Theorem 4.4.17A7E357C38C8E

Proof. For a dense domain remove an interior closed ball. Solve the continuous datum $f$ on the true boundary and the constant $M+1$ on the artificial sphere, where $|f|\le M$. Paste $M+1$ inside by taking the minimum with that constant on a collar. The resulting bounded superharmonic function is a global upper competitor. The lower construction uses $-M-1$. At every regular true boundary point of the punctured nondense domain, these give the corresponding boundary limits for the global upper/lower solutions. Their difference is bounded harmonic, with zero boundary limits outside the polar Kellogg set, and is zero by the preceding lemma. Thus $f$ is resolutive. Those same upper/lower bounds show each such punctured-domain regular point is regular for $\Omega $, proving Kellogg for the general domain. Local barriers and the capped-constant construction give the equivalent local criterion as before.

The full measurable-data proof now uses precisely the steps established for bounded domains: closed boundary indicators by continuous decreasing approximation, the pi-lambda argument, nonnegative simple approximation, and monotone continuity of the upper envelope via summable positive errors at a fixed pole. The only local harmonic operations needed are lowering and monotone convergence, available for both interior and exterior balls. For clarity, the necessity step also survives unchanged: countably many upper/lower competitors approximating a finite solution at a pole give Borel boundary liminf/limsup sandwiches. Their integrals have the same value at that pole, so their difference is zero almost everywhere there; positive harmonicity and Harnack transfer this common null set throughout the component. The sandwiches identify the datum in the completed measure class and give its absolute integrability. Conversely positive and negative integrable parts follow from monotone simple functions, and null changes are removed by the sandwich comparison. Apply this separately in the countably many components. Epsilon comparison with the constructed pole functions proves polar boundary nullity. This establishes the same full equivalence as the bounded Perron theorem, now with the compactified boundary condition at infinity included explicitly. $\square$