9
votes

### Boolean algebra of ambiguous Borel class

This is a very interesting question whose answer depends on dimension properties of the spaces $X,Y$.
First we introduce a suitable terminology. A function $f:X\to Y$ between topological spaces is ...

4
votes

Accepted

### Are all quasi-regular points on Polish spaces generic points?

$A_n \varphi (x) = \int_X \varphi \mathrm{d} \mu_n$ for a Borel probability measure $\mu_n$, in fact a measure with finite support
$\{ T^j(x) \mid j=0,1,\ldots,n-1\}$.
By the assumption about the ...

4
votes

Accepted

### A different version of Besicovitch Covering Theorem involving balls of half radius

Are you allowing $b$ to depend on the manifold (as it appears to me from your statement)? In that case, Besicovitch is overkill, and this statement holds in much more generality than Besicovitch. One ...

3
votes

Accepted

### Is there a "smooth Kantorovich-Rubinstein duality" for Wasserstein distances on smooth/Euclidean space?

$\newcommand{\K}{\mathcal K}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$No, the formula
\begin{equation*}
W_1(\mu,\nu) = \max\Big\{ \int f \, d\mu - \int f \, d\nu \, : \, f \in \K_r\Big\}, ...

3
votes

### $\sigma$-algebra generated by analytic sets

"Inverse image of an analytic set is analytic" would imply ${\cal B}^1/ {\cal B}^1$-measurable, so let's try that. [Check my argument.]
Let $f : \mathbb R \to \mathbb R$ be continuous. Let ...

2
votes

Accepted

### Coarea formula for measure of epsilon neighbourhood

Your formula is true (up to appropriate constants) if you take $\nu$ to be the Minkowski content, assuming that the volume of the tubular neighborhood is locally Lipschitz as a function of the tube ...

2
votes

Accepted

### Can we say that there exists a measurable function $f$ such that $ \nu=f_{\#}\mu$?

Yes: if $\pi_x=\delta_{f(x)}$, then
$$\nu(A)=\pi(X\times A)=\int_X\mu(dx)\pi_x(A) \\
=\int_X\mu(dx)\,1(f(x)\in A)=\int_X\mu(dx)\,1(x\in f^{-1}(A))
=\mu(f^{-1}(A))$$
for all Borel subsets $A$ of $X$, ...

2
votes

### Qualitative difference between "continuous" and "discontinuous" states on $M(G)$

Your definition (2) explicitly uses the Fourier transform, so let's think about this.
For any locally compact group $G$ we can turn both $L^1(G)$ and $M(G)$ into Banach $*$-algebras, as you do. Then ...

1
vote

Accepted

### Regular Lagrangian flow for explicit ODE with discontinuous right-hand side

$\newcommand{\Om}{\Omega}\newcommand{\om}{\omega}\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$As stated in my previous comment, in Theorem 3.1 of the paper linked by the OP about the existence (...

1
vote

Accepted

### What does $\mu$ and $\nu$ "dependent" mean?

It makes no sense to say that two probability measures $\mu$ and $\nu$ are "completely dependent". Dependence (or lack thereof) is a property of random elements, not of probability measures.
...

1
vote

Accepted

### Dimension-preserving non-linear map

Dimension Theory by Ryszard Engelking provides the answer I was looking for. Specifically, theorem 1.12.8 which I have provided below, although there are other results in this same area that may be ...

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