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I guess one can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\Gamma$. Define the linear functional $L$ on $C_0^\infty(K)$ by $$ L(f):= \frac{1}{2i}\int_\Gamma \left(\mathcal{C}f\right)(z)dz, $$ where $\mathcal{C}$ is the Cauchy transform $$ \left(\mathcal{C}f \right)(z)=\frac{1}{2\pi i}\int_K\frac{f(w)}{z-w}dw\wedge d\bar{w}. $$ Clearly, $L$ is a continuous functional on $C_0^\infty(K)$. Moreover, by Hölder inequality, the Cauchy transform is continuous, say, from $L^3(K)$ to $C(K)$. Hence the generalised function $L$ can be represented as integration with respect to a density: $$ L(f)=\int_K f(z)\psi(z)dz\wedge d\bar{z}. $$ Since for any $\varphi\in C_0^\infty$, we have $\mathcal{C}\bar{\partial}\varphi\equiv \varphi$, it remains to show that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside. (Here we use that $\Gamma$ is rectifiable and thus has zero area.) By testing against a mollifier, this boils down to checking that $$ \frac{1}{2\pi i}\int_\Gamma\frac{d\zeta}{\zeta-z}=\cases{1,\ z \text{ is inside}\\ 0,\ z \text{ is outside}.} $$ But the integral can be interpreted as the increment of $\log (\zeta - z)$ along $\Gamma$, and thus it is equal to the winding number of $\Gamma$. Therefore, the result follows from the topological statement that any Jordan curve has winding number $\pm 1$ w. r. t. a point inside it, and $0$ w. r. t. a point outside. Some references for this statement are discussed here.

One can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\Gamma$ and any $p>2$. Given a function $f\in L^p(K)$, we can define its Cauchy transform $$ \left(\mathcal{C}f \right)(z)=\frac{1}{\pi}\int_K\frac{f(w)}{z-w}. $$ By Hölder inequality, this is a continuous operator from $L^p(K)$ to $C(K)$. This operator inverts the $\bar{\partial}$ operator, namely, $\mathcal{C}\bar{\partial}\varphi\equiv\varphi$ for any $\varphi \in C^1_0(K)$; see the computation after the equation (8) in Chapter 5 of "Lectures on quasiconformal mappings" by Ahlfors (1966).

Now, the composition of $\mathcal C$ and integration over $\Gamma$ is a continuous linear functional on $L^p(K)$: $$ L(f):= \frac{1}{2i}\int_\Gamma \left(\mathcal{C}f\right)(z)dz=\int_K f(z)\psi(z), $$ for some $\psi \in L^{p'}(K)$. It follows that for any $\varphi \in C^1_0(K)$, we have $$ \frac{1}{2i}\int_\Gamma \varphi(z) dz = \int_K\psi(z)\bar{\partial}\varphi(z). $$Hence, it remains to check that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside; since $\Gamma$ itself is rectifiable, its area is zero. To this end, it suffices to check that $L(f)=\int_{\text{inside }\Gamma} f$ for any $f\in C_0(K\setminus\Gamma)$. For such an $f$, we can exchange the integrals: $$ L(f)=\frac{1}{2\pi i}\int_\Gamma dz\int_K \frac{f(w)}{z-w}=\int_K f(w)\frac{1}{2\pi i} \int_\Gamma \frac{dz}{z-w}. $$ The inner integral can be interpreted as the increment of $\log (z-w)$ along $\Gamma$, and thus it is equal to the winding number of $\Gamma$ w. r. t. $w$. Therefore, the result follows from the topological statement that any Jordan curve has winding number $\pm 1$ w. r. t. a point inside it, and $0$ w. r. t. a point outside. Some references for this statement are discussed here.

I guess one can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\Gamma$. Define the linear functional $L$ on $C_0^\infty(K)$ by $$ L(f):= \frac{1}{2i}\int_\Gamma \left(\mathcal{C}f\right)(z)dz, $$ where $\mathcal{C}$ is the Cauchy transform $$ \left(\mathcal{C}f \right)(z)=\frac{1}{2\pi i}\int_K\frac{f(w)}{z-w}dw\wedge d\bar{w}. $$ Clearly, $L$ is a continuous functional on $C_0^\infty(K)$. Moreover, by Hölder inequality, the Cauchy transform is continuous, say, from $L^3(K)$ to $C(K)$. Hence the generalised function $L$ can be represented as integration with respect to a density: $$ L(f)=\int_K f(z)\psi(z)dz\wedge d\bar{z}. $$ Since for any $\varphi\in C_0^\infty$, we have $\mathcal{C}\bar{\partial}\varphi\equiv \varphi$, it remains to show that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside. By testing against a mollifier, this follows from the Cauchy integral formula for $\Gamma$. But the integral of $(z-\zeta)^{-1}$ only depends on the winding number of $\Gamma$, thus, the result follows from the topological statement that any Jordan curve has winding number $\pm 1$ around a point inside it, and $0$ for a point outside.

