Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 ?

share|cite|improve this question

closed as too localized by Gerry Myerson, Mark Meckes, George Lowther, Andrés Caicedo, Anton Petrunin Apr 28 '11 at 15:54

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

More appropriate for –  camomille Apr 28 '11 at 11:33
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). –  Mark Apr 28 '11 at 11:36
OK, I will post it there instead –  Cosmonut Apr 28 '11 at 11:55
Since it has now been posted to (it's question 35606 there), I guess we can close it here. –  Gerry Myerson Apr 28 '11 at 12:10
I posted an answer over at…. 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. –  Joel David Hamkins Apr 28 '11 at 13:01

1 Answer 1

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.