2
votes
1answer
38 views

Does Vitali set imply the axiom of choice

I know that the construction of Vitali set needs the axiom of choice, but this only states that $AC \implies V$. Is it also true that $V \implies AC$? If $\neg AC \implies \neg V$, then what ...
5
votes
0answers
66 views

Connectedness of parts used in the Banach–Tarski paradox

A quote from the Wikipedia article "Axiom of choice": One example is the Banach–Tarski paradox which says that it is possible to decompose the 3-dimensional solid unit ball into finitely many ...
0
votes
1answer
94 views

“Hidden” axiom of choice?

Let $\mu$ be a measure on $S$ such that: $\mu\left(\emptyset\right)=0$ and $\mu(S)=1$ if $X\subseteq Y$, then $\mu(X)\leq\mu(Y)$ $\mu\left(\{a\}\right)=0$ for all $a\in S$ if $X_n$, ...
7
votes
1answer
117 views

Can Tarski's circle squaring problem be solved with measurable sets and/or without the Axiom of Choice?

Tarski asked whether a disk can be decomposed into finitely many pieces which can be rearranged into a square (necessarily of the same area by the failure of the Banach-Tarski paradox in two ...
3
votes
2answers
188 views

Why is the measure of the reals not zero?

I have followed the argument that rationals, being countable and ordered, can be covered by a convergent sequence of decreasing intervals. I am trying to understand why the same argument can’t be ...
4
votes
3answers
213 views

We cannot write this function

I'm a new user so if my question is inappropriate, please comment (or edit maybe). If we accept axiom of choice, we can find a choice function for $\mathbb{R} / \mathbb{Q} $ , this is obvious. But we ...
8
votes
1answer
175 views

Nonatomic vs. Continuous Measures

Here is an old measure theory exercise I remember solving, but I'm now a bit fuzzy on the details. Let $(X,\Sigma,\mu)$ be a finite measure space. Call $\mu$ nonatomic if for any $A\in\Sigma$ ...
13
votes
2answers
257 views

Is every vector space basis for $\mathbb{R}$ over the field $\mathbb{Q}$ a nonmeasurable set?

The existence of subsets of the real line which are not Lebesgue measurable can be argued using the Axiom of Choice. For example, define an equivalence relation on $[0, 1]$ by $a \thicksim b$ if and ...
9
votes
2answers
330 views

The “it's not possible” statement in math and the Axiom of Choice

This question actually consists 3 related pieces of text, which I've gathered under this title about which I would like your opinion (they rather contain the implicit question "is this the right way ...
2
votes
1answer
404 views

Lebesgue measure, Borel sets and Axiom of choice

I cannot proceed my study on measure theory since it seems my measure theory is really unstable. I desperately need someone to briefly answer below 3 questions... **For convenience, i will write ...
2
votes
0answers
99 views

Does the existence proof of $\mu$-completion require choice?

Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mathfrak{M}^*=\{E\subset X|\exists A,B\in\mathfrak{M} \text{ such that} A\subset E \subset B \text{ and} \mu(B\setminus A)=0\}$. How do i show that ...
1
vote
2answers
160 views

If the axiom of choice is false and every set is measurable, do $\sigma$-algebras still have a purpose?

Let's say we construct measure theory without the axiom of choice in our pocket. Since we don't have to worry about unmeasurable sets anymore (see this thread), is there any good reason one might ...
7
votes
1answer
141 views

Given a model of ZF where $ \mathbb{R} $ is the countable union of countable sets, does every subset of $ \mathbb{R} $ have measure zero?

The question basically says it all. It is a well-known result that there exists a model $ \mathcal{M} $ of ZF with the property that $ \mathbb{R}^{\mathcal{M}} $ (here, $ \mathbb{R}^{\mathcal{M}} $ is ...
3
votes
1answer
218 views

Lebesgue theory and axiom of choice

I have been told that the existence of non-Lebesgue-measurable sets on $\mathbb R$ is impossible without axiom of choice. Do any other well-known results in Lebesgue theory depend on the axiom of ...
8
votes
1answer
411 views

Intuitionistic Banach-Tarski Paradox

While the Banach-Tarski paradox is a counter-intuitive result which requires the Axiom of Choice, leading some people to argue specifically against Choice, and others to argue for constructive ...
1
vote
2answers
395 views

Advantage of accepting non-measurable sets

What would be the advantage of accepting non-measurable sets? I personally feel that non-measurable sets only exist because of infamous Banach-Tarski paradox...
12
votes
1answer
207 views

Are sets constructed using only ZF measurable using ZFC?

Suppose $S$ is a subset of $\mathbb{R}$ which can be defined without using the axiom of choice, i.e. which can be proved to exist using only the axioms of ZF. Does it follow that $S$ is measurable? ...
9
votes
1answer
737 views

Can one construct a non-measurable set without Axiom of choice?

Is axiom of choice required to show the existence of non-measurable sets? Is there a Lebesgue non-measurable set that can be constructed without axiom of choice? Related question on MO says it is ...
6
votes
2answers
269 views

Is there a Lebesgue measurable choice function?

A mapping $f$ from $\mathbb R$ to $\mathbb R$ is called a choice function if, for any $x, y \ {\rm in}\ \mathbb R$, $f(x)-x \in\mathbb Q$ and $f(x)=f(y)$ whenever $x-y$ is rational. My questions is: ...
7
votes
1answer
214 views

Existence of the measure without Axiom of Choice

Let $X$ be any set and $\mathscr F$ be a $\sigma$-algebra of its subsets, so $(X,\mathscr F)$ is a measure space. The function $$ \mu:\mathscr F\to[0,\infty] $$ is called a measure if $\quad 1.$ ...
10
votes
3answers
1k views

Axiom of choice, non-measurable sets, countable unions

I have been looking through several mathoverflow posts, especially these ones http://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice , ...
2
votes
2answers
245 views

Dense subset of the plane that intersects every rational line at precisely one point?

It seems there should exist a non-measurable bijection $f: \mathbb{R}\rightarrow \mathbb{R}$. And thus we can obtain a non-measurable graph on $\mathbb{R}^2$ which intersects every horizontal or ...
3
votes
2answers
280 views

Lebesgue Unmeasurable Sets in $\mathbb{R}$

I have seen a proof showing that there are subsets of $\mathbb{R}$ which are not Lebesgue measurable. If I recall correctly it uses the axiom of choice. My first question is, are there sensible sets ...