# Binary words starting with arbitrarily long squares

What is the measure of the following set of infinite binary words?

$$S=\{w\in\{0,1\}^\omega\ \text{such that},\ \text{for every}\ N\in\mathbb{N},\, w\ \text{has a prefix of the form}\ pp\ \text{with}\ |p|\ge N \}$$.

• What context does this problem arise in? – Carl-Fredrik Nyberg Brodda Dec 14 '20 at 12:55
• I am trying to understand what can be reasonably taken as the domain of the map $\mathcal{B}$ described in this other question: mathoverflow.net/q/377105/167834 (although the link may be not completely obvious, it is not too difficult to see that this question here is basic for the other one). – Alessandro Della Corte Dec 14 '20 at 13:58

For any given $$p\in\{0,1\}^*$$ (finite word), the measure of the set of $$w\in\{0,1\}^\omega$$ starting with $$pp$$ is $$2^{-2|p|}$$. So, summing over all the $$p$$ with $$|p|=L$$ (there are $$2^L$$ of them), given $$L\in\mathbb{N}$$, the measure of the set $$S_L$$ of $$w\in\{0,1\}^\omega$$ starting with $$pp$$ for some $$p$$ with $$|p|=L$$ is $$2^{-L}$$. So the measure of $$\bigcup_{L=N}^{+\infty} S_L$$ is $$\leq \sum_{L=N}^{+\infty} 2^{-L} = 2^{1-N}$$, and the measure of $$S = \bigcap_{N=0}^{+\infty} \bigcup_{L=N}^{+\infty} S_L$$ is $$0$$.