Normal $0,1$-sequence with infinitely many frequent finite substrings

Question

Let $\mathbb{N}$ denote the set of non-negative integers. We can identify every bitstream, i.e. a function $s:\mathbb{N}\to \{0,1\}$, with some $A\in{\cal P}(\mathbb{N})$: take $A = s^{-1}(\{1\})$.

Given any $A\subseteq \mathbb{N}$ we set $$\mu^{+}(A)= \lim \sup_{n\to\infty}\frac{|A \cap\{1,\ldots,n\}|}{n+1}.$$

We say that a bitstream $s$ is normal if every finite $01$-string appears infinitely often. For any finite $01$-string $\sigma$ we let $\text{Start}(\sigma)$ be the set of starting points of $\sigma$ inside $s$, and $\sigma$ is said to be frequent is $s$ if $\mu^+(\text{Start}(\sigma))>0$.

Question. Is there a normal bitstream $s$ with infinitely many frequent finite $01$-strings?

Answer 1

Yes. Enumerate the set of all finite $0$-$1$-sequences as $\langle\sigma_n:n\in\omega\rangle$ such that each sequence is listed infinitely often. Define $s$ to be the sequence that starts with $a_0$ copies $\sigma_0$, $a_1$ copies of $\sigma_1$, $a_2$ copies of $\sigma_2$, $\dots$, $a_n$ copies of $\sigma_n$, $\dots$, where the $a_n$ are determined, recursively as follows: $a_0=1$. When you ready to add copies of $\sigma_n$ let $a_n$ be the length of $s$ thus far and let $l_n$ be the length of $\sigma_n$. Append $a_n$ copies of $\sigma_n$ to $s$. The new sequence has length $a_n(1+l_n)$ (so $a_{n+1}=a_n(1+l_n)$) and it has at least $a_n$ points where a copy of $\sigma_n$ starts. This shows that in the end for a sequence $\sigma$ of length $l$ we have $\mu^+(\mathrm{Start}(\sigma))\ge 1/(1+l)$.