All Questions
Tagged with axiom-of-choice measure-theory
80
questions
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The relation between the cardinality of Bore $\sigma$-algebra and axiom of choice
Under axiom of choice, the cardinality of Borel $\sigma$-algebra $B$ is $\mathfrak{c}$.
In this proof axiom of choice is used three times: To prove
$\omega_1$-times recursion is sufficient,
each $|B_{...
1
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1
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86
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Finite Content Not Continuous on the Empty Set
I have come across a lemma which states:
Let $\mu$ be a finite content on an algebra $\mathcal{A}$.
$\mu$ is a pre-measure if and only if $\mu$ is continuous on the empty set, i.e., for every sequence ...
12
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1
answer
445
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To what extent will Radon-Nikodym's theorem hold if we do not admit axiom of choice?
In the proof of the Radon-Nikodym theorem, the function $f = \frac{\mathrm{d} \nu}{\mathrm{d} \mu}$ is constructed as the supremum of measurable functions satisfying $\int_A f \mathrm{d} \mu \leq \...
- 658
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Where is Choice invoked in the following recursive definition?
Lemma: Let $\rho$ be a signed measure that omits the value $+\infty$. If $\rho(G)>0$, then there exists a subset $E\subseteq G$ such that $E$ is totally positive and $\rho(E)>0$.
Proof: Let $E_1=...
- 1,467
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2
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95
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How do we reconcile Solovay’s Model with the Hanh Banach Theorem in ZF
From Wikipedia, I have the following two facts:
Solovay’s Model is a model of ZF in which all sets are measurable. In this way Solovay showed that in the proof of the existence of a non-measurable ...
- 33
0
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1
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161
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Does "Lebesgue null sets is closed under countable union" need the axiom of choice?
In this question, it has been asked whether the fact that the countable union of Lebesgue null sets is null, uses the Axiom of Choice or not. My opinion is that it is not needed. Here's a proof of ...
- 872
5
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Choice requirement in defining Lebesgue measure
In this MO post, the answers point out that the statement, "The reals are a countable union of countable sets" is consistent with $\textsf{ZF}$ (without choice). This question points out ...
- 1,249
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2
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71
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Choosing a "good" partition given that non-null measurable sets have a "good" subset with positive measure
I came up with this lemma when solving some other problem:
Given a measure space $(X, \mu)$ with $\mu(X) < +\infty$. There is a collection $\mathcal G \subseteq 2^X$ of "good" subsets, ...
- 4,867
2
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0
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86
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Measure and Integration theory based on intuitionism
Measure and integration theory uses the axiom of choice extensively, for example the idea behind σ-algebra is that there are non-measurable sets (in the sense of lebesgue) like Vitali set, but on the ...
- 143
2
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1
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158
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Is it possible to find Bernstein set without Axiom of Choice?
Statement(True/False) : Without Axiom of of choice there doesn't exists any set $B$ such that both $B$ and $B^c$ intersect every closed uncountable subsets Or every perfect sets in $\Bbb{R}$.
Any set ...
- 13k
2
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1
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280
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Vitali covering lemma and choice
Wikipedia proof of Vitali covering lemma uses Zorn's lemma. I was able to replace part of the argument which uses Zorn's lemma with another argument(I′ll provide the idea without the details in the ...
- 2,104
7
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2
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648
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Defining a function without using Axiom of Choice
I have a situation where I do not know if I need the axiom of choice:
Let $\mathcal{B}(\mathbb{R})$ be the collection of Borel measurable subsets of $\mathbb{R}$.
I have a (possibly non-Borel) subset $...
- 25.1k
0
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33
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Non-measurable sets construction [duplicate]
In explanation of non-measurable sets we take $Q_1=Q \cap [-1,1]$ (Q is set of rational numbers). Then we define equivalence relation on $[0,1]$
$x,y \in [0,1]$ as $x\sim y$ iff $x-y \in Q_1$. This ...
1
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1
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34
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Confused about axiom of choice applied to set of orbits for developing a non-measurable set
On p441 of Real Mathematical Analysis by Pugh, but also in this question and this blog post, the axiom of choice is applied to a set of irrational orbits $O(x)=\{R^k(x):k\in \mathbb{Z}\}$ to create a ...
- 13
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134
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Usage of the Axiom of choice in Terence Tao's measure theory (Exercise 1.2.19)
Let $E \subseteq \Bbb R^d$. Show that the following are equivalent. 1. $E$ is Lebesgue measurable. 2. $E$ is a $G_\delta$ set with a null set removed.
Let $E$ be Lebesgue measurable. By outer ...
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