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1 vote
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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 ...
B-S's user avatar
  • 39
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 $\...
Joris Wk's user avatar
  • 243
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 $\alpha=\beta+1$, ...
Hannes Jakob's user avatar
  • 1,799
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}...
Zhang Yuhan's user avatar
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 ...
Anweshi's user avatar
  • 7,442