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Let $f: R^n \rightarrow R^m$ be any function. Will the graph of f always have Lebesgue measure zero ?

1) I could prove that this is true if f is continuous.

2) I suspect it is true if f is measurable, but I'm not sure. (My idea was to use Fubini's theorem to integrate the indicator function of the graph, but I don't know if I'm using the theorem properly).

If 2) is incorrect, what would be a counterexample where the graph of f has positive measure ?

If 2) is correct, can we prove the existence of a non-measurable function whose graph has positive measure ?

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    $\begingroup$ More appropriate for math.stackexchange.com $\endgroup$
    – camomille
    Commented Apr 28, 2011 at 11:33
  • $\begingroup$ Note that if $f$ is non-measurable then the graph of $f$ is not a measurable set, so in this case you should probably ask if it has positive inner measure (or outer measure). $\endgroup$
    – Mark
    Commented Apr 28, 2011 at 11:36
  • $\begingroup$ OK, I will post it there instead $\endgroup$
    – Cosmonut
    Commented Apr 28, 2011 at 11:55
  • $\begingroup$ Since it has now been posted to m.se (it's question 35606 there), I guess we can close it here. $\endgroup$
    – Gerry Myerson
    Commented Apr 28, 2011 at 12:10
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    $\begingroup$ I posted an answer over at math.stackexchange.com/questions/35606/…. One can build a function using the axiom of choice, whose graph is not contained in any $G_\delta$ set with less than full measure. Thus, the graph has full outer measure. Meanwhile, the inner measure must always be zero, since there are uncountably many disjoint vertical translations. $\endgroup$
    – Joel David Hamkins
    Commented Apr 28, 2011 at 13:01

1 Answer 1

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If $f$ is a measurable function, then its graph is a measurable subset of $\mathbb R^{n+m}$. By Fubini's theorem, the measure of the graph is 0.

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  • $\begingroup$ (Hoping that someone will see this) why does that follow from Fubini? $\endgroup$
    – Amir Sagiv
    Commented Jun 12, 2018 at 21:07

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