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I can show that if $K \subset \mathbb{C}$ is compact, then for any $a\in K,$ there exists a probability measure $\mu$ supported on $\partial K$ such that $f(a)=\int f\, \text{d}\mu$ for any $f\in A(K) = \{ f\in C(K): f \text{ analytic on } K^{\circ}\}.$ Now I want to find this measure $\mu$ given that $K$ is the closed unit disc. I don't know much complex analysis, and from what I can gather, it doesn't seem like we can use Cauchy's here. Any help would be appreciated!

Oh, and to clarify the title, the existence of this measure follows from applying Hahn-Banach to extend the functional $L(f) = f(a)$ to continuous functions on $\partial K$ and subsequently applying the Riesz Representation Theorem. There are other details to be filled in, but that's the idea.

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The Poisson kernel is what you want: $$ f(re^{i\theta}) = \frac{1}{2\pi}\int_{0}^{2\pi}f(e^{i\theta'}) \frac{1-r^{2}}{1-2r\cos(\theta-\theta')+r^{2}}d\theta',\;\;\;\;\; 0 \le r < 1,\;\; 0 \le \theta \le 2\pi. $$ The Poisson representation holds for any function $f$ which is harmonic in the open unit disk, and continuous on the closed unit disk, which includes a function $f$ which is holomorphic in the open unit disk and continuous on the closed unit disk. The Poisson kernel is non-negative and is a probability measure because the Poisson representation holds for $f\equiv 1$.

The Poisson representation can be derived for holomoprhic functions by starting with the Cauchy integral formula for $0 < |z| < 1$: $$ f(z)=\frac{1}{2\pi i}\oint_{|w|=1}\frac{f(w)}{w-z}\,dw = \frac{1}{2\pi i}\oint_{|w|=1}f(w)\left[\frac{1}{w-z}-\frac{1}{w-1/\overline{z}}\right]dw. $$ Because $1/w=\overline{w}$ for $|w|=1$, the above may be written as $$ \begin{align} f(z) & =\frac{1}{2\pi i}\oint_{|w|=1}f(w)\left[\frac{1}{w-z}-\frac{\overline{zw}}{\overline{z}-\overline{w}}\right]dw \\ & = \frac{1}{2\pi i}\oint_{|w|=1}f(w)\frac{1-|z|^{2}}{|w-z|^{2}}\overline{w}dw \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}f(e^{i\theta'})\frac{1-|z|^{2}}{|e^{i\theta'}-z|^{2}}d\theta'. \end{align} $$ The final result extends to $z=0$ as well. Letting $z=re^{i\theta}$ gives the stated Poisson representation because $$ |e^{i\theta'}-re^{i\theta}|^{2}=|1-re^{i(\theta-\theta')}|^{2}=1-2r\cos(\theta-\theta')+r^{2}. $$ Because the Poisson kernel is real, the representation extends separately to the real and imaginary parts of any such $f$; but you don't need the extended representation for what you're doing.

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