Let's ask for such a set S as a subset of the unit interval instead of the entire real line. Then S = S intersect [0,1], and so S must have measure 1/2. So if S can't be measurable, that completes the proof. But why is S nonmeasurable? -- Gene From: James Propp <jamespropp@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, October 22, 2016 12:46 PM Subject: Re: [math-fun] 50% gray Are you using the axiom of choice? Or is your set constructible? (Clearly it's not measurable.) Jim Propp .On Saturday, October 22, 2016, Bill Gosper <billgosper@gmail.com> wrote:
I thought I read somewhere that there was no subset of the Reals whose intersection with every interval has measure density ½. Yet I think I see a nifty one. What gives? --rwg