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Let $D \subset \mathbb{R}^d$ and let $f:D \rightarrow \mathbb{R} $ be measurable function. Let $G=\{(x_1,x_2,\ldots,x_d,f(x_1,x_2,\ldots,x_d))\in \mathbb{R}^{d+1}:(x_1,x_2,\ldots,x_d)\in D \} $ be the graph of $f$.

Show that $d+1$-dimensional Lebesgue measure of $G$ equals $0$.

I honestly don't know how to bite that. I thought of using an integral and applying Fubini's theorem, but I don't even know how to start.

I would appreciate any hints.

Michael Hardy
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Mateusz
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    Well, you could have a look here: http://mathoverflow.net/questions/63284/lebesgue-measure-of-the-graph-of-a-function – kummerer94 Jan 06 '15 at 16:11
  • @kummerer94 Thanks, but it is more general question and no direct answer is given (as I assume f is measurable). There is a comment about Fubini's theorem, but I would like to know how to use it here (because I can't think of a way to do it). – Mateusz Jan 06 '15 at 16:14
  • Related (I found it by reading the MO question linked in the other comment): http://math.stackexchange.com/questions/35606/lebesgue-measure-of-the-graph-of-a-function. – PhoemueX Jan 06 '15 at 16:20
  • @PhoemueX I found it too, but again, not really helpful. – Mateusz Jan 06 '15 at 16:21
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    I first posted an answer to this question and then noted that the proof is just the same as I gave here: http://math.stackexchange.com/questions/954589/graph-of-continuous-function-has-measure-zero-by-fubini. The proof is just the same, even without the continuity assumption on $f$ (as long as $f$ is measurable). – PhoemueX Jan 06 '15 at 16:25
  • @PhoemueX thanks for the proof as it is very helpful and clear. One more question: how exactly do I prove that G is measurable? It seems intuitive, but I don't think that intuition suffices here. – Mateusz Jan 06 '15 at 16:43
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    First, you should convince yourself that $g : \Bbb{R}^{n+1} \to \Bbb{R}, (x_1, \dots, x_{n+1}) \mapsto f(x_1, \dots, x_n)$ is measurable. This implies that also $h : \Bbb{R}^{n+1} \to \Bbb{R}, x= (x_1, \dots, x_{n+1}) \mapsto x_{n+1} - g(x)$ is measurable. But then $G = h^{-1}({0})$ and inverse images of intervals (${0} = [0,0]$ is an interval) are measurable. – PhoemueX Jan 06 '15 at 17:12

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