Lemma 2.9.21DC992DBF55C7

Proof. Write $|\cdot |$ for Lebesgue measure on $[0,2\pi ]$, and put $\ell (t)=|F_ t|$ and $A_ t(b)=|F_ t\cap [0,b]|$ for $b\in [0,2\pi ]$. We first show that, for fixed $b$, the function $t\mapsto A_ t(b)$ is Borel; with $b=2\pi $, so is $\ell $. Since $T$ and $[0,2\pi ]$ are separable metric spaces, the Borel sets of $T\times [0,2\pi ]$ form the $\sigma $-algebra generated by the products of Borel sets (lemma 2.2.8). Hence the sections $G_ t$ of a Borel set $G$ are Borel, because the sets all of whose sections are Borel form a $\sigma $-algebra containing these products. The class of Borel sets $G\subset T\times [0,2\pi ]$ for which $t\mapsto |G_ t\cap [0,b]|$ is Borel contains the products $B_1\times B_2$ of Borel sets, for which this function is $\mathbf1_{B_1}(t)|B_2\cap [0,b]|$. It is closed under complements, since the complement of $G$ gives $b-|G_ t\cap [0,b]|$, and under countable disjoint unions, which give sums. Since the products of Borel sets are closed under finite intersections, lemma 2.2.6(1) shows that the class contains every Borel set.

For fixed $t$, the function $A_ t$ is nondecreasing and $1$-Lipschitz, with $A_ t(0)=0$ and $A_ t(2\pi )=\ell (t)>0$. Define

Because $A_ t$ is continuous and nondecreasing and $0<\ell (t)/2<\ell (t)$, the set of all $b\in [0,2\pi ]$ with $A_ t(b)\geq \ell (t)/2$ is an interval $[b_ t,2\pi ]$ with $0<b_ t<2\pi $ and $A_ t(b_ t)=\ell (t)/2$. The rational numbers are dense in this interval, so $a(t)=b_ t$ and $A_ t(a(t))=\ell (t)/2$. The map $a$ is Borel, since for every $c\in (0,2\pi ]$

Finally, suppose that $a(t)\notin F_ t$. Since $F_ t$ is closed and $a(t)>0$, some interval $[a(t)-\varepsilon ,a(t)]$ with $\varepsilon >0$ misses $F_ t$, and $A_ t$ is constant on it. A rational number $b$ in this interval with $b<a(t)$ then satisfies $A_ t(b)=\ell (t)/2$, contradicting the definition of $a(t)$. Hence $a(t)\in F_ t$. $\square$