# 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?

• What is $m$ here? I guess the Lebesgue measure because $m(E')=abm(E)$ is not true for every measure $m$. Also if $a<0$ and $b>0$ then both are not $0$ but $abm(E)$ might take negative values and I don't think you are dealing with signed measures. – drhab Feb 11 at 9:42

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 $$E' = f(E)$$.
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 $$\mathbb R$$