New answers tagged measure-theory
1
vote
For proper group action on closed Riemannian manifold, must the union of orbits with non-unique closest points to a given point be of 0 volume measure
Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has ...
7
votes
Accepted
For a closed Riemannian manifold $M$, must the set of points with non-unique closest points to a closed submanifold $S$ of $M$ be of 0 volume measure?
More generally, for any closed subset $S$ of a complete manifold $M$, the set of points $x$ at whose minimal distance to $S$ is attained at more than point has measure $0$.
Indeed, consider the ...
4
votes
Accepted
Expectation on a Polish space
If $(X,|\cdot|)$ is a Hilbert space and $Z$ is a random vector in $X$ with $E|Z|^2<\infty$, then $EZ$ is the unique minimizer of $E|Z-a|^2$ in $a\in X$.
So, more generally, for any random element $...
3
votes
Product of low dimensional Hausdorff measures
I assume that your measure is defined as
$$ \mathcal H^s\otimes\mathcal H^r(S) := \inf\left\{\sum_{k=0}^\infty\mathcal H^s(A_k)\mathcal H^r(B_k)\middle|A\in\mathcal B(\mathbb R^n)^{\mathbb N},B\in\...
7
votes
Accepted
Sufficient condition for the graph of a measurable map to be measurable
The assumption that $Y$ is countably separated cannot be meaningfully weakened. The following is Proposition 2.1 of [Musial, Kazimierz. "Projective limits of perfect measure spaces." Fund. ...
1
vote
Accepted
Lipschitz approximation of a probability measure with finite $1$-st moment by the ones with finite $p$-th moment
$\newcommand{\R}{\mathbb R}$This is only a partial answer in the sense that it will provide $\frac 1p$-Hölder maps, not Lipschitz. I still hope it can help.
Fix a big radius $R>0$ (I like $R\to\...
0
votes
Is the total variation of a vector measure $\mu$ a (classical) measure, even when $\mu$ is not of bounded variation?
I won't say there is anything wrong. You just end up with $\infty = \infty $. It is valid in the classic measure theory. The thing is that when we write sup in the definition of variation we already ...
3
votes
Accepted
Concentration of measure on spheres with respect to a unitary of trace approximately zero
This is indeed true. Assume WLOG that the matrices are diagonal. A useful way to handle the uniform measure on the sphere is that if you let $X_1,\ldots,X_n$ be iid (complex) Gaussians, then the ...
5
votes
Accepted
Integral means vs infinite convex combinations
No.
Let $(X, \cal A, \mu)$ be $[0,1]$ with Lebesgue measure.
Let $E = L^2[0,1]$ with inner product $\langle \alpha,\beta\rangle := \int \alpha(t)\overline{\beta(t)}\;dt$.
Define $f : [0,1] \to L^2[0,1]...
4
votes
Integral means vs infinite convex combinations
I don't think so. Consider the functions $f(x,y)=e^{ixy}, -1<x<1, y\in \mathbb{R}$. Then,
$$ \int_{-1}^1 f(x,y) \frac{dx}{2} = \frac{\sin(y)}{y}. $$
The question is if this is representable as
$...
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