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7 votes
2 answers
448 views

Uncountable collections of distinct subsets of an interval (existence)

Throughout, $\mu$ is just the Lebesgue measure. Question: does there exist an uncountable family of distinct subsets of $[-1, 1]$, denoted by $(U_j)_{j \in [-1, 1]}$, with $\mu(U_j) > 0$ for each $...
Stepan Plyushkin's user avatar
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
4 votes
1 answer
203 views

Generalized limits in Boolean algebras

Let $\mathbb{B}$ be an infinite $\sigma$-complete Boolean algebra. By $\mathbb{B}^\omega$ we denote the countable product of $\mathbb{B}$ with the coordinate-wise operations. Let us call a ...
Damian Sobota's user avatar
5 votes
0 answers
364 views

Computing the infinite dimensional Lebesgue measure of "cubes"

There is no Lebesgue measure in infinite dimensions—this slogan is familiar to every student interested in analysis. One possible, precise statement of this result may be as follows: if $X$ is an ...
truebaran's user avatar
  • 9,330
3 votes
2 answers
236 views

Can a continuous surjection from a Hilbert cube to a segment behave bad wrt Lebesgue measures?

Suppose $\hat{I}=[0,1]^\mathbb{N}$ is a Hilbert cube and $I=[0,1]$. Consider Lebesgue measures $m_1$ and $m_2$ on $\hat{I}$ and $I$ correspondingly. By Lebesgue measure on the Hilbert cube I mean the ...
Dmitry Todorov's user avatar
4 votes
1 answer
2k views

Lebesgue measure of boundary of Caccioppoli set

Can anything be said about the measure of the topological boundary of a Cacciopoli set in $R^n$? Of course, the reduced boundary has finite (n-1)-dimensional Hausdorff measure, but this does not say ...
Martijn's user avatar
  • 320