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Apr 28, 2011 at 15:54 history closed Gerry Myerson
Mark Meckes
George Lowther
Andrés E. Caicedo
Anton Petrunin
too localized
Apr 28, 2011 at 13:01 comment added Joel David Hamkins 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.
Apr 28, 2011 at 12:10 comment added Gerry Myerson Since it has now been posted to m.se (it's question 35606 there), I guess we can close it here.
Apr 28, 2011 at 12:06 answer added Stefan Geschke timeline score: 5
Apr 28, 2011 at 11:55 comment added Cosmonut OK, I will post it there instead
Apr 28, 2011 at 11:36 comment added Mark 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).
Apr 28, 2011 at 11:33 comment added camomille More appropriate for math.stackexchange.com
Apr 28, 2011 at 11:09 history asked Cosmonut CC BY-SA 3.0