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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.