= Solution
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 $r\downarrow0$.
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 $\nu$,
$$
\mathbf L_U\cap\mathbf L_V=\mathbf L_W.
$$
<Inclusion-exclusion> for the finite deficits gives
$$
\nu(D_U\cap D_V)=A(U)+A(V)-A(W).
$$
Taking $V=rD$ and subtracting from $\nu(H_r)$ gives the useful recovery formula
$$
\boxed{\nu(H_r\cap\mathbf L_U)=A(W)-A(U),\qquad
W=\text{the component of }rD\cap U\text{ containing }0.}
$$
The 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>.
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