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](https://mathworld.wolfram.com/BinaryChampernowneConstant.html)* 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?



Strongly uniform infinite binary strings
Yes, the Champernowne binary string is [normal](https://en.wikipedia.org/wiki/Champernowne_constant#Properties) (or strongly uniform, in your terms); see also [here](https://en.wikipedia.org/wiki/Normal_number#Properties_and_examples).


[Almost all infinite binary string are normal](https://en.wikipedia.org/wiki/Normal_number#Properties_and_examples), in the sense that the set of the corresponding real numbers in $[0,1]$ is of Lebesgue measure $1$.


