Lebesgue measure of sets are equal using Fubini's theorem

Question

Let $f : \mathbb{R}_2 → \mathbb{R}$ be continuous. Let $E$ be lebesgue measurable and $E_f = \{(x,y,z + f(x,y)) : (x,y,z) ∈E\}$. Show that $m_3(E) = m_3(E_f)$ directly from Fubini’s theorem, where $m_3$ is the $3$ dimensional lebesgue measure.

I am unsure how to use Fubini's theorem to show that the measures are equal. How would one go about doing this? Let $(E_f)_{z}=E_z$ be the $z$ section of of $E_f$. Then

$m_3(E_f)=\int \chi_{(E_f)_{z}}m(f(x,y)+E)dm(x)dm(y)=\int \chi_{E_z}m(E)dm(x)dm(y)=m_3(E)$

How to do this? I know this attempt is way wrong, but I cannot start.

It seems $E_f=E+(0,0,f(x,y))$ which would imply $m_3(E_f)=m_3(E)$ since the graph of a continuous function has measure zero. But how to show this using Fubini?

@thewatcher do you mean $E_f$ is graph of $f$ shifted by $z$ units? — Rene, Oct 30 '21 at 03:03

score 0 · Accepted Answer · answered Oct 30 '21 at 05:05

0

The section of $E_f$ by $z$, $(E_f)_z$, can be written as $\{(x,y): (x,y,z)\in E-(0,0,f(x,y))\}$ (where $E-u$ denotes $\{v-u: v \in E\}$). Take the two dimensional Lebesgue measure ($m_2$) of this. By translation invariance of $m_2$ you get $m_2 (\{(x,y): (x,y,z)\in E\}$. Now integrate w.r.t. $z$ to finish the proof.

answered Oct 30 '21 at 05:05

Kavi Rama Murthy

311,013

Thanks so much for the help. – Rene Oct 30 '21 at 05:29
Hi Kavi, am I writing this correctly? We have $\int m_2((E_f)_z)dm(z)=\int m_2({(x,y) \mid (x,y,z \in E})dm(z)=m_3(E)$? I do not know how much I need to show. – Rene Oct 30 '21 at 21:33
@Rene Yes, what you have written is fine. – Kavi Rama Murthy Oct 30 '21 at 23:12

Lebesgue measure of sets are equal using Fubini's theorem

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