Is "All Subsets of R Measurable" Consistent w/ ZF Axioms?

[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 existence 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-FraenkelAxioms.html

 

[Jimmy Snyder]

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.)

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]

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

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.

 

[Jimmy Snyder]

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?

 

[mathman]

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?

No - they are independent of each other.

Could you back up your first statement?

 

[wofsy]

No - they are independent of each other.

Could you back up your first statement?

Sure.

Assume the Continuum Hypothesis and map the real numbers from 0 to 1 bijectively onto the first uncountable ordinal.

In the unit square, define the function f(x,y) = 1 if x<y in the induced ordering, 0 otherwise.

f is the characteristic function of the set (x,y) with x<y so this set is measurable if f is a measurable function. If f is measurable then we can integrate it and find the measure of the set. By Fubini's theorem we can calculate the integral as a double integral and can integrate along either the x or y-axis first and get the same answer.

Intgerating along the x-axis first:

For each y, there are only countable many x < y since the ordering is onto the first uncountable ordinal. Thus the integral of f over the x-axis is zero since any countable set has measure zero. The double integral is thus zero.

Integrating along the y-axis first:

For each x, there are only countable many y less that it so the measure of the set of y greater than x is 1. Thus the integral along the y-axis is 1 for each fixed x. Thus the double integral equals 1.

This contradicts Fubini's Theorem so f is not measurable.

 

[Hurkyl]

Sure.

Assume the Continuum Hypothesis and map the real numbers from 0 to 1 bijectively onto the first uncountable ordinal.

That is a well-ordering of R. Why does it exist?

 

[wofsy]

Can we do that without the well-ordering theorem? Without it, there exist cardinal numbers that are not bijective with ordinal numbers...

That is a good question. I am not sure of the answer. I think though that the first uncountable ordinal can be constructed inductively and does not require the Well Ordering Hypothesis. It inherits a well ordering from its construction. The bijection of the reals onto it - i.e. the assumption of the Continuum Hypothesis - automatically transfers this ordering to the reals.I think the well ordering Hypothesis really says that any set can be mapped bijectively onto one of the ordinals - countable or uncountable. My reading of Cantor's construction of the ordinals is that the well ordering is an automatic by product of the way he constructs them. For instance, the inductive construction of the integers automatically gives you a well ordering of them. But... I am not sure.

 

[Hurkyl]

The bijection of the reals onto it - i.e. the assumption of the Continuum Hypothesis - automatically transfers this ordering to the reals.

The continuum hypothesis says the first uncountable cardinal is bijective with R. The first uncountable ordinal may be bigger, I think. (Although I'm not sure)

 

[wofsy]

The continuum hypothesis says the first uncountable cardinal is bijective with R. The first uncountable ordinal may be bigger, I think. (Although I'm not sure)

The first uncountable ordinal can not be bigger because it has only countable many predecessors. There is no predecessor that is uncountable.

 

[Hurkyl]

The first uncountable ordinal can not be bigger because it has only countable many predecessors.

Almost by definition, the first uncountable ordinal has uncountably many predecessors.

Did you mean all of its predecessors are countable? That's not a contradiction, if there are no ordinal numbers with cardinality |R|...

 

[wofsy]

Almost by definition, the first uncountable ordinal has uncountably many predecessors.

Did you mean all of its predecessors are countable? That's not a contradiction, if there are no ordinal numbers with cardinality |R|...

You are right. I meant that any of its predecessors is countable and has only countably many predecessors.