Lemma 1

Let A be a balanced word. Then for each integer k such that $0 \leqslant k < l(A)$ there exists exactly one balanced segment S of A which begins at the $(k + 1)$th place.

The uniqueness of $S$ is an immediate consequence of the following remark : if $T$ is a balanced word, then by definition no proper initial segment of $T$ is balanced. Let us prove the existence of $S$. Write $A = BC$ where $l(B) = k$. For each $i$ such that $0 \leqslant i \leqslant q = l(C)$, let $C_i$ be the initial segment of $C$ of length $i$. Since $B$ is a proper initial segment of $A$, we have

On the other hand, we have $0 = l(C_0) \leqslant n(C_0) = 0$. Let $i$ be the largest of the integers $j < q$ such that $l(C_h) \leqslant n(C_h)$ for $0 \leqslant h \leqslant j$; then we have $l(C_i)\leq n(C_i)$ and $l(C_{i+1})\geq n(C_{i+1})+1$. We show that $C_{i+1}$ is balanced: the condition relating to proper initial segments is satisfied by reason of the definition of $i$; on the other hand, we have

so that $l(C_{i+1})=n(C_{i+1})+1$, and the proof of Lemma 1 is complete.