# Tagged Questions

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### 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 ...
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### 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 ...
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### “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 $X_n$, ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 $A\in\Sigma$ ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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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 ...
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### 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...
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### 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? ...
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### 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 ...
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### 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: ...
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### 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 $\quad 1.$ ...
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### 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 , ...
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 ...
### 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 ...