Newtonian potentials solve the distributional Poisson equation

Choose a finite-valued representative $f_0$ vanishing outside $K$. By [F2] and [A1], apply the completed-measurable representative theorem separately to the real and imaginary parts of $f_0$, reset any nonfinite exceptional values to zero, and combine them using [F16] to obtain a finite Borel representative equal to $f_0$ almost everywhere. Reset it to zero off $K$ and call it $\tilde{f}$. Set $\mathrm{\Phi }(0)=0$; its displayed radial formula is continuous away from the singleton pole, so it is Borel. By [F3], $H(x,y)=\mathrm{\Phi }(x-y)\tilde{f}(y)$ is Borel, hence product-measurable for ${\beta }_n\otimes {\beta }_n$. The open balls $B(0,m)$ are Borel and exhaust ${\mathbb{R}}^n$; they have finite ${\beta }_n$-measure by [F4], so ${\beta }_n$ is sigma-finite.

For compactly supported $L^p$ data with $1<p<\mathrm{\infty }$, Hölder [F8] on the finite-measure set $K$ gives ${\int }_K\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }{\mathbf{1}}_K{\mathrm{\parallel }}_q=\mathrm{\parallel }f{\mathrm{\parallel }}_p{\lambda }_n(K{)}^{1\mathrm{/}q}<\mathrm{\infty },$ where $1\mathrm{/}p+1\mathrm{/}q=1$. The case $p=1$ is immediate, and the endpoint $p=\mathrm{\infty }$ uses the endpoint clause of [F8]; moreover the bounded-data result [F14] gives everywhere absolute convergence there. A continuous compactly supported $g\in C_c$ is bounded on its compact support by [F11], whose measure is finite by [F4], so it too belongs to $L^1$. This verifies the stated $C_c$ and full $1\le p\le \mathrm{\infty }$ inclusions.

The compact set $K$ is closed and bounded, hence Borel, and has finite ${\lambda }_n$-measure by [F4]. If $K=\mathrm{\varnothing }$ or ${\lambda }_n(K)=0$, then $\tilde{f}$ vanishes off a null set, so [F9] gives $Nf(x)=0$ with absolute convergence for every $x$; the $L^1$ class is zero. Assume henceforth $K\mathrm{\ne }\mathrm{\varnothing }$ and choose $R>0$ with $K\subset B(0,R)$.

If $g$ is any other finite-valued measurable representative of the same $L^1$ class, then for each fixed $x$ the functions $y\mapsto \mathrm{\Phi }(x-y)g(y)$ and $y\mapsto \mathrm{\Phi }(x-y)\tilde{f}(y)$ agree outside a null set; their absolute values are measurable by [F16]. For nonnegative measurable functions agreeing off a null set, split each integral over that set and its complement; [F9] shows the two extended absolute integrals agree. Thus absolute finiteness is equivalent for the two representatives. When finite, their difference is integrable with integral zero by [F9], so linearity gives equal potential values. The pole assignment also changes the integrand only on the null singleton $\{x\}$. Therefore $Nf$ depends only on the $L^1$ class, with equality at every point where the integrals are defined.

Fix an integer $m\ge 1$ and $y\in K$. For $x\in B(0,m)$, the Euclidean triangle inequality gives $x-y\in B(0,m+R)$. Translation by $-y$ preserves Lebesgue integrals by [F6], so ${\int }_{B(0,m)}\mathrm{\mid }\mathrm{\Phi }(x-y)\mathrm{\mid }dx={\int }_{B(0,m)-y}\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }dz\le {\int }_{B(0,m+R)}\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }dz=:C_{m,R}<\mathrm{\infty },$ where finiteness follows from [F1] and the last inequality from [F8].

Let $x_0\mathrm{\notin }K$. If $K=\mathrm{\varnothing }$, then $Nf=0$ and the claim holds. Otherwise, since $K$ is closed, choose $r>0$ such that $B(x_0,2r)\cap K=\mathrm{\varnothing }$. For $x\in \bar{B}(x_0,r)$ and $y\in K$, the triangle inequality and boundedness of $K$ give $r\le \mathrm{\mid }x-y\mathrm{\mid }\le M$ for some finite $M$. Choose $y_0\in K$. Since $B(x_0,2r)\cap K=\mathrm{\varnothing }$, $\mathrm{\mid }{x}_0-y_0\mathrm{\mid }\ge 2r>r$ and the upper bound gives $\mathrm{\mid }{x}_0-y_0\mathrm{\mid }\le M$, so $x_0-y_0\in A:=\{z:r\le \mathrm{\mid }z\mathrm{\mid }\le M\}$ and $A$ is nonempty. The annulus $A$ is closed because the norm is continuous [F11] and $[r,M]$ is closed; it is bounded by $M$, hence compact. Every continuous derivative $D^{\alpha }\mathrm{\Phi }$ is therefore bounded on $A$ [F11]. Thus for each multi-index $\alpha$ there is $C_{\alpha }<\mathrm{\infty }$ with $\mathrm{\mid }{D}^{\alpha }\mathrm{\Phi }(x-y)\tilde{f}(y)\mathrm{\mid }\le C_{\alpha }\mathrm{\mid }\tilde{f}(y)\mathrm{\mid }{\mathbf{1}}_K(y)\mathrm{.}$ The majorant is integrable since $\tilde{f}=f$ almost everywhere and ${\int }_K\mathrm{\mid }\tilde{f}\mathrm{\mid }=\mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$ by [F2, F9]. Applying differentiation under the integral sign coordinate by coordinate on a small box about $x_0$, and dominated convergence for continuity of each derivative, proves $Nf\in C^{\mathrm{\infty }}$ near $x_0$. Since ${\mathrm{\Delta }}_x\mathrm{\Phi }(x-y)=0$ for $x\mathrm{\ne }y$ by [F10], differentiating twice yields $\mathrm{\Delta }Nf(x)=0$ there. Therefore $Nf$ is smooth and harmonic on ${\mathbb{R}}^n\setminus K$.

