Showing that a solid of revolution is Lebesgue measurable

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Let $E \subset \mathbb{R}^2$ be Lebesgue measurable in $\mathbb{R}^2$, bounded and in the right half-plane: $(r,z) \in E \Rightarrow r >0.$ $R(E)$ is rotating around the $z$-axis.

$$R(E):=\{(x,y,z)\in \mathbb{R}^3 \space| \space(\sqrt{x^2+y^2},z) \in E \}$$

How do I show that $R(E)$ is Lebesgue-measurable in $\mathbb{R}^3$?


I think I have to do something with characteristic functions and the transformation theorem. Then I could choose a smart diffeomoprhism but I'm not really sure how.

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Consider the map $F:\Bbb{R}\times (0,\infty)\times\Bbb{R}\to\Bbb{R}^3$ defined by $F(\phi,r,z):= (r\cos\phi,r\sin\phi,z)$. Then, $F$ is $C^{\infty}$, hence locally Lipschitz, so it sends Lebesgue measurable sets to Lebesgue measurable sets. What can you deduce about the revolved set $R(E)$ in relation to $F$?