Proof, [Proposition VIII.1.1, Con78]. Via fattening, let $V \in \cn_{\complex}^{o}(K)$ precompact with $\ol V \subset U$. Identify $\complex = \real^{2}$, then since $V$ is precompact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)}\subset U$ such that:

1. (1)

   For each $1 \le j \le n$, $R_{j} = [x_{j}, x_{j} + \delta] \times [y_{j}, y_{j} + \delta] \subset U$.

2. (2)

   $\bigcup_{j = 1}^{n} R_{j} \supset V$.

3. (3)

   $x_{j}, y_{j} = 0 \mod \delta$.

In other words, $V$ is covered with a grid of squares with side-length $\delta$ and corners $\seqf{(x_j, y_j)}$. Now, for each $1 \le j \le n$ and $t \in [0, 1]$, let

and $\gamma_{j} = \gamma_{j, \downarrow}\cdot \gamma_{j, \leftarrow}\cdot \gamma_{j, \uparrow}\cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_{j}$,

by Cauchy’s Integral Formula. Let $\seqf[N]{\mu_j}$ be an enumeration of

then for each $z_{0} \in \bigcup_{j = 1}^{n} R_{j}^{o}$ and $f \in H(U; E)$,

From here, it is sufficient to eliminate line segments that intersect $K$ and ensure that the remaining segments form a collection of loops. Let $1 \le j \le k \le N$, then $(\mu_{j}, \mu_{k})$ is redundant if for each $t \in [0, 1]$, $\mu_{j}(t) = \mu_{k}(1 - t)$. If $(\mu_{j}, \mu_{k})$ are redundant, then

1. (4)

   There exists no $1 \le l \le N$ with $l \ne j, k$ such that $(\mu_{j}, \mu_{l})$ or $(\mu_{k}, \mu_{l})$ is redundant.

2. (5)

   $\int_{\mu_j}\frac{f(z)}{z - z_{0}}dz + \int_{\mu_k}\frac{f(z)}{z - z_{0}}dz = 0$.

so every line segment either cannot form a redundant pair, or forms a unique one.

By relabeling, let $\bracs{\mu_j|1 \le j \le m}$ such that for each $1 \le j \le m$, there exists no $1 \le k \le n$ such that $(\mu_{j}, \mu_{k})$ is redundant. By (5), for each $z_{0} \in \bigcup_{j = 1}^{n} R_{j}^{o}$ and $f \in H(U; E)$,

Let $1 \le j \le N$ such that $\mu_{j}([0, 1]) \cap K \ne \emptyset$. Since $V \in \cn_{\complex}(K)$, there exists $1 \le k \le N$ such that $(\mu_{j}, \mu_{k})$ are redundant by (2) and (3). Therefore $\bigcup_{j = 1}^{m} \mu_{j}([0, 1]) \cap K = \emptyset$. Since $\bigcup_{j = 1}^{n} R_{j}^{o}$ is dense in $K$, the above also holds for all $z \in K$.

Finally, let $1 \le j \le m$. Since $\mu_{j}$ does not form a redundant pair, $\mu_{j}(1)$ intersects at most two distinct squares by (3). In which case, there must exist $1 \le k \le m$ such that $\mu_{k}(0) = \mu_{j}(1)$. Therefore $\seqf[m]{\mu_j}$ forms a collection of closed rectifiable curves.$\square$