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Suppose that $f: R^2 \to [0, \infty)$ is measurable, $\Omega_1 \subseteq R^2$ is Lebesque measurable and $\Omega = \{(x, y, z) \in R^3 | (x, y) \in \Omega_1, 0 \leq z \leq f(x, y)\}$. Show that $\Omega$ is Lebesgue measurable in $R^3$ and $|\Omega|_3 = \int_{\Omega_1} fdxdy$

What i tried:

I tried to cover the set $\Omega$ with an open set, to verify the definition of lebesgue-measurable...I tried to use the measure of the function $f$ but I got nothing, then after a few hours I stuck

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The graph of a measurable function to the reals (endowed with the Borel $\sigma$-algebra) is always measurable in the product $\sigma$-algebra. – Michael Greinecker Nov 18 '12 at 19:28
so the idea is to find the open set that contains the graph.... and join it with an open set that contains the rest of the solid... ? Please help – LFRC Nov 18 '12 at 19:40
Actually, what I wrote was a bit misleading. You have to use the measurability of the hypograph of $f$. I don't know which results to that effect you covered in your course. – Michael Greinecker Nov 18 '12 at 19:46
My cuestion is that if use the fact that the hypergraph is measurable, is possible to find an open that contains $\Omega$, that satisfices the definition, I mean that if I take an open set that exist for the hypergraph with other open set is possible to find the desired open set? – LFRC Nov 18 '12 at 20:01
possible duplicate of About Lebesgue measure – Martin Argerami Nov 20 '12 at 3:12

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