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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 outer measure?

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Originally posted to MO (mathoverflow.net/questions/63284/…), now posted here in response to a suggestion there. –  Gerry Myerson Apr 28 '11 at 12:11
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1 Answer

No function can have a graph with positive measure or even positive inner measure, since every function graph has uncountably many disjoint vertical translations, which cover the plane.

Meanwhile, using the axiom of choice, there is a function whose graph has positive outer measure. The construction is easiest to see if one assumes that the Continuum Hypothesis is true, so let me assume that.

To begin, note first that there are only continuum many open sets in the plane, since every such set is determined by a countable union of basic open balls with rational center and rational radius. Next, it follows that the number of $G_\delta$ sets is also continuum, since any such set is determined by a countable sequence of open sets, and $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$.

Thus, we may enumerate the $G_\delta$ sets in the plane as $A_\alpha$ for $\alpha\lt \aleph_1$ (using CH). Build a function $f:\mathbb{R}\to\mathbb{R}$ by transfinite induction. At any stage $\alpha\lt \aleph_1$, we have the approximation $f_\alpha$ to $f$, and we assume that it has been defined on only $\alpha$ many points. Given $f_\alpha$, consider the $G_\delta$ set $A_\alpha$. If we can extend $f_\alpha$ to a function $f_{\alpha+1}$ by defining it on one more point $x$, so that $(x,f_{\alpha+1}(x))$ is outside $A_\alpha$, then do so. Otherwise, $A_\alpha$ contains the complement of countably many vertical lines in the plane, and thus has full measure.

After this construction, extend the resulting function if necessary to a total function $f:\mathbb{R}\to\mathbb{R}$. It now follows that the graph of $f$ is not contained in any $G_\delta$ set with less than full measure. Thus, the graph has full outer measure.

Now, finally, the same construction works without CH, once you realize that any $G_\delta$ set containing the complement of fewer than continuum many vertical lines has full measure.

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This is a very interesting construction, what happens when one removes the axiom of choice from the theory? Are there any results in that area related to such existence of non-existence of such functions? –  Asaf Karagila May 4 '11 at 13:58
    
A mathematician's answer is so blatantly different from the ones given by undergrad/grad students. –  mez May 19 '13 at 20:59
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