Let $f$ be an analytic function in $D$ and continuous on $\partial D$, where $D\subseteq\mathbb{C}$ is an open set, bounded by a Jordan curve.
Is it true that $f\left(z\right)=\frac{1}{2\pi i}\int\limits _{\partial D}\frac{f\left(\xi\right)}{\xi-z}d\xi$?
I know how to prove it for a disk D, using uniform continuity of $\frac{f\left(\xi\right)\cdot\xi}{\xi-z}$, and perhaps if it is half a circle with a straight line connecting the edges, but it seems to fail when $D$ is any open set.
[EDIT 1: Refined the boundary to be bounded by a Jordan curve]
[EDIT 2: Refined terms]
An outline of the above comment explicited:
Let $U$ a domain bounded by a rectifiable Jordan curve $J$; then $U$ is simply connected so we can pick $\phi$ a Riemann map from the unit disc $\mathbb D$ to $U$; then Caratheodory theorem says that $\phi$ extends to a continuous (bijective) map from the unit circle $C$ to $J$, call it still $\phi$, and a corollary of the Riesz brothers theorem on analytic measures says that $J$ is rectifiable if and only if $\phi$ is absolutely continuous on $C$ (or equivalently $\phi' \in H^1$) and then $\phi$ gives a change of variables on the boundary hence $$\int_J \frac {f(\zeta)d\zeta}{z-\zeta}=\int_C \frac {f(\phi(\eta))\phi'(\eta)d\eta}{\phi (w)-\phi(\eta)}, z=\phi(w)$$ and by the usual Cauchy (or residue theorem if you prefer as the integrand has a pole at $w$ with residue $f(\phi(w))$) $$\int_C \frac {f(\phi(\eta))\phi'(\eta)d\eta}{\phi (w)-\phi(\eta)}=2\pi i f(\phi(w))=2\pi if(z)$$