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?



Normal $0,1$-sequence with infinitely many frequent finite substrings
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 $\mu^+(\mathrm{Start}(\sigma))\ge 1/(1+l)$.


