I'm now taking a course in complex analysis and in wikipedia it was said that Cauchy Integral formula is true also for a function which is "holomorphic in the open region enclosed by the path and continuous on its closure."
More precisely, How can I wonder how to prove the following theorem
given $f$ analytic in open set $\Omega\subset\mathbb{C}$ and continious in $\overline{\Omega}$, $$\forall z\in\Omega, f(z)=\frac 1 {2\pi i}\int_{\partial\Omega}\frac{f(\xi)}{\xi-z}d\xi$$
using only Cauchy integral formula for open subsets of the complex plane?
First of all, you need some assumption on $\partial\Omega$ so that the integral in the RHS makes sense. Say, $\partial D$ consists of finitely many smooth, or at least rectifiable curves. The difficulty of the proof depends on the exact assumption you make.
The complete proof is somewhat technical. (See, for example, Shabat, Complex Analysis, volume 1). The idea is the following. You consider a sequence $$D_1\subset D_2\subset\ldots$$ of regions with nice boundaries which exhaust $\Omega$ and whose boundaries tend to $\Omega$. Then you apply the usual Cauchy theorem to the $\Omega_n$ and pass to the limit. The difficult part is to justify this limit, and this depends on your exact assumptions about $\partial\Omega$.