7 votes
Accepted

Subset of the reals with zero inner measure and "full" outer measure in $\mathsf{ZF}+\mathsf{DC}$

I'll start as in Monroe Eskew's answer: Assume $X$ isn't measurable and get Borel sets $A,B$, with $A\subseteq X\subseteq B$, such that the measure of $A$ (resp. $B$) is the inner (resp. outer) ...
Andreas Blass's user avatar
5 votes
Accepted

Graph on $\mathbb{N}$ where almost every vertex is shy

All vertices can be shy. You may add edges to your graph recursively, on $n$-th step fixing all edges from $1,2,\ldots,n$ and possibly some other (finitely many) edges, so that $1,2,\ldots,n$ already ...
Fedor Petrov's user avatar
5 votes
Accepted

Does weak convergence in $L^2$ imply convergence a.e. of a subsequence?

No, but it does imply something that might suffice for your purposes. By the Banach-Saks Theorem, $(u_n)$ admits a subsequence $(u_{n(k)})$ whose Cesàro means $$ {1\over N}\sum_{k=1}^N u_{n(k)} $$ ...
John Dawkins's user avatar
  • 1,874
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]...
Gerald Edgar's user avatar
  • 40.2k
5 votes
Accepted

How should the "measure theoretic" Jacobians of a dynamical map be understood in Lai-Sang Young's "Recurrence Times and Rates of Mixing"

To motivate the terminology, let $m$ be Lebesgue measure on $\mathbb{R}^d$, let $U\subset \mathbb{R}^d$ be open, and let $F\colon U \to \mathbb{R}^d$ be a diffeomorphism onto its image. Recall from ...
Vaughn Climenhaga's user avatar
4 votes

How should the "measure theoretic" Jacobians of a dynamical map be understood in Lai-Sang Young's "Recurrence Times and Rates of Mixing"

I believe the book Nonuniformly Hyperbolic Attractors by Jose F. Alves covers the measure theoretic Jacobian in detail. Namely section 2.2, on page 17.
Nate River's user avatar
  • 4,822
4 votes

Graph on $\mathbb{N}$ where almost every vertex is shy

In a star graph, all but one vertex is shy, so a simple construction is to build a star graph from $\{1,\ldots,4\}$, another from $\{5,\ldots,12\}$, etc., doubling the size each time.
Peter Taylor's user avatar
  • 6,516
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 $...
an_ordinary_mathematician's user avatar
4 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?

My answers: yes does not apply, see 1. I did not think sufficiently about it in order to make a precise statement, but I would guess, the answer is positive as well, probably with the same proof as ...
Bernd Ammann's user avatar
4 votes
Accepted

A question about the maximal function

The answer is no. Let $m$ be a very large positive constant. You can find smooth $f$ that equals $m$ on a small ball $B_R(0)$ and still satisfy $\int_{B_6(0)}|f|<\delta$. You can do it with $R$ ...
Piotr Hajlasz's user avatar
3 votes

On the definition of symmetric rearrangement

Your approach works. I think the only thing one should be careful about is $$ \int_{\mathbb{R}^n} f (|u^*|) = \int_{\mathbb{R}^n} f (|u|), $$ which can fail for example if $f(t)=(1 -t^2)^2$.
Jean Van Schaftingen's user avatar
3 votes
Accepted

Decay rate of $\lim_{|z| \to \infty} \|1_{B(z, 1)} f\|_{L^p} =0$

Such a bound does not exist: Consider $f=f_z:=1_{B(z,1)}$.
Iosif Pinelis's user avatar
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 ...
Marcus M's user avatar
  • 900
3 votes
Accepted

On the existence of a complicated fractal-like set of finite perimeter

The answer is yes. In what follows I will refer to the book [EG]. By Example on p.198 we know wthat if $U\subset\mathbb{R}^n$ is open with smooth boundary and $\mathcal{H}^{n-1}(\partial U)<\infty$,...
Piotr Hajlasz's user avatar
2 votes
Accepted

From convergence of sequences to uniform convergence in probability

The answer is no in general. Indeed, suppose that $I_n=\{2^n+1,\dots,(n+1)2^n\}$, $X_{n,i}=Y_i$, $P(Y_i=1)=1/i=1-P(Y_i=0)$, the $Y_i$'s are independent, and $c=0$. Then all your conditions hold. ...
Iosif Pinelis's user avatar
2 votes

Graph on $\mathbb{N}$ where almost every vertex is shy

Take any locally finite countable graph with infinitely many shy vertices, e.g., the disjoint union of $\aleph_0$ copies of $K_{1,2}$. Identify the vertex set with $\mathbb N$ in such a way that the ...
bof's user avatar
  • 11.5k

Only top scored, non community-wiki answers of a minimum length are eligible