The azimuthal line field ${\rho }^{-1}{e}_{\phi }$: local integrability and distributional curl on the axis

The axis $Z=\{(0,0,x_3):x_3\in \mathbb{R}\}$ is closed, hence Borel. By [F3] in the product ${\mathbb{R}}^2\times \mathbb{R}$, ${\lambda }_3(Z)={\int }_{\mathbb{R}}{\lambda }_2(\{(0,0)\})d{x}_3=0$, because the singleton in ${\mathbb{R}}^2$ is null by [F5].

On ${\mathbb{R}}^3\setminus Z$ the components of $f$ are quotients of polynomials with nonvanishing denominator $x_1^2+x_2^2$, so by [F6] the partial derivatives exist and satisfy ${\mathrm{\partial }}_1{f}_1=\frac{2x_1{x}_2}{{\rho }^4}$, ${\mathrm{\partial }}_2{f}_2=-\frac{2x_1{x}_2}{{\rho }^4}$, ${\mathrm{\partial }}_1{f}_2=\frac{x_2^2-x_1^2}{{\rho }^4}$, ${\mathrm{\partial }}_2{f}_1=\frac{x_2^2-x_1^2}{{\rho }^4}$ and ${\mathrm{\partial }}_{3}f=0$. Summing by [F7] gives $\mathrm{div}f={\mathrm{\partial }}_1{f}_1+{\mathrm{\partial }}_2{f}_2=0$, and the curl coordinates are ${\mathrm{\partial }}_1{f}_2-{\mathrm{\partial }}_2{f}_1=0$, ${\mathrm{\partial }}_3{f}_1-{\mathrm{\partial }}_1{f}_3=0$ and ${\mathrm{\partial }}_2{f}_3-{\mathrm{\partial }}_3{f}_2=0$; hence $\mathrm{div}f=\mathrm{curl}f=0$ pointwise on ${\mathbb{R}}^3\setminus Z$.

Fix $r>0$ and a height $h\in \mathbb{R}$, and let $\gamma (t)=(r\cos t,r\sin t,h)$, $t\in [0,2\pi ]$, a closed $C^1$ path whose velocity is ${\gamma }^{\mathrm{\prime }}(t)=r{e}_{\phi }(\gamma (t))$ by [F8]. On the trace, $f(\gamma (t))=r^{-1}{e}_{\phi }(\gamma (t))$, so the integrand of the vector line integral of [F12] is the constant $f(\gamma (t))\cdot {\gamma }^{\mathrm{\prime }}(t)=r^{-1}\cdot r=1$, and the circulation is ${\int }_0^{2\pi }1dt=2\pi$.

Let $U={\mathbb{R}}^3\setminus Z$ and define $\tilde{f}:{\mathbb{R}}^3\to {\mathbb{R}}^3$ by $\tilde{f}=f$ on $U$ and $\tilde{f}=0$ on $Z$. For every open $O\subseteq {\mathbb{R}}^3$, continuity of $f$ on $U$ makes $f^{-1}(O)$ open in $U$, hence open in ${\mathbb{R}}^3$ because $U$ is open. The inverse image ${\tilde{f}}^{-1}(O)$ is $f^{-1}(O)$ if $0\mathrm{\notin }O$, and $f^{-1}(O)\cup Z$ if $0\in O$; both are Borel, since $Z$ is closed. The class $\mathcal{C}=\{B\subseteq {\mathbb{R}}^3:{\tilde{f}}^{-1}(B)\text{ is Borel}\}$ is a sigma-algebra: inverse images preserve complements and countable unions. It contains every open set, hence every Borel set by [F5]. Thus $\tilde{f}$ is Borel measurable without applying the cited full-space continuity result to $f$. For its local integrability, let $K\subset {\mathbb{R}}^3$ be compact and choose $R>0$, $H>0$ with $K\subset \{\mathrm{\mid }{x}_3\mathrm{\mid }\le H,\rho \le R\}$. By Tonelli [F3] and the polar formula [F4] in dimension two, applied to the nonnegative Borel function $u(x)={\rho }^{-1}{\mathbf{1}}_{\{\mathrm{\mid }{x}_3\mathrm{\mid }\le H,\rho \le R\}}(x)$ off $Z$, with $u=0$ on $Z$ (the same open-inverse-image argument proves measurability of ${\rho }^{-1}$ extended by zero, and restriction to this closed cylinder preserves it), ${\int }_{{\mathbb{R}}^3}ud{\lambda }_3={\int }_{-H}^H\left({\int }_{\rho \le R}{\rho }^{-1}d{\lambda }_2\right)d{x}_3={\int }_{-H}^{H}R\sigma (S^1)d{x}_3=2HR\sigma (S^1)$, and this is finite because $\sigma$ is a finite measure [F4]. Since $\mathrm{\mid }\tilde{f}\mathrm{\mid }={\rho }^{-1}$ off $Z$ and $Z$ is null by step 1.1, ${\mathbf{1}}_K\mathrm{\mid }\tilde{f}\mathrm{\mid }\le u$ off a null set, so ${\int }_K\mathrm{\mid }\tilde{f}\mathrm{\mid }d{\lambda }_3<+\mathrm{\infty }$. Hence $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^3)$ and $T_f$ is a well-defined distribution by [F1], [F2].

