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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-...
David Fernandez-Breton's user avatar
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 ...
Giafazio's user avatar
  • 205
1 vote
1 answer
183 views

Topological analog of the Lusin-N property

$A\subset \Bbb{R}$ is meager if $A$ can be expressed as a countable union of nowhere dense sets. Let $f:[a, b]\to \Bbb{R}$ is absolutely continuous, i.e., for every $\epsilon>0$, there exists $\...
SoG's user avatar
  • 307
16 votes
2 answers
1k views

Is it known how the Sigma Algebra generated by Jordan measurable sets compares to universally measurable sets and analytic sets?

Unlike the collection $L$ of Lebesgue measurable sets, the collection $J$ of Jordan measurable sets do not form a Sigma algebra. (A set is Jordan measurable if and only if its characteristic function ...
Keshav Srinivasan's user avatar
5 votes
2 answers
488 views

Uncountable atomless subalgebras of the Boolean algebra of all Jordan measurable sets in [0,1]

Definition: Suppose $\mathcal A$ is the Boolean algebra of all Jordan measurable sets in $I=[0,1]$ (i.e $\mathcal A=\{A\subseteq I: \mu(\partial(A))=0\}$, where $\mu$ is the Lebesgue measure and $\...
Ameen's user avatar
  • 103