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In an exercise for class we were asked to prove that the graph of a continuous measurable function has measure zero.

Ok, so let us just look at some measurable function that is not necessarily continuous. For example the characteristic function on the set of irrationals. Then the graph is an uncountable union of points. I know that each point has measure zero, but how do we deal with the fact that we have uncountably many such points ?

I think the graph of such a function would then have measure equal to the measure of the irrationals which is not zero.

user7090
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  • I'm guessing that you are measuring the graph in the Lebesgue measure on $\mathbb{R}^2$. Since the graph is a 1-dimensional topological manifold, it makes sense that it has $0$ area. The graph of a characteristic function on any compact set in $\mathbb{R}$ is contained in rectangles of arbitrarily small area. – J. David Taylor Oct 21 '14 at 04:32
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    The graph always has measure zero as long as the function is measurable, see my answer here http://math.stackexchange.com/questions/954589/graph-of-continuous-function-has-measure-zero-by-fubini/954598#954598. If the function is not measurable, the graph will not be measurable in general. – PhoemueX Oct 21 '14 at 04:36
  • @Krzysztof: This question is similar to yours. – Alphonse Aug 21 '16 at 20:58

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If $f:\mathbb{R}^{d}\to\mathbb{R}$, then in the referenced exercise, you most likely proved $\mathcal{G}(f)$ (read: graph of $f$) has $d+1$-dimensional measure $0$, which is an important distinction, because while it does not make sense in the framework of Lebesgue measure, $\mathcal{G}(f)$ has in general non-zero $d$-dimensional measure (think arc-length and surface area; this leads into the subject known as geometric measure theory).

So even for your discontinuous counter-example we have $\mu^{d=2}(\mathcal{G}(f))=0,$ which should be obvious.

Since the graph of your example is just a "countably punctured" horizontal line in $\mathbb{R}^{2}$, you could in principle compute $\mu^{d=1}(\mathcal{G}(f))$ without invoking geometric measure theory to see that it has non-zero measure, but you ought to refrain from going there since that line of thinking is somewhat misguided

I'm not actually familiar with any function where $\mu^{d+1}(\mathcal{G}(f))\neq0$. But see space filling curve to get a sense of what such a function (if one even exists) must exhibit.

Sargera
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