The regular-distribution pairings of [F1] and [F2] give $⟨\mathrm{curl}{T}_f,\psi ⟩={\int }_{{\mathbb{R}}^3}f\cdot \mathrm{curl}\psi d{\lambda }_3$ and $⟨\mathrm{div}{T}_f,\phi ⟩=-{\int }_{{\mathbb{R}}^3}f\cdot \mathrm{\nabla }\phi d{\lambda }_3$, the integrands being dominated by $\mathrm{\parallel }\mathrm{curl}\psi {\mathrm{\parallel }}_{\mathrm{\infty }}{\rho }^{-1}{\mathbf{1}}_{\mathrm{supp}\psi }$ and $\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}{\rho }^{-1}{\mathbf{1}}_{\mathrm{supp}\phi }$, which are integrable by step 2.1. Fix $R>0$ with $\mathrm{supp}\psi \cup \mathrm{supp}\phi \subset \{\rho <R,\mathrm{\mid }{x}_3\mathrm{\mid }<R\}$ and put $V_{\epsilon }:=\{\epsilon <\rho <R,\mathrm{\mid }{x}_3\mathrm{\mid }<R\}$ for $0<\epsilon <R$. The functions ${\mathbf{1}}_{V_{\epsilon }}f\cdot \mathrm{curl}\psi$ are dominated by $\mathrm{\mid }f\cdot \mathrm{curl}\psi \mathrm{\mid }$ and converge pointwise off the null axis $Z$ of step 1.1 to $f\cdot \mathrm{curl}\psi$; by [F11], ${\int }_{{\mathbb{R}}^3}f\cdot \mathrm{curl}\psi d{\lambda }_3={\lim }_{\epsilon \to 0}{\int }_{V_{\epsilon }}f\cdot \mathrm{curl}\psi d{\lambda }_3$, and the same argument gives ${\int }_{{\mathbb{R}}^3}f\cdot \mathrm{\nabla }\phi d{\lambda }_3={\lim }_{\epsilon \to 0}{\int }_{V_{\epsilon }}f\cdot \mathrm{\nabla }\phi d{\lambda }_3$.

On the neighborhood $\{\rho >\epsilon \mathrm{/}2\}$ of the compact closure $\overline{V_{\epsilon }}$ the field $\psi \times f$ is $C^1$, since $f$ is smooth there, and by [F7] its divergence is $\mathrm{div}(\psi \times f)=f\cdot \mathrm{curl}\psi -\psi \cdot \mathrm{curl}f$; here $\mathrm{curl}f=0$ on $V_{\epsilon }$ by step 1.2. Applying the divergence theorem [F9] on $V_{\epsilon }$ gives ${\int }_{V_{\epsilon }}f\cdot \mathrm{curl}\psi d{\lambda }_3={\int }_{\mathrm{\partial }{V}_{\epsilon }}(\psi \times f)\cdot \nu dS$. The faces at $\rho =R$ and at $\mathrm{\mid }{x}_3\mathrm{\mid }=R$ contribute $0$ because $\mathrm{supp}\psi \subset \{\rho <R,\mathrm{\mid }{x}_3\mathrm{\mid }<R\}$; on the inner cylinder the outward normal is $-e_{\rho }$ and $f=e_{\phi }\mathrm{/}\epsilon$, so by [F8] the remaining term is $(\psi \times e_{\phi })\cdot (-e_{\rho })\mathrm{/}\epsilon =-\psi \cdot (e_{\phi }\times e_{\rho })\mathrm{/}\epsilon ={\psi }_3\mathrm{/}\epsilon$, and the surface integral of [F10] over the chart $X(\theta ,z)=(\epsilon \cos \theta ,\epsilon \sin \theta ,z)$ gives ${\int }_{\rho =\epsilon ,\mathrm{\mid }{x}_3\mathrm{\mid }<R}\frac{{\psi }_3}{\epsilon }dS={\int }_0^{2\pi }{\int }_{-R}^R{\psi }_3(\epsilon \cos \theta ,\epsilon \sin \theta ,z)dzd\theta \mathrm{.}$

