Given a compact $K\subset\mathbb{C}$, Let $A(K)=\mathcal{C}^0(K)\cap \mathcal{H}(\text{int}(K))$. I am asked to prove that, for every $a\in K$, there is a probability (Borel) measure $\mu_a$ on $\partial K$ such that $\int_{\partial K} fd\mu_a=f(a)$ for every $f\in A(K)$.
My idea was: first, note that $A(K)$ is a closed subset of $\mathcal{C}^0(\partial K)$ equipped with the usual norm, bu Cauchy inequalities. Now the evaluation in $a$ is a bounded linear functional on $A(K)$, which can be extended to a bounded linear functional on $\mathcal{C}^0(\partial K)$ by Hahn-Banach. By Riesz's representation theorem, there's a Borel regular measure $\mu$ on $\partial K$ such that $f(a)=\int fd\mu$. Since the evalutaion, as a linear functional, is easily seen to have norm $1$, $||\mu||=1$ (where $||\mu||=|\mu|(X)$, i.e. the total variation norm). The only thing left to prove is that $\mu$ is a real unsigned measure. I have no idea on how to proceed, an hint would be very helpful.
In the particular case $K=\mathbb{D}$, the result is easy: the measure is induced by the poisson kernel, i.e. $$f(a)=\int_{0}^{2\pi}f(e^{it})\text{Re}\left(\frac{e^{it}+z}{e^{it}-z}\right)\frac{dt}{2\pi}$$ However, I don't think this approach is useful in general (one could use it if $\partial K$ where sufficiently regular and $K$ simply connected maybe, using the conformal map to the disk to translate the poisson kernel, butit does not seem to work in general)
I think I got it: instead of thinking about holomorphic functions, let us think about harmonic functions (and so about $\Re(f),\Im(f)$). Let $h(K)$ be the set of functions continuous on $K$ harmonic on $\text{int}(K)$. The maximum principle implies that $h(K)$ is a closed subspace of $\mathcal{C}^0(K,\mathbb{R})$. As before, the evaluation $e_a$ is a continuous (real) linear functional on a $K$, with norm $1$, and it is positive. We can now apply the real version of Riesz-Markov representation theorem, giving us a positive probability measure $\mu_a$ such that
$$u(a)=\int_{\partial K}u(t)d\mu_a(t)\\ v(a)=\int_{\partial K}v(t)d\mu_a(t)$$
And thus, considering $f(z)=u(z)+iv(z)$, the result follows.