The regularized Perron envelope is harmonic

By [L1], the Perron family $\mathcal{P}(\phi ,\mathrm{\Omega })$ is nonempty and locally bounded above. Applying [L6] shows that $H_{\phi }$ is finite-valued and subharmonic on $\mathrm{\Omega }$, with $m\le U_{\phi }\le H_{\phi }\le M$.

Fix $a\in \mathrm{\Omega }$ and a disc $D=D(a,R)$ with $\bar{D}⋐\mathrm{\Omega }$. For each $v\in \mathcal{P}(\phi ,\mathrm{\Omega })$ write $h_v=(P_{D}v){\mathrm{\mid }}_D$. By [L2], $h_v$ is harmonic on $D$, the globally defined $P_{D}v$ is again a Perron lower function, and $v\le h_v\le U_{\phi }\le H_{\phi }\le M$ on $D$. The family of these lifts is nonempty; the constant lower function $m$ gives $h_m=m$.

The Poisson modification is order preserving: if $v\le w$ are two lower functions, take any decreasing continuous boundary approximations ${\alpha }_n\downarrow v{\mathrm{\mid }}_{\mathrm{\partial }D}$ and ${\beta }_n\downarrow w{\mathrm{\mid }}_{\mathrm{\partial }D}$ supplied by [L4]. The finite minima ${\gamma }_n:=\min ({\alpha }_n,{\beta }_n)$ are continuous, decrease to $v{\mathrm{\mid }}_{\mathrm{\partial }D}$, and satisfy ${\gamma }_n\le {\beta }_n$. Positivity of the Poisson kernel in [L4] makes the harmonic Poisson integrals satisfy $P[{\gamma }_n]\le P[{\beta }_n]$ on $D$. Taking their decreasing limits in the intrinsic definition of Poisson modification yields $h_v\le h_w$. This uses two witnesses and their finite minima, with no countable family of choices.

Given $v,w$ in the Perron family, $t=\max (v,w)$ belongs to the family by [L3], and step 3.1 gives $h_t\ge h_v,h_w$. Thus the lifted family is upward directed. Define $h(z):={\sup }_v{h}_v(z)$ on $D$; by step 2.1 and the constant member $h_m=m$, it is finite and satisfies $m\le h\le H_{\phi }$.

We prove that $h$ is harmonic without choosing a sequence of lifts. Fix $b\in D$ and a closed disc $\overline{D(b,R_b)}\subset D$. For any $ϵ>0$, the supremum defining $h(b)$ gives one lift $h_v$ with $h_v(b)>h(b)-ϵ$. For any other lift $h_w$, step 4.1 gives a common upper lift $h_t\ge h_v,h_w$. The difference $h_t-h_v$ is nonnegative harmonic on $D$ and has value less than $ϵ$ at $b$. Apply [L5] to $h_t-h_v+\delta$ and let $\delta \downarrow 0$: for every $r<R_b$ there is a finite constant $C_{b,r}$, independent of $v,w,t$, such that $0\le h_t(z)-h_v(z)\le C_{b,r}ϵ$ on $\overline{D(b,r)}$. Hence $0\le h(z)-h_v(z)\le C_{b,r}ϵ$ there after taking the supremum over $w$. One harmonic lift therefore approximates $h$ uniformly on each smaller disc to any prescribed error, using only one existential witness per error.

We show $h(a)=H_{\phi }(a)$. Step 4.1 already gives $h(a)\le H_{\phi }(a)$. Fix $ϵ>0$. By [L1], $m\le H_{\phi }(a)\le M$. Choose a radius $r>0$ small enough that $\overline{D(a,r)}\subset D$ and the Harnack upper factor $C(r)$ from [L5] for the larger disc $D$ satisfies $(C(r)-1)(M-m+ϵ)<ϵ\mathrm{/}2$. The limsup definition in [L1] gives one point $z\in D(a,r)$ with $U_{\phi }(z)>H_{\phi }(a)-ϵ\mathrm{/}4$, and the supremum defining $U_{\phi }(z)$ gives one lower function $v$ with $v(z)>U_{\phi }(z)-ϵ\mathrm{/}4$. By step 2.1, $h_v(z)\ge v(z)>H_{\phi }(a)-ϵ\mathrm{/}2$. The function $M-h_v$ is nonnegative harmonic on $D$; [L5], applied to $M-h_v+\delta$ and then with $\delta \downarrow 0$, gives $M-h_v(a)\le C(r)(M-h_v(z))$. Since $M-h_v(z)<M-m+ϵ\mathrm{/}2$, rearrangement gives $h_v(a)>H_{\phi }(a)-ϵ$. Consequently $h(a)\ge h_v(a)>H_{\phi }(a)-ϵ$ for every $ϵ>0$, so $h(a)=H_{\phi }(a)$. The point and lower function are chosen only for this one $ϵ$; no countable choice is used.

The uniform approximation in step 5.1 makes $h$ continuous: given a tolerance, choose one harmonic $h_v$ uniformly close on a neighbourhood, then use continuity of that $h_v$ and the triangle inequality. On any circle whose closed disc lies in $D$, approximate $h$ uniformly on that closed disc by one $h_v$, use the circle mean-value identity for $h_v$ from [L7], and let the error tend to $0$. Thus $h$ has the local circle mean-value property. By the converse in [L7], $h$ is harmonic on $D$. This epsilon proof does not assemble the individual witnesses into a sequence.

The difference $H_{\phi }-h$ is subharmonic on $D$: $H_{\phi }$ is subharmonic by step 1.1, $-h$ is harmonic and hence subharmonic by step 6.1, and [L3] preserves the sum. It is nonpositive by step 4.1 and vanishes at the interior point $a$ by step 5.2. The maximum principle [L8] forces it to be identically zero on $D$. Therefore $H_{\phi }=h$ on $D$ and is harmonic near $a$. Since $a$ was arbitrary, $H_{\phi }$ is harmonic on $\mathrm{\Omega }$. The empty-family case is excluded by [L1], $m$ handles the constant lower bound, and every approximation uses only finite existential choices; no choice axiom has been added to the theorem.