May somebody show me the proof of the theorem: Lebesgue measure of a graph of continuous function is equal to $0$. I will be really grateful.
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First restrict the domain your function to a compact interval and think about what happens when you translate the graph up and down. – Tim kinsella Jun 19 '13 at 21:55
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Duplicate? http://math.stackexchange.com/questions/108660/measure-zero-of-the-graph-of-a-continuous-function?rq=1 – DCT Jun 19 '13 at 22:18
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As for a hint, all the essential ideas are in the case where the function is from [0,1] to R. Then, use uniform continuity to cover the graph with boxes of small height (by partitioning [0,1]). – DCT Jun 19 '13 at 22:19
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@Dtseng Yeah this is a duplicate, although I'm surprised the proof I mentioned above isn't suggested in the duplicate post. Maybe I'm missing something. – Tim kinsella Jun 19 '13 at 22:27
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1I think it's a good idea (more clever than what I suggested), as I think it shows that if the graph is measurable, it has measure 0 (without continuity hypothesis). The only thing I think is that you first need to know the graph is measurable. The post http://math.stackexchange.com/questions/35606/lebesgue-measure-of-the-graph-of-a-function?rq=1 seems to be relevant. – DCT Jun 19 '13 at 22:39
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It rather depends on what you can use. If you know that the graph of a continuous function is closed, then it's obviously measurable. Then you can apply Fubini to its characteristic function to conclude that the measure is 0, simply because singletons have measure 0.
Niels J. Diepeveen
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I know this is pretty silly, but how exactly do you use Fubini's Theorem there? The argument makes perfect sense, but I can't seem to write it down appropriately. – José Siqueira Oct 01 '14 at 11:25
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$\lambda(\Gamma) = \int \chi_\Gamma d\lambda = \int\int \chi_\Gamma(x, y) ,dy dx = \int 0 ,dx = 0$ – Niels J. Diepeveen Oct 02 '14 at 20:52