20 votes

Is the area of the Mandelbrot set known?

New Approximations for the Area of the Mandelbrot Set gives the state of the art from 2014, and a related paper from 2015 is On a numerical approximation of the boundary structure and of the area of ...
Carlo Beenakker's user avatar
17 votes
Accepted

Understanding the condition $\frac{1}{p} + \frac{1}{q} = 1$ in the estimate $xy \le \frac{1}{p}x^p + \frac{1}{q}y^q$

This is related to the concept of Fenchel-Legendre convex conjugate. More precisely, the functions $$ \Phi(x)=\frac 1p x^p,\qquad \Psi(y)=\frac 1q y^q $$ are convex conjugate from each other if and ...
leo monsaingeon's user avatar
15 votes

Understanding the condition $\frac{1}{p} + \frac{1}{q} = 1$ in the estimate $xy \le \frac{1}{p}x^p + \frac{1}{q}y^q$

One reasonable explanation, to me, is that the above is nothing but the convexity inequality $f\big(t u+(1-t)v\big)\le tf(u)+(1-t)f(v)$ for the function $-\log$, changing the names of the variables ...
Pietro Majer's user avatar
  • 54.5k
15 votes
Accepted

Steinhaus theorem and Hausdorff dimension

The answer to the question is negative. Körner in Hausdorff dimension of sums of sets with themselves and Schmeling-Shmerkin in On the dimension of iterated sumsets showed that for any increasing ...
Jarosław Błasiok's user avatar
10 votes

Understanding the condition $\frac{1}{p} + \frac{1}{q} = 1$ in the estimate $xy \le \frac{1}{p}x^p + \frac{1}{q}y^q$

You may think the following way. Ignore the sharp constants and study when $xy\le C(x^p+y^q) $ for some $C>0$ and arbitrary positive $x, y$. This is equivalent to asking when $xy\le C\max(x^p, y^q)$...
Fedor Petrov's user avatar
  • 96.9k
9 votes
Accepted

How "correct" is Knuth's fast addition $(a,b) \mapsto (a \oplus b) \oplus ((a\land b) \ll 1)$?

Note that we have $$a + b = (a \oplus b) + ((a \land b) \ll 1). \tag{1}$$ So asking whether $a + b = a +_K b$ is asking when $x \oplus y = x + y$ where $x = a \oplus b$ and $y = ((a\land b) \ll 1)$. ...
Jarosław Błasiok's user avatar
6 votes

Understanding the condition $\frac{1}{p} + \frac{1}{q} = 1$ in the estimate $xy \le \frac{1}{p}x^p + \frac{1}{q}y^q$

This is actually the weighted AM-GM inequality for $n = 2$ in disguise. Recall that this inequality says that if $w_1, w_2$ are two non-negative weights such that $w_1 + w_2 = 1$ and $x_1, x_2 \ge 0$, ...
Qiaochu Yuan's user avatar
5 votes

Steinhaus theorem and Hausdorff dimension

Here is a quite short example to show that you question cannot have a positive answer. Assume that $V$ is a Sacks extension of constructible universe $L$. Then the set of constructible reals $A=(\...
喻 良's user avatar
  • 4,076
3 votes

The category Prob of finite measure spaces does not admit all products

$\newcommand{\pr}{\mathsf{Prob}}\newcommand{\id}{\mathrm{id}}\newcommand{\tX}{\tilde X}\newcommand{\tx}{\tilde x}\newcommand{\tmu}{\tilde\mu}\newcommand{\tnu}{\tilde\nu}\newcommand{\tpi}{\tilde\pi}\...
Iosif Pinelis's user avatar
3 votes

Generalization of the concept of a measure

I do not have reference for your type of measure, but I have a version of the Riesz representation theorem for all such measures when the lattice is distributive. In this post, we shall not assume ...
Joseph Van Name's user avatar
3 votes

Signed measures and poset inequalities

SAGE says it's not true for non-partionable but Cohen-Macaulay complexes. Problem of finding $\omega$ can be reformulated as a LP feasibility problem (as explained by Ilya) and for an example from &...
Raman Sanyal's user avatar
2 votes

Steinhaus theorem and Hausdorff dimension

Lemma 2.7 of [1] says that if $A$ is a compact subset of a separable compact abelian group $G$ with Haar measure 0, then there is a compact set $B\subset G$ with positive Haar measure such that $A+B$ ...
John Griesmer's user avatar
2 votes

An example that the sum of two Borel sets which is not a Borel set in n-dimensional Euclidean space

I'm not sure if it's easy, but this answer provides another example in $\mathbb{R}^n$ where $n\ge3$. The core idea is to exploit a non-Borel analytic set $A'_1\subset[-1,1]$ which is also the image of ...
Ano2Math5's user avatar
2 votes

How do sets with unit fractional Hausdorff measure of dimension $>1$ look like?

Let $s>1$ be aby number. The unit interval $[0,1]$ with the metric $d(x,y)=|x-y|^{1/s}$ has poisitive and finite $s$-dimensional Hausdorff measure. While, it is an abstract metric space, it can be ...
Piotr Hajlasz's user avatar
2 votes

Generalization of the concept of a measure

The OP asked "has it been discussed in the literature". The two mentioned below are, indeed, known in the literature. But they are perhaps not precisely the same as the OP. Sion, Maurice, ...
Gerald Edgar's user avatar
  • 39.9k
1 vote
Accepted

Sufficient conditions for the graph measurability of a multivalued function

Graph measurability of $P$ is not sufficient. Let $E\subseteq[0,1]^2$ be a Borel set whose projection $\pi(E)$ onto the first coordinate is not Borel. Let $X=\mathbb{R}$ and let $B$ have the constant ...
Michael Greinecker's user avatar
1 vote

Measurability of two hitting times at the stopped $\sigma$-algebra

Yes. By definition, we must show that for all $t \geq 0$, the event $A_t := \{\tau_b < \tau_a\} \cap \{\tau_b \leq t\}$ is $\mathcal F_t$ measurable. To do so, it suffices to show that $\mathbf 1_{...
Nate River's user avatar
  • 3,414
1 vote

Is there a Borel measurable $f:\mathbb{R}^d \to \mathbb{R}^d$ such that $f(x) \in \partial \varphi (x)$ for all $x$?

A Borel measurable function $f:\mathbb{R}^d\to\mathbb{R}^d$ such that $f(x)\in\partial\varphi(x)$ for all $x\in\mathbb{R}^d$ exists. This follows from the Federer-Morse Theorem. The following ...
Piotr Hajlasz's user avatar
1 vote

A possible measure-theoretic pathology

Suppose that $S$ is the graph of a continuous, strictly increasing function $\psi$. Then $S(W) = \psi(W)$, and the question asks if there is $\psi$ and $W$ such that $|W| = 1$, but $|\psi(W)| = 0$. ...
Mateusz Kwaśnicki's user avatar
1 vote

A possible measure-theoretic pathology

Let $N$ be a dense null set and $S = (([0,1] \setminus N) \times N) \cup (N \times [0,1])$ and let $W = [0,1] \setminus N$. Then $N(W) = N$ so $W$ is as needed. Let $A$ and $B$ be open intervals. ...
user1138k's user avatar

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