Finding Lebesgue measure using Fubini's theorem

Let $f : \mathbb{R}^2 → \mathbb{R}$ be continuous. Let $E$ be Lebesgue measurable and

$$E_f = \{(x,y,z + f(x,y)) : (x,y,z) \in E\}.$$

Show that $m_3(E) = m_3(E_f)$ directly from Fubini’s theorem, where $m_3$ is the $3$ dimensional Lebesgue measure.

Attempt:

\begin{align*} &\iiint_{\{(x,y,z): (x,y,z)\in E+(0,0,f(x,y))\}\}} \, dxdydz \\ &=\int m_2(\{(x,y): (x,y,z)\in E-(0,0,f(x,y))\}\}) \, dz\\ &=\int_{\{z\mid (x,y,z)\} \in E} m_2(\{(x,y): (x,y,z)\in E\})\\ &=m_3(E) \end{align*}

Using Fubini, the translation-invariance of $m_1$ for each $\textit{fixed}\ (x,y),$ and the definitions of $E$ and $E_f,$ calculate (spoiler in the middle, if you get really stuck),

$\displaystyle m_3(E_f)=\int_{\mathbb R^3}\chi_{E_f}(x,y,z_1)dm_3(x,y,z_1)=\int_{\mathbb R^2}\left(\int_{\mathbb R}\chi_{E_f}(x,y,z_1)dm(z_1)\right)dm_2(x,y)$

$\displaystyle=\int_{\mathbb R^2}\left(\int_{\mathbb R}\chi_E(x,y,z+f(x,y))dm(z+f(x,y))\right)dm_2(x,y)$

$=\displaystyle \int_{\mathbb R^2}\left(\int_{\mathbb R}\chi_E(x,y,z)dm(z)\right)dm_2(x,y)=\int_{\mathbb R^3}\chi_E(x,y,z)dm_3(x,y,z)=m_3(E).$