The set of convergent points on $\partial D(a,R)$ is measurable Today in our complex analysis class, the professor talked about an interesting theorem by words but didn't give a name, and would someone like to give me a reference? Here is the sketch:
Consider a power series $f(z)=\sum_{0}^{\infty}a_n (z-a)^n$ with radius of convergence $R$. We know that $f$ converges for $|z|<R$ and diverges for $|z|>R$. We have no clue when $|z|=R$, but the set of convergent points on $|z|=R$ is measurable.
I would like to know the reference for the last sentence. Appreciate any help!
 A: Consider first a sequence $\{f_n\}_{n=0}^{\infty}$ of $[-\infty,\infty]$ valued functions defined on some measurable space $(X,\mathfrak{M})$. Then, $\sup f_n$ is measurable (with respect to $\mathfrak{M}$ on the domain and the usual Borel $\sigma$-algebra on the target). Why? For each $t\in [-\infty,\infty]$, we have,
\begin{align}
\left(\sup\limits_{n\geq 0}f_n\right)^{-1}([-\infty,t])=\bigcap_{n=0}^{\infty}f_n^{-1}([-\infty,t]).
\end{align}
Each set on the right belongs to $\mathfrak{M}$ since each $f_n$ is measurable, therefore, the countable intersection also belongs to $\mathfrak{M}$. Finally, since the intervals generate the Borel $\sigma$-algebra, this proves countable supremums are measurable.  As a simple corollary, we have that $\inf\limits_{n\geq 0}f_n$ is also measurable (since $\inf f_n = -\sup(-f_n)$). As a further corollary, $\limsup\limits_{n\to\infty} f_n=\inf\limits_{k\geq 0}\sup\limits_{n\geq k}f_n$ and $\liminf\limits_{n\to\infty} f_n=-\limsup\limits_{n\to\infty}(-f_n)$ are also measurable functions.
Next, suppose $f,g:X\to [-\infty,\infty]$ are measurable functions. Then the set $[f<g]:=\{x\in X\,:\, f(x)<g(x)\}$ is also measurable. This follows easily because
\begin{align}
[f<g]&=\bigcup_{q\in \Bbb{Q}}[f(\cdot)<q]\cap [q<g(\cdot)].
\end{align}
On the right, measurability of $f$ and $g$ imply the sets $[f<q]$ and $[q<g]$ are measurable. Hence, so is their intersection, and therefore so is the countable union. Now that we know $[f<g]$ is a measurable set, we know that its complement $[f\geq g]$ is as well. So, by interchanging the roles of $f$ and $g$, we know $[f<g],[f>g], [f\leq g],[f\geq g]$ are all measurable, so finally $[f=g]=[f\leq g]\cap [f\geq g]$ is measurable as well.
With these remarks in mind, it easily follows that the set of points where the limit exists is measurable, because the limit exists if and only if the $\limsup$ and $\liminf$ are equal. In symbols, $\{x\in X\,:\, \text{$\lim\limits_{n\to\infty}f_n(x)$ exists in $[-\infty,\infty]$}\}=[\limsup f_n = \liminf f_n]$, and the latter we know is a measurable set.
Now, for complex-valued functions, we have measurability if and only if the real and imaginary parts are measurable (and complex numbers/functions have limits if and only if their real and imaginary parts do), so the set of points where the limit of a sequence of complex-valued functions exist is again a measurable subset of $X$. Finally, a series is defined as the limit of partial sums. Can you finish this off?
