Would the following not be a correct proof of Cauchy's Integral Theorem?

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I'm somewhat confused as to how the general version of Cauchy's Theorem does not follow (almost) immediately from its version in a disk. At least Ahlfors as well as Stewart & Tall prove the general theorem with arguments more complicated than the one I have in mind, which leads me to ponder: is the proof below wrong?

Cauchy's Integral Theorem: let $U\subseteq\mathbb{C}$ be an open set, with $\gamma$ be a piecewise smooth curve homologous to $0$ in $U$, and $f$ holomorphic in $U$. Then $$\int_\gamma f = 0.$$

Proof: by the Paving Lemma, there is a finite collection of open balls $B_j$ centered at points $\gamma(t_j)$ of $\text{im}(\gamma)$ such that $$\text{im}(\gamma)\subset\bigcup_j B_j.$$ WLOG we may assume $t_1<\cdots<t_m$. By Cauchy's Theorem in a Disk and the Gradient Theorem there are antiderivatives $F_j:B_j\to\mathbb{C}$ of $f$. As $[t_j,t_{j+1}]$ is connected, so is $B_j\cap B_{j+1}$, an open set where both $F_j$ and $F_{j+1}$ are defined and differ up to a constant. Adjusting $F_{j+1}$ by adding a constant, we find an antiderivative $F_{j,j+1}:B_j\cup B_{j+1}\to\mathbb{C}$ of $f$. Repeating this process as necessary we find an antiderivative $F:\bigcup_j B_j\to\mathbb{C}$ of $f$. The result now follows by the Gradient Theorem.


The following is Theorem 6.44 in Stewart's & Tall's Complex Analysis. As it seems to me a form of the Gradient Theorem for $\mathbb{R}^2$, I call it by that name:

Gradient Theorem: Let $f$ be a continuous complex function defined on a domain $D$. Then the following conditions are equivalent:

  1. $f$ has an antiderivative in $D$,
  2. $\int_\gamma f = 0$ for every closed contour $\gamma$ in $D$,
  3. $\int_\gamma f$ depends only on the end points of $γ$ for any contour $γ$ in $D$.