non measurable set

Hurkyl

Is the statement "All subsets of R are measurable" consistent with the axioms of ZF? (Specifically, of course, the omission of the axiom of choice)

I've never really gotten a straight answer to this question, and I'd be very happy to hear it.

matt grime

I cannot provide any definite answer I'm afraid, but I suspect the answer is yes it is consistent, as is its negation.

Since R may not be a set in some model of ZF, I don't know how this works. But, I seem to remember reading somewhere that in order to construct a non-measurable set you need to use the axiom of choice.

fourier jr

yes it's impossible to construct a nonmeasurable set without the axiom of choice. solovay proved it in 1970

mathwonk

so, to repeat the obvious: it seems that the existence of a non measurable set is an axiom equivalent to the axiom of choice?

now that i like. the axiom of choice is a stupidly obvious, highly useful, property I can barely live without. (e.g every surjective function has a right inverse: duhhh.)

measurability and non measurable sets are interesting and not obvious at all though!

Hurkyl

It would surprise me if it was equivalent -- I can't even begin to imagine how I could get from the existance of a single nonmeasurable set to being able to, say, construct a well-ordering on any (arbitrarily large!) set.

mathman

From the above discussion, it seems that without the axiom of choice, existence of non-measurable sets cannot be proven. However, it still doesn't follow that you can prove all subsets of R are measurable.

Palindrom

For the ignorant (me)- what is ZF?

fourier jr

ZF := zermelo-fraenkel axioms of set theory
http://mathworld.wolfram.com/Zermelo...kelAxioms.html

jimmysnyder

Jerry Bona once said,

The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma? ;)

Palindrom

Interesting. Thanks.

Hurkyl

I don't know what people have against the well-ordering principle. I wield transfinite induction far more easily than I wield Zorn's lemma for proving things!

The funny thing is that sometimes I don't even notice that the problem is practically already in the form needed for Zorn's lemma until I'm done with the transfinite induction proof.

jimmysnyder

They're equivalent. ;)

mathwonk

i thought it was obvious that was the content of post 3, but i could be oversimplifying.

futurebird

I was thinking about this same question today. Though, to be very honest I'm yet to find a construction of a non-measurable set that I "get" -- I can follow the steps of the proof but I don't see the big picture. You know?

It is very interesting to me that something sensible like the axiom of choice is linked to something annoying and not intuitive like non-measurable sets.

wofsy

I don't know anything about the axiom of choice but it seems that whenever you are dealing with infinite sets it would have to come up. Measurability is just a question about an infinite set - is it not?

Also measurable or non-measurable seems to mean Lebesque measurable on R^n - is that what we are talking about here?

The Continuum Hypothesis implies the existence of a non-lebesque measurable subset of the plane. Is the Continuum Hypothesis equivalent to the Axiom of Choice?


A friend of mine once tried to resurrect St. Anselm's proof of the existence of God using the Hausdorff Maximal Principle. The set of virtues are partially ordered by inclusion. The union of any set of virtues is a set of virtues so any chain in the partial ordering has a maximal element. Therefore there is a set of virtues greater than which none can be conceived. So the existence of the idea of God as consumately virtuous depends upon the Axiom of Choice

wofsy

For the real line a non-measurable set is a basis for the reals over the rationals. The existence of this basis is an application of the Axiom of Choice. But is there a proof that every non-measurable subset of the line is obtained this way?