We provide a proof of Morera's theorem under the hypothesis that for any circuit $\Gamma$ contained in $G$ . This is apparently more restrictive, but actually equivalent, to supposing for any triangle $\Delta\subseteq G$ , provided that $f$ is continuous in $G$ .

The idea is to prove that $f$ has an antiderivative $F$ in $G$ . Then $F$ , being holomorphic in $G$ , will have derivatives of any order in $G$ ; but $F^{(n)}(z)=f^{(n-1)}(z)$ for all $z\in G$ , $n\in\mathbb{N}$ , $n\geq 1$ .

First, suppose $G$ is connected. Then $G$ , being open, is also pathwise connected.

Fix $z_0\in G$ . For any $z\in G$ define $F(z)$ as \begin{equation} \label{eq:antid} F(z) = \int_{\gamma(z_0,z)}f(w)dw\;, \end{equation}where $\gamma(z_0,z)$ is a path entirely contained in $G$ with initial point $z_0$ and final point $z$ .

The function $F:G\to\mathbb{C}$ is well defined. In fact, let $\gamma_1$ and $\gamma_2$ be any two paths entirely contained in $G$ with initial point $z_0$ and final point $z$ ; define a circuit $\Gamma$ by joining $\gamma_1$ and $-\gamma_2$ , the path obtained from $\gamma_2$ by ``reversing the parameter direction''. Then by linearity and additivity of integral \begin{equation} \label{eq:wd} \int_\Gamma f(w)dw =\int_{\gamma_1}f(w)dw+\int_{-\gamma_2}f(w)dw =\int_{\gamma_1}f(w)dw-\int_{\gamma_2}f(w)dw\;; \end{equation}but the left-hand side is 0 by hypothesis, thus the two integrals on the right-hand side are equal.

We must now prove that $F'=f$ in $G$ . Given $z\in G$ , there exists $r>0$ such that the ball $B_r(z)$ of radius $r$ centered in $z$ is contained in $G$ . Suppose $0<|\Delta{z}|<r$ : then we can choose as a path from $z$ to $z+\Delta{z}$ the segment $\gamma:[0,1]\to G$ parameterized by $t\mapsto z+t\Delta{z}$ . Write $f=u+iv$ with $u,v:G\to\mathbb{R}$ : by additivity of integral and the mean value theorem, \begin{equation} \label{eq:ir} \frac{F(z+\Delta{z})-F(z)}{\Delta{z}} =\frac{1}{\Delta{z}}\int_{\gamma}f(w)dw =\frac{1}{\Delta{z}}\int_0^1f(z+t\Delta{z})\Delta{z}dt =u(z+\theta_u\Delta{z})+iv(z+\theta_v\Delta{z}) \end{equation}for some $\theta_u,\theta_v\in(0,1)$ . Since $f$ is continuous, so are $u$ and $v$ , and $\lim_{\Delta{z}\to 0}\frac{F(z+\Delta{z})-F(z)}{\Delta{z}}=u(z)+iv(z)=f(z)$ .

In the general case, we just repeat the procedure once for each connected component of $G$ .