While reading on Cauchy's integral formula, I found the following two theorems on the Cauchy integral operator:
The first theorem states: Let $f$ be a complex function that is holomorphic in a region $U$. Let $\Delta_r(c)$ be a disk, which along with its boundary $\delta \Delta_r(c)$ is contained in $U$. For all $z \in \Delta_r(c)$ the following is true: \begin{equation*} f(z) = \frac{1}{2 \pi i} \int_{\delta \Delta_r(c)} \frac{f(\zeta)}{\zeta - z} d\zeta \end{equation*}
The second theorem states: Let $f: \overline{\Delta_1(0)} \rightarrow \mathbb{C}$ be a continous function which is also holomorphic in $\Delta_1(0)$. It holds for every $z \in \Delta_1(0)$ that \begin{equation*} f(z) = \frac{1}{2 \pi i} \int_{\delta \Delta_1(0)} \frac{f(\zeta)}{\zeta - z} d\zeta \end{equation*}
The second theorem is given without proof. My question is whether the second statement holds for general disks and whether it can be proven using the first one?
Suppose $f \colon \overline{\Delta_R(c)} \rightarrow \mathbb{C}$ is continuous and holomorphic on $\Delta_R(c)$. Then by the first theorem, we have $$f(z) = \frac{1}{2\pi i} \int_{\delta \Delta_r(c)} \frac{f(\zeta)}{\zeta-z} d\zeta$$ for all $r<R$. It should be clear that the right hand side depends continuously on $r$ so that we can take the limit $r \rightarrow R$ to get $$f(z) = \frac{1}{2\pi i}\int_{\delta \Delta_R(c)} \frac{f(\zeta)}{\zeta-z} d\zeta.$$