On the same neighborhood the field $\phi f$ is $C^1$, and $\mathrm{div}(\phi f)=f\cdot \mathrm{\nabla }\phi +\phi \mathrm{div}f$ by [F7]; here $\mathrm{div}f=0$ on $V_{\epsilon }$ by step 1.2. The divergence theorem [F9] gives ${\int }_{V_{\epsilon }}f\cdot \mathrm{\nabla }\phi d{\lambda }_3={\int }_{\mathrm{\partial }{V}_{\epsilon }}\phi f\cdot \nu dS$. On both cylinder faces $f\cdot \nu =0$ because $f\perp e_{\rho }$, and on the two cap annuli $\phi =0$ because $\mathrm{supp}\phi \subset \{\mathrm{\mid }{x}_3\mathrm{\mid }<R\}$; hence ${\int }_{V_{\epsilon }}f\cdot \mathrm{\nabla }\phi d{\lambda }_3=0$ for every $\epsilon$.

With step 4.2, the identity ${\int }_{{\mathbb{R}}^3}f\cdot \mathrm{\nabla }\phi d{\lambda }_3={\lim }_{\epsilon \to 0}{\int }_{V_{\epsilon }}f\cdot \mathrm{\nabla }\phi d{\lambda }_3$ of step 3.1 gives ${\int }_{{\mathbb{R}}^3}f\cdot \mathrm{\nabla }\phi d{\lambda }_3=0$, and therefore $⟨\mathrm{div}{T}_f,\phi ⟩=0$ for every scalar test $\phi$ by step 3.1.

The double integral of step 4.1 has the limit ${\int }_0^{2\pi }{\int }_{-R}^R{\psi }_3(0,0,z)dzd\theta =2\pi A(\psi )$ as $\epsilon \to 0$. Indeed, ${\psi }_3$ is continuous on ${\mathbb{R}}^3$, hence uniformly continuous on the compact set $\{x:\rho \le R,\mathrm{\mid }{x}_3\mathrm{\mid }\le R\}$ by [F11], so for $\eta >0$ there is $\delta >0$ with $\mathrm{\mid }{\psi }_3(\epsilon \cos \theta ,\epsilon \sin \theta ,z)-{\psi }_3(0,0,z)\mathrm{\mid }\le \eta$ whenever $0<\epsilon <\delta$ and $(\theta ,z)\in [0,2\pi ]\times [-R,R]$; dominated convergence [F11] on the finite-measure rectangle, or equivalently the estimate with the constant majorant $\eta$, gives convergence of the double integrals; the limit integrand does not depend on $\theta$, and finite linearity [F11] gives the iterated integral $\left({\int }_0^{2\pi }1d\theta \right)({\int }_{-R}^R{\psi }_3(0,0,z)dz)=2\pi A(\psi )$.

Combining steps 3.1, 4.1 and 5.2, $⟨\mathrm{curl}{T}_f,\psi ⟩={\int }_{{\mathbb{R}}^3}f\cdot \mathrm{curl}\psi d{\lambda }_3={\lim }_{\epsilon \to 0}{\int }_{V_{\epsilon }}f\cdot \mathrm{curl}\psi d{\lambda }_3={\lim }_{\epsilon \to 0}{\int }_0^{2\pi }{\int }_{-R}^R{\psi }_3(\epsilon \cos \theta ,\epsilon \sin \theta ,z)dzd\theta =2\pi A(\psi )$ for every vector test $\psi$, which is the claimed identification $\mathrm{curl}{T}_f=2\pi e_3{\delta }_Z$.

Steps 5.1 and 6.1 prove the two distributional identities of (c), step 1.2 proves the pointwise vanishing (b), and step 1.3 proves the circulation (d); step 2.1 is (a). Every pairing above is an integral of a locally integrable function against a smooth test, so no product of two distributions is formed. Countable Choice is used only through the measure and distribution interfaces [F13], namely Tonelli, polar coordinates, the nullity of countable sets and the regular-distribution embedding; no stronger choice principle occurs.