I'm reading the article "A Note on Dixon's Proof of Cauchy's Integral Theorem". It is stated, that it is easy to see that the map $g$ in Proposition 2 is analytic in $G\setminus \{ \gamma \} $
without using Fubini and the joint continuity of $\varphi$. How does that work?
I know only proofs where one shows the joint continuity of $\varphi$ and then applies the Leibniz integral rule or Morera's theorem, both involving Fubini.
The situation is as follows:
$G\subset \mathbb{C}$ is an open set, $f$ is analytic on $G$, $\gamma$ is a closed curve in $G$ (partially continuously differentiable), $\varphi(w,z)=\frac{f(w)-f(z)}{w-z}$ for $w\neq z$ and $\varphi(z,z)=f^{\prime}(z)$, $g$ is defined by $g(z)=\int_{\gamma}\varphi(w,z) dw$
Thank you very much and please excuse my poor English.
The explanation you seek can be found in the proof of the Proposition. The displayed equation at the beginning of the proof of Proposition 2 shows that $g(z)$ is the difference of two expressions that are analytic in $z$ (because they are constructed out of (i) the analytic function $f(z)$ and (ii) an integral that is the Cauchy transform of a function that is continuous on the path of integration.
P.S. The displayed equation itself is justified by an appeal to a reference [3] that was not included in your posting. If your question pertains to that, I suggest you give more details about [3]