Theorem 4.4.209873BDAF3E0

Proof. The finite superharmonic members of an upper class form a nonempty saturated family whenever that class has a member finite somewhere in the component: minimum and ball lowering preserve the boundary inequalities and the lower bound. The saturated-family lemma gives its harmonic or $-\infty $ infimum; if there is no such member the infimum is infinity. For $u$ in the lower and $v$ in the upper class, $v-u$ is superharmonic, bounded below, with nonnegative boundary lower limits, including where $f=\pm \infty $ (use the global bounds to avoid undefined subtraction). The bounded minimum principle gives $u\le v$. Order, constant translation, positive scaling, sign reversal and upper subadditivity follow from the defining classes, whenever sums are defined. For finite resolutive functions these prove linearity. A uniform change of boundary data by at most $\varepsilon $ changes each solution by at most $\varepsilon $, hence uniform limits of resolutive data are resolutive. The checked Foundations perron resolutivity supplies equality of the continuous upper and lower subsolution envelopes for continuous $f$. These classes are contained in our classes, so their common harmonic value lies between our lower and upper solutions; the reverse inequalities from class inclusion force equality here. Equivalently, approximate continuous data by smooth functions and split each smooth function as a difference of continuous subharmonic functions by adding a large multiple of $|x|^2$.

We supply the required monotone limit argument. If $f_ j\uparrow f$ and $\overline H_{f_1}>-\infty $, then $\overline H_{f_ j}\uparrow \overline H_ f$. Work on one component. If the limit is infinity, order proves the claim. Otherwise the limit $h$ is harmonic by monotone harmonic convergence. At a fixed $x_0$ choose $v_ j\in \mathcal U_{f_ j}$ with $v_ j(x_0)<\overline H_{f_ j}(x_0)+\varepsilon 2^{-j}$. The nonnegative superharmonic series

is finite at $x_0$, hence superharmonic, and dominates every $v_ j$. It is bounded below by one $v_ j$ and has boundary lower limits at least $f$, so belongs to $\mathcal U_ f$. Its value at $x_0$ is at most $h(x_0)+\varepsilon $. Let $\varepsilon \downarrow 0$; harmonicity and the minimum principle, or repetition at each $x_0$, gives the claim. The negative-data version gives the corresponding decreasing limit. If all $f_ j$ are resolutive, their increasing limit has equal upper/lower solutions, and is resolutive when the common value is finite.

For continuous data, $f\mapsto H_ f(x)$ is positive linear, sends $1$ to $1$, and has norm one. Compact-space Riesz representation gives the unique probability $\omega _ x$. Decreasing continuous approximations to indicators of closed subsets, followed by the monotone limit just proved, give $H_{1_ F}(x)=\omega _ x(F)$ for closed $F$. The class of boundary sets for which this equality and resolutivity hold contains the closed sets, is closed under proper differences and increasing unions, and therefore contains their generated sigma algebra. For clarity, the elementary pi-lambda argument proves the last assertion: first fix a closed set and close the class of sets whose intersections with it lie in the class; then repeat with an arbitrary member. This gives closure under finite intersections, hence under finite unions and, with increasing unions, under all countable unions. Thus all Borel indicators, bounded simple functions, and by monotone approximation all nonnegative Borel functions have the integral formula.

For a Borel null set $N$ at $x$, positivity and harmonicity show $H_{1_ N}=0$ on its component. Every subset of $N$ has upper and lower solutions zero there, by order. Thus the formula extends to the completion of $\omega _ x$, and componentwise to the intersection of all these completed sigma algebras. Simple approximation proves the nonnegative formula; positive/negative parts prove sufficiency for integrable data. A positive harmonic function is finite everywhere in a component if finite at one point, so integrability at one point per component suffices.

For necessity suppose $f$ is resolutive and work on one component. Choose decreasing $v_ j\in \mathcal U_ f$ whose values at $x_0$ tend to $H_ f(x_0)$, and put $F_ j(z)=\liminf _{D\ni x\to z}v_ j(x)$. These are lower semicontinuous extended Borel functions, bounded below, decreasing and at least $f$. Their harmonic solutions satisfy $H_ f\le H_{F_ j}\le v_ j$; they are finite at $x_0$ and hence finite harmonic everywhere. Similarly lower-class approximants yield increasing Borel integrable $G_ j\le f$, with $H_{G_ j}(x_0)\to H_ f(x_0)$. Let $F=\inf F_ j$, $G=\sup G_ j$. The inequalities $G_1\le G\le f\le F\le F_1$ imply that $F,G$ are integrable for every $\omega _ x$. Monotone integration gives $\int (F-G)d\omega _{x_0}=0$. The nonnegative harmonic integral of $F-G$ is then zero throughout the component. Thus $F=f=G$ almost everywhere for every $\omega _ x$, proving completed measurability and integrability.

Finally Harnack applied to $H_{1_ E}$ gives $C^{-1}\omega _ y(E)\le \omega _ x(E)\le C\omega _ y(E)$ for fixed interior $x,y$ in one component. Integration yields $C^{-1/p}\| f\| _{p,y}\le \| f\| _{p,x}\le C^{1/p}\| f\| _{p,y}$. Equivalent norms give the same continuous dual and weak topology; strong topologies also agree. These statements apply to the $p$th powers of norms, which are harmonic integrals; the norms themselves need not be harmonic. Weak continuity of measures follows since $\int f d\omega _ x=H_ f(x)$ is continuous for every continuous $f$. $\square$