All Questions
10 questions
5
votes
0
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95
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Is there an equivalent condition for Borel projections being Borel?
Let $X,Y$ be Polish spaces, and $B\subseteq X \times Y$ a Borel subset. The projection $B_X$ is not necessarily Borel in $X$. I have seen a few sufficient conditions for the projection to be Borel, ...
-1
votes
1
answer
132
views
What is an "open Baire set"?
In Measures Which Agree on Balls by Hoffmann-Jørgensen, it is stated that if $\varphi$ is a Baire function (which I presume means a pointwise limit of continuous functions), then $\{a<\varphi\}$ is ...
1
vote
1
answer
120
views
How to characterize the Borel sets of product between finite and uncountable space?
Consider the product space $Z=X\times Y$, where $X$ is a finite set with discrete topology and $Y$ is an uncountable compact subset of $\mathbb{R}^n$ with the usual subspace topology. Denote with $\...
2
votes
1
answer
852
views
The Borel sigma-algebra of a product of two topological spaces
The following problem arose while trying to justify some "known results" in abstract harmonic analysis on noncommutative groups, for which I couldn't find explicit statements in the ...
6
votes
1
answer
353
views
A strong Borel selection theorem for equivalence relations
In Kechris' book "Classical Descriptive Set Theory" there is the following theorem (12.16):
Let $X$ be a Polish space and $E$ an equivalence relation such that every equivalence class is ...
4
votes
1
answer
718
views
Is every element of $\omega_1$ the rank of some Borel set?
It is well known that we can obtain the $\sigma$-algebra of Borel subsets of $2^{\omega}$ in the following way: Let $B_0$ be the collection of all open subsets of $2^{\omega}$. For $\alpha=\beta+1$, ...
5
votes
1
answer
310
views
Abstract transverse measure theory
After reading Noncommutative Geometry book (see here) I came across the notion of the so called abstract transverse measure theory which is a generalization of standard measure theory well adapted to ...
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 ...
4
votes
3
answers
688
views
Does every separated measurable space embed into a power of $\{0,1\}$?
Let $(X,\Sigma)$ be a measurable space of arbitrary cardinality. I would like to understand under which conditions this space is isomorphic to a measurable subset of $\{0,1\}^\kappa$ for some cardinal ...
75
votes
4
answers
24k
views
Non-Borel sets without axiom of choice
This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel sets would be a Hamel basis, which needs axiom of ...