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 ...
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 ...
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 ...
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 ...
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 $xy\le C\max(x^p, y^q)$...
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 $y = ((a\land b) \ll 1)$. ...
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 $x_1, x_2 \ge 0$, ...
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=(\...
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}\...
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 ...
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 &...
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 $A+B$ ...
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 ...
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 ...
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, ...
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 ...
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_{...
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 ...
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 $|\psi(W)| = 0$. ...
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. ...
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