All Questions
8 questions
6
votes
1
answer
290
views
Subset of the reals with zero inner measure and "full" outer measure in $\mathsf{ZF}+\mathsf{DC}$
Working in $\mathsf{ZF}+\mathsf{DC}$ (that is, we are allowed to use Dependent Choice but not full choice), suppose that there exists a non-measurable subset of the unit interval $[0,1]$ (just non-...
1
vote
1
answer
258
views
What is the measure of two sets which partition the reals into subsets of positive measure?
This is a follow up to this question, where I wish to partition the reals into two sets $A$ and $B$ that are dense (with positive measure) in every non-empty sub-interval $(a,b)$ of $\mathbb{R}$.
(In ...
- 63
4
votes
1
answer
351
views
$\sigma$-algebra generated by analytic sets
The Borel $\sigma$-algebra $\cal B$ on real numbers has many good properties. For instance, all continuous functions are $\cal B/\cal B$-measurable. On the other side, not only $\cal B$ is not ...
- 205
15
votes
2
answers
530
views
Nontrivial signed measure on Lebesgue measurable sets being trivial on Borel sets
Let $\mathfrak{L}(\mathbb{R})$ be the collection of Lebesgue measurable sets and $\mathfrak{B}(\mathbb{R})$ be the Borel sets.
Question: Is there a nontrivial signed measure on $\mathfrak{L}(\mathbb{R}...
- 807
14
votes
1
answer
596
views
On the existence of a family of countably additive extensions of Lebesgue measure
Let $m$ be Lebesgue measure on $\mathbb R$, and let $m_i$ and $m_o$ be the inner and outer measures respectively.
Is it the case that for all $A \subset \mathbb R$ and all $x \in [m_i(A), m_o(A)]$ ...
- 869
10
votes
1
answer
326
views
Partition into sets of positive outer measure
Let $\mu^{\star}$ denote Lebesgue outer measure. Suppose $X \subseteq [0, 1]$ and $\mu^{\star}(X) > 0$. Can we divide $X$ into uncountably many sets $\{X_i : i \in I\}$ such that for every $i \in I$...
- 121
15
votes
1
answer
572
views
Does the existence of a non-principal measure on ω imply that of a non Lebesgue measurable set?
A non-principal [probability] measure on a set X is a function $\mu$ defined on all subsets of $X$, with values in $[0,1]$, which is finitely additive, satisfies $\mu(X)=1$, and vanishes on singletons....
10
votes
1
answer
3k
views
Axiom of choice and non-measurable set
We know that existence of a Lebesgue non-measurable set follows from the Axiom Of Choice. Is the converse true? That is, does the existence of a Lebesgue non-measurable set imply the Axiom Of Choice?...
- 101