A proof about Boundary Asymptotics of the Cauchy Transform using the Dominated Convergence Theorem

Question

In this paper, on page 8, the authors make the remark that, for a complex, finite borel measure $\mu$ on the circle/unit interval, and for the Cauchy Transform: $$K\left\{ \mu\right\} \left(z\right)\overset{\textrm{def}}{=}\int_{0}^{1}\frac{d\mu\left(t\right)}{1-e^{-2\pi it}z},\textrm{ }\forall\left|z\right|<1$$ one can use the Dominated Convergence Theorem to show that: $$\lim_{r\uparrow1}\left(1-r\right)K\left\{ \mu\right\} \left(re^{2\pi it}\right)=\mu\left(\left\{ t\right\} \right)$$ I would like to see this argument written out in full detail (no handwaving, no skipped steps, all objects explicitly constructed whenever possible, no skipped computations), seeing as it is directly related to what I'm researching, and I want to be able to have a baseline proof that I can refer to without needing to worry about whether or not I might have made a mistake in writing up said baseline proof.

Answer 1

Write $\mu=h|\mu|$ for the polar decomposition of $\mu$ with a measurable function $h\colon [0,1]\to \mathbb{T}$. You want to evaluate the limit $$\lim_{r\nearrow 1}\int_0^1\frac{1-r}{1-re^{2\pi i(t-s)}}h(s)\,d|\mu|(s).$$ By the reverse triangle inequality, $|1-re^{2\pi i(t-s)}|\geq 1-r$. Thus the absolute value of the integrand is bounded by $1$, which is integrable since $\mu$ is finite.

For $s=t$, the integrand is $1$. For $s\neq t$, the numerator of the integrand tends to $0$ as $r\nearrow 1$, while the denominator converges to $1-e^{2\pi i(t-s)}\neq 0$. Altogether $$ \lim_{r\nearrow 1}\frac{1-r}{1-re^{2\pi i(t-s)}}h(s)=1_{\{t\}}(s)h(s). $$ Hence $$ \lim_{r\nearrow 1}\int_0^1\frac{1-r}{1-re^{2\pi i(t-s)}}\,h(s)d|\mu|(s)=\int_0^1 1_{\{t\}}(s)\,h(s)d|\mu|(s)=\mu(\{t\}) $$ by the dominated convergence theorem.