2.9.20 Circle selection and high-value sets965F35E04B2D

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 $a(t)\in F_ t$. $\square$

Proof. We obtain $\mu $ by choosing, in a Borel way, a point of $E$ on each circle of radius $r\in [R/2,R]$ centered at $x$ in a complex line through $x$, and then we estimate its energy. For $\xi \in \mathbb P^{n-1}(\mathbb C)$ we first choose a unit vector $e_0(\xi )\in \xi $ in a Borel way: if $j$ is the first index with $(P_\xi )_{jj}>0$, put $e_0(\xi )=P_\xi e_ j/\sqrt{(P_\xi )_{jj}}$. Since $|P_\xi e_ j|^2=\langle P_\xi e_ j,e_ j\rangle =(P_\xi )_{jj}$, this is a unit vector in $\xi $. The set of $\xi $ with a given first index $j$ is Borel, and $e_0$ is continuous on it.

Let $r\in [R/2,R]$ and $\xi \in \mathbb P^{n-1}(\mathbb C)$. The circle $\theta \mapsto x+r\mathrm e^{\mathrm i\theta }e_0(\xi )$ lies in the closed disc $\{ x+\zeta e_0(\xi ):|\zeta |\leq r\} $, which is connected, contains $x$ and is contained in $D$; hence it lies in the connected component containing $x$ of the section of $D$ by the line $x+\xi $. On this component $u$ is subharmonic; in particular it is upper semicontinuous and bounded above on the circle, and by theorem 3.4.5(2) and the arc-length formula for circle means at the beginning of this section, it satisfies the submean inequality

Let

This set is closed by upper semicontinuity, and it has positive length. If $A=-\infty $, it is the whole interval $[0,2\pi ]$. If $A$ is finite and $F_{r,\xi }$ were a null set, then $A-u$ would be positive almost everywhere on the circle, so the circle mean of $u$ would be strictly less than $A$, contradicting the submean inequality.

The set of $(r,\xi ,\theta )$ with $\theta \in F_{r,\xi }$ is Borel, because $u$ is Borel and the map $(r,\xi ,\theta )\mapsto x+r\mathrm e^{\mathrm i\theta }e_0(\xi )$ into $\overline{B(x,R)}$ is Borel, being continuous on each of the finitely many Borel sets on which $e_0$ is continuous. Since $[R/2,R]\times \mathbb P^{n-1}(\mathbb C)$ is a compact, hence separable, metric space, lemma 2.9.21 gives a Borel map $(r,\xi )\mapsto \alpha (r,\xi )$ with $\alpha (r,\xi )\in F_{r,\xi }$. Put $e(r,\xi )=\mathrm e^{\mathrm i\alpha (r,\xi )}e_0(\xi )$ and $\Psi (r,\xi )=x+re(r,\xi )$. Then $\Psi $ is Borel, hence product-measurable (lemma 2.2.8), and it takes its values in $E$. Let $\mu $ be the image under $\Psi $ of the product of normalized Lebesgue measure on $[R/2,R]$ and $\sigma $. It is a Borel probability measure carried by $E$.

It remains to estimate the energy; the estimate does not depend on the choice of the phases $\alpha (r,\xi )$. For $r,r'\in [R/2,R]$ and unit vectors $e,e'$ spanning $\xi ,\xi '$,

where the identity follows by expanding both sides with $|e|=|e'|=1$ and $|e-e'|^2=2-2\operatorname{Re}\langle e,e'\rangle $, and the inequality holds since $rr'\geq R^2/4$ and $\operatorname{Re}\langle e,e'\rangle \leq |\langle e,e'\rangle |$. With $a=d(\xi ,\xi ')\in [0,\pi /2]$, we have $2-2|\langle e,e'\rangle |=4\sin ^2(a/2)\geq 4a^2/\pi ^2$, because $\sin y\geq 2y/\pi $ for $0\leq y\leq \pi /2$ by concavity of the sine on this interval. Hence $|re-r'e'|^2\geq (r-r')^2+R^2a^2/\pi ^2$.

Write $r=Rt$ with $t\in [1/2,1]$. Equip $[1/2,1]\times \mathbb P^{n-1}(\mathbb C)$ with the metric $\rho $ defined by $\rho ((t,\xi ),(t',\xi '))^2=(t-t')^2+d(\xi ,\xi ')^2$, and with the product probability $m$ of normalized Lebesgue measure and $\sigma $, which is defined on all Borel sets of this separable metric space by lemma 2.2.8. Then $\mu $ is the image of $m$ under $(t,\xi )\mapsto \Psi (Rt,\xi )$, and the lower bound for $|re-r'e'|^2$, together with $\pi >1$, gives

A $\rho $-ball of radius $s\leq 1$ is contained in the product of an interval of length $2s$, whose normalized measure is at most $4s$, and a ball of radius $s$ for the metric $d$. Since $s\leq 1<\pi /2$, lemma 2.9.19 shows that its $m$-measure is at most $4s^{2n-1}$. In particular $m$ has no atoms, so the diagonal, a closed and hence product-measurable set (lemma 2.2.8), is null for $m\times m$ by Tonelli's theorem. Fix $z\in [1/2,1]\times \mathbb P^{n-1}(\mathbb C)$. Splitting the integral into the set where $\rho (z,z')\geq 1$ and the shells where $2^{-k-1}\leq \rho (z,z')<2^{-k}$, $k\geq 0$, we obtain

Since $\mu $ is the image of $m$, the product $\mu \times \mu $ is the image of $m\times m$ under the product map, because both measures agree on rectangles (proposition 2.2.7). Therefore the lower bound for $|\Psi (Rt,\xi )-\Psi (Rt',\xi ')|$, valid off the $m\times m$-null diagonal, and an integration in $z$ give

This is the assertion, with $C_ n=\pi ^{2n-2}(1+2^{2n+1})$. $\square$