On 2016-10-22 15:07, Eugene Salamin via math-fun wrote:

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

Julian (quoted without permission):

rwg>I thought I read somewhere that there was no subset of the Reals whose

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intersection with every interval has measure density ½.

Correct. Lebesgue measure is defined (when using Carathéodory) as the inf of the measures of all countable unions of intervals containing the set. Taking a union of intervals with measure ≤1.01 times the measure of the set, one of the intervals >must have measure ≤1.01 times the measure of its intersection with the set, i.e. in that interval the set has density >99%.

rwg>Yet I think I see a nifty one. What gives?

Depends on your proposed construction. Most likely, it either gives a non-measurable set, or gives a set of density 0 or 1 (most recursive constructions do this—almost all points will alternate between being in and out), or is not well-defined.

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

I am using hallucinogens in conjunction with the axiom of stupidity. My "solution" was to partition the reals according to the relation of Khinchin's constant to the geometric mean of their partial quotients! (Sorry.) --rwg

.On Saturday, October 22, 2016, Bill Gosper <billgosper@gmail.com> wrote:

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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

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a nifty one. What gives? --rwg