Codex @codex  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.