All Questions
5 questions
1
vote
0
answers
154
views
Polish spaces and analytic sets
Can we conclude that an analytic subset $A$ of a Polish space $X$ is also Polish?
Let $\mathcal{M}(R^d)$ denotes the family of Borel probability measures on $R^d$ equipped with the Lévy-Prokhorov ...
10
votes
3
answers
1k
views
How to prove that the Lebesgue $\sigma$-Algebra is not countably generated?
How to prove that there can't exist a countable set $\{A_1,A_2,\dots\}\subset \mathcal{L}(\mathbb{R})$ (where $\mathcal{L}(\mathbb{R})$ denotes the family of all Lebesgue measurable sets) such that $\...
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 $\alpha=\beta+1$, ...
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}...
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 ...