Apply Tonelli [F5] to the nonnegative Borel function $\mathrm{\mid }H(x,y)\mathrm{\mid }{\mathbf{1}}_{B(0,m)}(x){\mathbf{1}}_K(y)$. Using step 3.1 gives ${\int }_{B(0,m)}{\int }_K\mathrm{\mid }\mathrm{\Phi }(x-y)\tilde{f}(y)\mathrm{\mid }dydx\le C_{m,R}{\int }_K\mathrm{\mid }\tilde{f}(y)\mathrm{\mid }dy=C_{m,R}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^1}<\mathrm{\infty },$ where the last equality uses [F2, F9] because $\tilde{f}=f$ almost everywhere and ${\beta }_n$ completes to ${\lambda }_n$. Thus the complex function $H$ is in $L^1(B(0,m)\times K)$ for every $m$.

Fubini [F5] on each such product shows that for almost every $x\in B(0,m)$ the section $y\mapsto H(x,y){\mathbf{1}}_K(y)$ is absolutely integrable, and its integral is an $L^1(B(0,m))$ function with integral of its absolute value at most the finite bound in step 4.1, by the integral triangle inequality [F8]. These section integrals agree with $Nf(x)$ wherever absolutely finite by [F13]. The balls $B(0,m)$ exhaust ${\mathbb{R}}^n$, so, writing $E_m$ for the measurable exceptional set in $B(0,m)$, [F15] gives ${\lambda }_n({\bigcup }_{m\ge 1}{E}_m)\le {\sum }_{m\ge 1}{\lambda }_n(E_m)=0$. Thus $Nf$ is finite almost everywhere on all of ${\mathbb{R}}^n$; assign it value zero on this null set. Each compact set lies in some $B(0,m)$ by [F4], proving $Nf\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$.

Let $\phi \in \mathcal{D}({\mathbb{R}}^n)$ and choose $m$ with $\mathrm{supp}\phi \subset B(0,m)$. The function $H(x,y)=\mathrm{\Phi }(x-y)\tilde{f}(y)$ is Borel by [F3]. The pullback of the Borel function $-\mathrm{\Delta }\phi$ by the first-coordinate projection is Borel: the preimage of a Borel set $E$ is $E\times {\mathbb{R}}^n$, which belongs to the product sigma-algebra and hence to the Euclidean Borel sigma-algebra by [F3]. The test function is smooth, so $\mathrm{\Delta }\phi$ is continuous; [F16] gives its Borel measurability. Thus $G(x,y)=H(x,y)(-\mathrm{\Delta }\phi (x))$ is Borel by [F16]. Since $\mathrm{\Delta }\phi$ is bounded and compactly supported, step 4.1 shows $G$ is integrable on the product. Fubini therefore gives ${\int }_{{\mathbb{R}}^n}Nf(x)(-\mathrm{\Delta }\phi (x))dx={\int }_K\tilde{f}(y)({\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(x-y)(-\mathrm{\Delta }\phi (x))dx)dy\mathrm{.}$ The inner integral is $\phi (y)$ by the translated point-source identity [F12]. Hence the right side is ${\int }_K\tilde{f}(y)\phi (y)dy$.

By [F7], step 6.1 is exactly $⟨-\mathrm{\Delta }{T}_{Nf},\phi ⟩=⟨T_f,\phi ⟩$ for every test $\phi$. The locally integrable embedding makes both sides distributions, so they are equal in ${\mathcal{D}}^{\mathrm{\prime }}$.

The logarithmic kernel at $n=2$ and power kernel at $n\ge 3$ are both covered by [F1], [F2] and [F12]; the distinct $n=1$ case is excluded by the statement. The zero source and empty or null support were handled in step 2.1; the Hölder endpoint cases $p=1,\mathrm{\infty }$ are explicit in step 1.2. Countable Choice is used to obtain a Borel representative, and is inherited by the published kernel identity and distribution embedding [F2, F7, F12]. No full Axiom of Choice or later result is used; the statement is not an iff.