Proof using tonelli's theorem I'm trying to learn how to use Tonelli's theorem and I'm having some trouble.
Let $E\in \mathbb{R^2}$ be borel measurable. Let $E'=\{(ax,by)|(x,y)\in E\}$ and $a,b$ are not $0$. Prove that: $m(E')=abm(E)$
I've tried integrating like so:
$m(E')=\int_\mathbb{R}\int_\mathbb{R}1_{E'}dm_xdm_y=\int_\mathbb{x\in E}(\int_\mathbb{y=ax/b}1_{E'}dm_x)dm_y$ but isn't that just $0$? because there is only one $y$ such that $y=ax/b$. Any chance you can help me understand these integrals better?
Attemted proof that $E'$ is measurable: $E$ is measurable, Thus $E_x,E^y$ are measurable. $E'=aE^y \times bE_x$ so I just need to prov that they are measurable. But $f(x)=ax$ is continuous and $aE^y=f^{-1}(E)$. Is this correct?
 A: Let $A\subseteq \mathbb{R}^2$ and let $A^y = \{x\in \mathbb{R}: (x,y) \in A\}$. Then
\begin{align*}
m(E') &= \int_\mathbb{R} \left(\int_\mathbb{R} 1_{E'}(x',y') dx' \right)dy' \\
      &= \int_\mathbb{R} m_x(E'^{,y'})dy'
\end{align*}
Notice that 
\begin{align*}
E'^{y'} &= \{x'\in \mathbb{R}:(x',y')\in E'\} \\
        &= aE^{y'/b}
\end{align*}
Therefore,
\begin{align*}
m(E') &= \int_\mathbb{R} |a| m_x(E^{y'/b})dy' \\
      &= \int_\mathbb{R} |ab| m_x(E^y) dy \\
      &= |ab| \int_\mathbb{R} \int_\mathbb{R} 1_E(x,y) dxdy\\
      &= |ab| m(E)
\end{align*}
Notice that $f:(x,y)\mapsto(ax,by)$ is a homeomorphism since $a,b\ne 0$ and thus must be bi-measurable, i.e., $f,f^{-1}$ are both measurable. Since $E$ is measurable, so must $E' = f(E)$.
A: You have to take $a,b >0$. (Measure of a set cannot be negative). $E'$ is measurable because it is the inverse image of $E$ under the continuous map $(x,y) \to (ax,by)$. Let $E_x=\{y:(x,y) \in E \}$ and $E^{y}=\{x:(x,y) \in E\}$. Then $m(E')=\int\int m((E')_x) dy dx$. Veirfy that $E'_x =\{by:y \in E_ax\}$. Hence $m(E')_x)=bm(E_x)$. Now use  a similar argument for integral w.r.t. $x$ to get $ab \int\int m(E)_x) dy dx=abm(E)$.
Note that I am using the same symbol $m$ for Lebesgue measure on $\mathbb R^{2}$ as well as $\mathbb R$
