Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 35 / 2 / b

Use finite cutoff windows to avoid subtracting two infinite masses. Write $D$ for the unit disc, $D_U={\mathbf{L}}_D\setminus {\mathbf{L}}_U$ on loops surrounding $0$, and $A(U)=\nu (D_U)$. For $0<r<1$, let $H_r=D_{rD}$. These increase to all loops in ${\mathbf{L}}_D$ surrounding $0$ as $r\downarrow 0$.

On the finite-mass windows, the domain-containment events form a pi-system. Indeed, if $U,V$ are simply connected and contain $0$, a simple loop surrounding $0$ that lies in both has its filled interior in both. It therefore lies in the component $W$ of $U\cap V$ containing $0$, which is simply connected. Thus, on the support of $\nu$,Inclusion-exclusion for the finite deficits givesTaking $V=rD$ and subtracting from $\nu (H_r)$ gives the useful recovery formulaThe assumed generation of the loop sigma-field by domain-containment events and the uniqueness theorem for measures now determine the finite restriction to $H_r$. Increasing these windows recovers ${\nu }_D$; conformal transport recovers its restriction to any simply connected domain containing the marked point. If a simply connected domain omits $0$, no loop in it can surround $0$.

The argument uses finite annular deficits. This is the usual local-finiteness convention for these loop measures and is explicitly ensured by the finite logarithmic formula assumed in the continuation. Bare sigma-finiteness must not be used as permission for formal $\infty -\infty$ subtraction. This is the recovery of a loop measure from conformal deficits.

Let ${\varphi }_i:D\to U_i$ be the normalized maps, and put $V={\varphi }_1(U_2)$. A simple loop in $U_1$ has its interior there, so applying ${\varphi }_1^{-1}$ preserves the property of surrounding $0$. There is a disjoint decompositionThe conformal restriction property identifies the measure of the second set with $\nu ({\mathbf{L}}_D\setminus {\mathbf{L}}_{U_2})$. ThereforeThe composition fixes $0$ and has positive derivative there, so it is exactly the normalized map onto $V$. No interchange of infinite differences is needed: the argument is additivity on disjoint sets. The later logarithmic expression is consistent with this identity because $({\varphi }_1\circ {\varphi }_2{)}^{\prime }(0)={\varphi }_1^{\prime }(0){\varphi }_2^{\prime }(0)$.

Let $\Gamma (B)$ be the outer boundary of a planar compact set applied to the Brownian loop. For the pinned Brownian measure here, its outer boundary is almost surely simple and surrounds its root. If $U$ is simply connected, thenThe forward implication uses the fact that filling a compact subset of a simply connected domain cannot leave that domain. For the converse, the Brownian trace is contained in the closed filled interior of its outer boundary, and that interior also lies in $U$. This is precisely why simply connected domains are used.

Let $\eta ={\Gamma }_{\ast }\rho$, a pushforward measure on simple loops surrounding $0$. Its deficits areThey agree with those of $\lambda \eta$ and ${\nu }_D$. The recovery argument from part (i), now with explicitly finite deficits, givesIf $\lambda =0$, the measure is zero on every cutoff window and hence zero. The rooted Brownian loop itself is not a simple loop; taking its outer boundary is essential.