal B^d\otimes\mathcal B_+/\mathcal B^2$ map $(x,y)\to (f(x),y)$ with the continuous map $\Bbb R^2\ni(u,v)\to u-v\in\Bbb R$. It follows that $g$ is $\mathcal B^d\otimes\mathcal B_+/\mathcal B$-measurable, hence $E\in \mathcal B^d\otimes\mathcal B_+$.

Similar considerations apply if $f$ is Lebesgue measurable.

$\endgroup$

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.