Linked Questions
16 questions linked to/from Can one construct a non-measurable set without Axiom of choice?
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Can we construct a non-measurable set using only the axiom of countable choice? [duplicate]
I know that it's established that non-measurable sets cannot be constructed without the axiom of choice but I was wondering if countable choice was enough? Since the ones that I have seen use ...
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Without AOC, every subset of $\mathbb{R}^n$ is measurable [duplicate]
My analysis prof constructed a non-measurable subset of $\mathbb{R}$ today. A student noticed he used axiom of choice and asked for a construction that doesn't use it. Then, another student responded ...
user975734
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Reference request for proof that there is no non-trivial translation invariant measure on the power set of $\Bbb R^n$ which does not invoke AC [duplicate]
The standard proof considers the equivalence relation $x\sim y\iff x-y\in\Bbb Q$, and from the quotient set $[0,1]/\sim$ we select one representative from each equivalence class, and place it in a set ...
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Equivalent between axiom of choice and existence of non-measure sets? [duplicate]
It can be shown that axioms of choice allows us to construct non-measurable sets but can I say the converse i.e. $ZF + non \; measure \; sets \; exist$ contains $ZFC$.
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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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Why do we not avoid the phrase "if we assume AC " and take it as granted?
This is slightly different question. First I need to mention that I am neither a mathematician nor a researcher. As an ordinary student the separation " with and without Axiom of choice " ...
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Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice?
Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice?
What would it imply if there was such an example?
EDIT: For instance, maybe this will ...
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Construct a subset in $\mathbb{R}^{2}$ that is not Lebesgue measurable
Construct directly a subset in $\mathbb{R}^{2}$ that is not Lebesgue measurable. (Don't use the corresponding result in $\mathbb{R}$).
Since I could not use the result in $\mathbb{R}$ where we can ...
- 1,987
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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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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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Lebesgue-measurable sets requiring the Axiom of Choice to construct
Every known construction of the Vitali set relies on the Axiom of Choice. It happens to not be Lebesgue-measurable.
Must every set whose construction relies on the Axiom of Choice not
be Lebesgue-...
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Measurability of $\sup_{[0,1]}X(t)$ with respect to continuity
Let $X\colon[0,1]\to\mathbb{R}$ be a stochastic process on some probability space $(\Omega, \mathcal{F}, P)$.
I was always told that
$$
f(\omega)=\sup_{t\in[0,1]} X(t, \omega)
$$
is not always a ...
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Can we pick representatives from the "difference is rational" equivalence classes?
Let us define an equivalence relation $\sim$ on $\mathbb{R}$ by saying that $x\sim y$ if $x-y\in \mathbb{Q}$? This equivalence relation partitions $\mathbb{R}$ into uncountably many equivalence ...
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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 ...
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Any rule of thumb that says any reasonable function I can write down is measurable?
Question arose in the context of probability: If $Y$ is $F-$measurable and $h$ is some function, $h(Y)$ is $F-$measurable if $h$ is a measurable function (Doob-Dynkin).
For example, let $f(x,y)$ be ...
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