Riemann sums of the Cauchy integral give rational approximation

Choose $\mathrm{\Gamma }$ as in [L1]. Because $n(\mathrm{\Gamma },z)=1$ on $K$ and ${\mathrm{\Gamma }}^{\ast }\subseteq U$, the cycle is null-homologous in $U$ and [L2] gives $f(z)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{f(\zeta )}{\zeta -z}d\zeta (z\in K)\mathrm{.}$

Decompose $\mathrm{\Gamma }$ into finitely many oriented line segments ${\gamma }_j:[a_j,b_j]\to \mathbb{C}$. For each $j$, the function ${\varphi }_j(t,z)=f({\gamma }_j(t)){\gamma }_j^{\mathrm{\prime }}(t)\mathrm{/}({\gamma }_j(t)-z)$ is continuous on the compact set $[a_j,b_j]\times K$, because ${\mathrm{\Gamma }}^{\ast }\cap K=\mathrm{\varnothing }$. By [L3], each ${\varphi }_j$ is uniformly continuous there, so a fine enough Riemann sum approximates ${\int }_{a_j}^{b_j}{\varphi }_j(t,z)dt$ uniformly in $z\in K$.

Summing those edgewise Riemann sums gives a rational function of the form $r(z)={\sum }_{\nu }{c}_{\nu }\mathrm{/}({\xi }_{\nu }-z)$ with sample points ${\xi }_{\nu }\in {\mathrm{\Gamma }}^{\ast }$. Choosing the mesh so that the total edgewise error is below $\epsilon$ and using step 1.1 yields ${\sup }_{z\in K}\mathrm{\mid }f(z)-r(z)\mathrm{\mid }<\epsilon$.