= Solution
For $|\zeta|=1$ and $|z|<1$, the <Poisson kernel on the circle> satisfies
$$
\operatorname{Re}\frac{\zeta+z}{\zeta-z}
=\frac{1-|z|^2}{|\zeta-z|^2}\geq0.
$$
For fixed $\zeta$, it is the real part of a <holomorphic function> of $z$, hence a <harmonic function>. On each compact subdisk, its derivatives are bounded uniformly in $\zeta$ because the denominator stays away from zero. Differentiation under the integral therefore proves that $P\phi$ is <harmonic>, with the word interpreted componentwise when $\phi$ is complex-valued.
The <geometric series>
$$
\frac{\zeta+z}{\zeta-z}=1+2\sum_{n\geq1}z^n\zeta^{-n}
$$
converges uniformly in $\zeta$ for a fixed compact subdisk. With normalized <Lebesgue measure>, integration of its real part gives kernel mass one. Fix a boundary point $\omega$ and choose $\delta>0$ so that $|\phi(\zeta)-\phi(\omega)|<\varepsilon$ when $|\zeta-\omega|<\delta$. Then
$$
|P\phi(z)-\phi(\omega)|
\leq\varepsilon+2\|\phi\|_\infty
\int_{|\zeta-\omega|\geq\delta}\frac{1-|z|^2}{|\zeta-z|^2}\,dm(\zeta).
$$
If $z\to\omega$ from inside the <unit disc>, the remaining denominator is at least $(\delta/2)^2$ for sufficiently close $z$, while $1-|z|^2\to0$. Hence \b[$P\phi(z)\to\phi(\omega)$ along every interior approach], not just radial ones.
Orient the unit circle counterclockwise in the <Cauchy transform>. Its kernel and every $w$-derivative are uniformly bounded for $w$ in any compact subset of $\mathbb C\setminus\mathbb T$. Thus differentiation under the integral gives
$$
\Phi'(w)=\frac1{2\pi i}\int_{\mathbb T}\frac{\phi(\zeta)}{(\zeta-w)^2}\,d\zeta,
$$
and proves that $\Phi$ is a <holomorphic function> on both components of its domain. Since $d\zeta=2\pi i\zeta\,dm(\zeta)$, for $0<r<1$ and $|\omega|=1$ the difference of its two kernels is
$$
\frac{\zeta}{\zeta-r\omega}-\frac{\zeta}{\zeta-r^{-1}\omega}
=\frac{1-r^2}{|\zeta-r\omega|^2}.
$$
Consequently the <paired radial jump of a continuous Cauchy transform> obeys the exact identity
$$
\boxed{\Phi(r\omega)-\Phi(r^{-1}\omega)=P\phi(r\omega)\quad(0<r<1)},
$$
and the established <Poisson integral> boundary convergence gives
$$
\boxed{\Phi(r\omega)-\Phi(r^{-1}\omega)\longrightarrow\phi(\omega)
\quad\text{as }r\uparrow1.}
$$
This proves the paired difference without assuming that the two individual boundary values exist for every continuous datum.
\b[The printed direction $r\downarrow1$ has the opposite sign.] It is present in the original PDF, not merely in the converted TeX. With that direction $r>1$, so substituting $s=1/r<1$ in the exact identity gives
$$
\boxed{\Phi(r\omega)-\Phi(r^{-1}\omega)=-P\phi(r^{-1}\omega)\longrightarrow-\phi(\omega).}
$$
The simplest counterexample is $\phi\equiv1$: the <Cauchy integral formula> gives $\Phi(w)=1$ inside and $\Phi(w)=0$ outside, so the printed difference is constantly $-1$. The requested positive jump is corrected by taking $r\uparrow1$, or by reversing the order of the difference when $r\downarrow1$.
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