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Apr 27, 2020 at 15:23 answer added Claudio Rea timeline score: 0
Apr 26, 2020 at 2:13 comment added jcdornano By the aswer of Nik Weaver we have the same result for any infinite $\Gamma$.
Apr 25, 2020 at 22:21 history became hot network question
Apr 25, 2020 at 16:37 answer added Nik Weaver timeline score: 14
Apr 25, 2020 at 14:56 answer added Josiah Park timeline score: 6
Apr 25, 2020 at 14:45 comment added Gerhard Paseman If there is a counterexample, E can be taken to be a union of orbits of Gamma without being of full measure. I think one can show a la Vitali that such an E is non-measurable, but I am unsure. If not, consider Bernstein sets. Gerhard "Taking Measure Of This Question" Paseman, 2020.04.25.
Apr 25, 2020 at 14:12 history asked Claudio Rea CC BY-SA 4.0