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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?
…
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?
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 …
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 …
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 …
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 …
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 $$\frac{\mu(A\cap[x,y])}{ …
0
votes
1
answer
52
views
Binarily universal members of $[0,1]$
Let $r\in[0,1]$. We look at the binary represenation of $r$ and say that $r$ is binarily universal if every finite binary string appears in at least one place in the binary representation of $r$. Let …
0
votes
1
answer
280
views
Upper density versus upper Banach density on $\omega$
For $A\subseteq\omega$ we define the upper density by $$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 $|a\setminus y|$ …
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 …
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 \ …
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 $\ …
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 …
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.
…
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 …