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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
19 votes
5 answers
10k views

Projection of Borel set from $R^2$ to $R^1$

This should be easy to prove but I have no idea how to do it: If $X \subseteq \mathbb{R}^2$ is borel then $f(X)$ is borel where $f(x,y) = x$ Thanks Tobias
Tobias Neukom's user avatar
10 votes
1 answer
3k views

Radon-Nikodym derivatives as limits of ratios

Let $\mu_1$ and $\mu_2$ be measures with $\mu_1 \ll \mu_2$. Suppose we can characterize (a version of) their Radon-Nikodym derivative this way: $$\frac{d\mu_1}{d\mu_2}(x) = \lim_{n \to \infty} \frac{\...
Neil Toronto's user avatar
6 votes
1 answer
396 views

Set of integral curves of a vector field

Let $V \colon [0,1]\times \mathbb R^d \to \mathbb R^d$ be a Borel vector field which is globally bounded, $V \in L^\infty$. I am looking for a reference for the following result (which I suppose it ...
Romeo's user avatar
  • 980
5 votes
1 answer
254 views

Boolean algebra of ambiguous Borel class

Suppose $X$, $Y$ are uncountable compact metric spaces and $\Delta^0_\xi(X)$, $\Delta^0_\xi(Y)$ ($2\le\xi\le\omega_1$) are the Boolean algebras of Borel sets of ambiguous class $\xi$. So for $\xi=2$ ...
Fred Dashiell's user avatar
4 votes
2 answers
261 views

Product of locally Borel sets locally Borel

Let $X$ be a locally compact Hausdorff space with a fixed Radon measure (= Borel measure that is finite on compact subsets, inner regular on open subsets and outer regular on Borel sets) $\mu$ . A ...
Andromeda's user avatar
  • 175
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