Lemma 4.5.11FDE7BF6A9927

Proof. Off a common polar set of irregular boundary points, closed sweeping is the identity on $F$ and the harmonic-measure integral off $F$, with zero on the original boundary. These kernels are Borel: for continuous compact tests their integrals are continuous in the pole; a countable monotone class gives indicators and all nonnegative Borel tests. Since the Euclidean second factor is second countable, its product Borel sigma algebra is generated by rectangles. Thus the reduced family is jointly Borel off that common polar set. Extend it by zero there and recover its actual canonical values by limits of ball averages; these averages are jointly Borel, so the recovered family is jointly Borel everywhere. The integral is lsc by Fatou and super-mean-valued by Tonelli, and is finite at $x_0$ because reduction is at most $u_ t$. Outside the common polar set, Tonelli in the harmonic-measure formula identifies it with the reduction of the integral. Canonical ball averages give equality everywhere. Only a sigma-finite effective parameter set is needed: the sets $\{ u_ t(x_0)>1/m\} $ have finite measure, and a nonnegative superharmonic function zero at $x_0$ vanishes identically. For potentials, lower harmonically along a regular exhaustion. Each minorant decreases to zero; its integral is dominated at $x_0$ by the integrable $u_ t(x_0)$. Dominated convergence gives zero greatest harmonic minorant of the integral. Interior uniqueness proves the measure identity. $\square$