For $A\subseteq \omega$ we let the lower and upper density be defined as $$\mu^-(A):= \lim\inf_{n\to\infty}\frac{|A\cap n|}{n+1} \text{ and } \mu^+(A):= \lim\sup_{n\to\infty}\frac{|A\cap n|}{n+1}$$ respectively.
Let $s:\omega\to\{0,1\}$ be an infinite binary string, $n\in\omega\setminus\{0\}$ a positive integer and $t\in\{0,1\}^n$ a finite binary string of length $n$. Then we define the set of starting points of $t$ in $s$ by $$\text{start}(t, s) = \{k\in \omega: (\forall i\in n)\; t(i) = s(k+i)\}.$$ We say that $s$ is uniform for $t\in\{0,1\}^n$ if $$\mu^-\big(\text{start}(t,s)\big) = \mu^+\big(\text{start}(t,s)\big) = 1/2^n.$$
(The $1/2^n$ part is motivated by the fact that there are $2^n$ binary strings of length $n$.) We say that $s:\omega\to\{0,1\}$ is strongly uniform if for all positive integers $n$ and for all $t\in\{0,1\}^n$ we have that $s$ is uniform for $t$.
It is not clear to me whether strongly uniform infinite binary strings exist. A candidate could be the Champernowne binary string which is obtained by concatenating the binary representations of the integers: $$0\; 1 \; 10\; 11\; 100\; 101\;\ldots.$$
Question. Is the the Champernowne binary string strongly uniform? If not, is there a strongly uniform infinite binary string?
Best Answer
Yes, the Champernowne binary string is normal (or strongly uniform, in your terms); see also here.
Almost all infinite binary string are normal, in the sense that the set of the corresponding real numbers in $[0,1]$ is of Lebesgue measure $1$.