I guess one can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\Gamma$. Define the linear functional $L$ on $C_0^\infty(K)$ by $$ L(f):= \frac{1}{2i}\int_\Gamma \left(\mathcal{C}f\right)(z)dz, $$ where $\mathcal{C}$ is the Cauchy transform $$ \left(\mathcal{C}f \right)(z)=\frac{1}{2\pi i}\int_K\frac{f(w)}{z-w}dw\wedge d\bar{w}. $$ Clearly, $L$ is a continuous functional on $C_0^\infty(K)$. Moreover, by Hölder inequality, the Cauchy transform is continuous, say, from $L^3(K)$ to $C(K)$. Hence the generalised function $L$ can be represented as integration with respect to a density: $$ L(f)=\int_K f(z)\psi(z)dz\wedge d\bar{z}. $$ Since for any $\varphi\in C_0^\infty$, we have $\mathcal{C}\bar{\partial}\varphi\equiv \varphi$, it remains to show that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside. (Here we use that $\Gamma$ is rectifiable and thus has zero area.) By testing against a mollifier, this boils down to checking that $$ \frac{1}{2\pi i}\int_\Gamma\frac{d\zeta}{\zeta-z}=\cases{1,\ z \text{ is inside}\\ 0,\ z \text{ is outside}.} $$ But the integral can be interpreted as the increment of $\log (\zeta - z)$ along $\Gamma$, and thus it is equal to the winding number of $\Gamma$. Therefore, the result follows from the topological statement that any Jordan curve has winding number $\pm 1$ w. r. t. a point inside it, and $0$ w. r. t. a point outside. Some references for this statement are discussed here.

I guess one can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\gamma$. Define the linear functional $L$ on $C_0^\infty(K)$ by $$ L(f):= \frac{1}{2i}\int_\Gamma \left(\mathcal{C}f\right)(z)dz, $$ where $\mathcal{C}$ is the Cauchy transform $$ \left(\mathcal{C}f \right)(z)=\frac{1}{2\pi i}\int_K\frac{f(w)}{z-w}dw\wedge d\bar{w}. $$ Clearly, $L$ is a continuous functional on $C_0^\infty(K)$. Moreover, by Hölder inequality, the Cauchy transform is continuous, say, from $L^3(K)$ to $C(K)$. Hence the generalised function $L$ can be represented as integration with respect to a density: $$ L(f)=\int_K f(z)\psi(z)dz\wedge d\bar{z}. $$ Since for any $\varphi\in C_0^\infty$, we have $\mathcal{C}\bar{\partial}\varphi\equiv \varphi$, it remains to show that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside. By testing against a mollifier, this follows from the Cauchy integral formula for $\Gamma$. But the integral of $(z-\zeta)^{-1}$ only depends on the winding number of $\Gamma$, thus, the result follows from the topological statement that any Jordan curve has winding number $\pm 1$ around a point inside it, and $0$ for a point outside.

I guess one can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\Gamma$. Define the linear functional $L$ on $C_0^\infty(K)$ by $$ L(f):= \frac{1}{2i}\int_\Gamma \left(\mathcal{C}f\right)(z)dz, $$ where $\mathcal{C}$ is the Cauchy transform $$ \left(\mathcal{C}f \right)(z)=\frac{1}{2\pi i}\int_K\frac{f(w)}{z-w}dw\wedge d\bar{w}. $$ Clearly, $L$ is a continuous functional on $C_0^\infty(K)$. Moreover, by Hölder inequality, the Cauchy transform is continuous, say, from $L^3(K)$ to $C(K)$. Hence the generalised function $L$ can be represented as integration with respect to a density: $$ L(f)=\int_K f(z)\psi(z)dz\wedge d\bar{z}. $$ Since for any $\varphi\in C_0^\infty$, we have $\mathcal{C}\bar{\partial}\varphi\equiv \varphi$, it remains to show that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside. By testing against a mollifier, this follows from the Cauchy integral formula for $\Gamma$. But the integral of $(z-\zeta)^{-1}$ only depends on the winding number of $\Gamma$, thus, the result follows from the topological statement that any Jordan curve has winding number $\pm 1$ around a point inside it, and $0$ for a point outside.