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Aug 27, 2019 at 16:30 comment added LSpice @TomLeinster, of course it's very silly, but, if you want your $f$ to be surjective, then $S$ had better be $[0, 1)$, not $[0, 1]$.
Mar 31, 2012 at 14:04 comment added gowers Does $f$ have to be continuous, or something like that? Otherwise, the result seems to be trivially false because you can mess about with the map on a set of measure zero.
Feb 20, 2012 at 11:40 comment added Tom Leinster Actually, I don't get it. Take n = 1 and S = [0, 1]. Take the map f: S --> S defined by f(x) = 2x (mod 1). Then f is measure-preserving (wrt the usual measure) and surjective, but not injective. Maybe you mean something different by "volume-preserving"...? I had a look in Agarwal and Pach's book, but couldn't find a clean statement of the result to which you're alluding.
Feb 18, 2012 at 9:03 history edited Erik Aas CC BY-SA 3.0
fixed grammar
Feb 17, 2012 at 21:51 history made wiki Post Made Community Wiki by François G. Dorais
Feb 17, 2012 at 21:07 history answered Erik Aas CC BY-SA 3.0