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Questions about abstract measure and Lebesgue integral theory. Also concerns such properties as measurability of maps and sets.

3 votes
2 answers
632 views

Lifting a probability measure to the power set

Let $X\neq\emptyset$ be a set and let $\mu:{\cal P}(X)\to [0,1]$ be a probability measure. Is there a probability measure $$\bar{\mu}:{\cal P}({\cal P}(X))\to [0,1]$$ with the following property? …
Dominic van der Zypen's user avatar
8 votes
2 answers
694 views

Exponential objects in the category of measurable spaces

Let $\text{Meas}$ be the category of measurable spaces with measurable functions as morphisms. Does $\text{Meas}$ have exponential objects?
Dominic van der Zypen's user avatar
1 vote
Accepted

About the Caratheodory class.

Given a set $X$ and outer measure $L:\mathcal{P}(X)\to [0,\infty]$, is $L$ an $\mathcal{A}_L$-regular map? The answer is No. Consider the following example: Let $X = \mathbb{N}$ and define $L:\mathc …
Dominic van der Zypen's user avatar
2 votes
1 answer
170 views

Density measure on $\mathbb{N}^2$

For $A\subseteq \mathbb{N}$ we define the upper density by $$\mu(A)=\limsup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$ A nice property of this map $\mu:{\cal P}(\mathbb{N})\to [0,1]$ is that it i …
Dominic van der Zypen's user avatar
1 vote
2 answers
123 views

Distribution of pre-images of the divisor function $\sigma$

If $A\subseteq\mathbb{N}$ is a subset of the positive integers, we let $$\mu^+(A) = \lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$$ be the upper density of $A$. For $n\in\mathbb{N}$ we let $\si …
Dominic van der Zypen's user avatar
0 votes
1 answer
131 views

Two notions of thinness of subsets of $\mathbb{N}$ [closed]

If $A$ is a subset of the set of positive integers $\mathbb{N}$, there are (at least) two notions of what it means for $A$ to be thin: $A$ is thin in the 1st sense if $\lim\sup_{n\to\infty}\frac{|A\c …
Dominic van der Zypen's user avatar
1 vote
1 answer
253 views

Does a subset of positive measure in $\mathbb{R}$ locally "almost" have density $1$? [closed]

Let $A\subseteq \mathbb{R}$ be a Lebesgue-measurable set. We say that $A$ is locally $\varepsilon$-dense if for any $\varepsilon > 0$, there are $x<y\in\mathbb{R}$ such that $$\frac{\mu(A\cap[x,y])}{ …
Dominic van der Zypen's user avatar
0 votes
1 answer
52 views

Binarily universal members of $[0,1]$

Let $r\in[0,1]$. We look at the binary represenation of $r$ and say that $r$ is binarily universal if every finite binary string appears in at least one place in the binary representation of $r$. Let …
Dominic van der Zypen's user avatar
0 votes
1 answer
280 views

Upper density versus upper Banach density on $\omega$

For $A\subseteq\omega$ we define the upper density by $$d_u(A) = \lim\sup_{n\to\infty}\frac{|A\cap n|}{n+1}.$$ For $y\in \omega$ we set $A - y:= \{|a\setminus y|:a\in A\}.$ Note that $|a\setminus y|$ …
Dominic van der Zypen's user avatar
17 votes
1 answer
1k views

Does the set of square numbers adhere to Benford's law in every base?

Does the set of squares $S = \{n^2: n\in\omega\}$ adhere to Benford's law for the first digit in every base $b\geq 2$? Precise formulation of what it means for a set $T\subseteq \omega$ to "adhere to …
Dominic van der Zypen's user avatar
2 votes
1 answer
393 views

Measurability and continuity for general topological spaces

Let $(X,\tau)$ be a topological space. We call $S\subseteq X$ saturated if $S=\bigcap\{U\in\tau: U\supseteq S\}$. Let $\sigma(X,\tau)$ be the $\sigma$-algebra generated by $\tau\cup\{K\subseteq X: K \ …
Dominic van der Zypen's user avatar
5 votes
1 answer
338 views

Golomb subsets of $\mathbb{N}$

A set $A\subseteq\newcommand{\N}{\mathbb{N}}\N$ is said to be Golomb if whenever $a<b \in A$ and $a'<b' \in A$ with $(b-a) = (b' - a')$, then $a=a'$ and $b=b'$. If $A\subseteq \N$ is Golomb, we let $\ …
Dominic van der Zypen's user avatar
2 votes
Accepted

If $f : [-a,a] \rightarrow \mathbb{IR}$ is Scott continuous, why are $f^-$ and $f^+$ measura...

First, we note that $\mathbb{IR}$ is ordered by $[a,b] \sqsubseteq [c,d]$ iff $[c,d] \subseteq [a,b]$, which makes $\mathbb{IR}$ into a domain. Now suppose $f: [-a,a]\to \mathbb{IR}$ is Scott-continuo …
Dominic van der Zypen's user avatar
1 vote

If $f : [-a,a] \rightarrow \mathbb{IR}$ is Scott continuous, why are $f^-$ and $f^+$ measura...

This answers @user3078439's question in his comment to my original answer. First, a short argument to show that if $P,Q$ are domains and $f:P\to Q$ is Scott-continuous, then $f$ is order preserving. …
Dominic van der Zypen's user avatar
5 votes
3 answers
456 views

Arithmetically random bitstreams

Motivation (informal). When trying to generate a random bit-stream, we expect that "half of the" bits are $0$, and the "other half" are $1$. So, how about $010101\ldots$? Well, we would also expect th …
Dominic van der Zypen's user avatar

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