= Solution
Define the <Radon transform> and its transverse <Hilbert transform> by
$$
Rf(\rho,\theta)=\int_{\mathbb R}F(\tau',\rho,\theta)d\tau',\qquad
Hh(\rho)=\frac1\pi\operatorname{PV}\int_{\mathbb R}\frac{h(\rho')}{\rho-\rho'}d\rho'.
$$
Use the <signed Cauchy boundary operators> $P^+=(I+iH)/2$ and $P^-=(-I+iH)/2$. Thus $P^+$ and $-P^-$ are the complementary idempotent projections; $P^-$ itself carries the signed lower-boundary convention. The PDF defines the outside limit $\lambda^-=(1+\varepsilon)e^{i\theta}$ but omits the definition of the inside limit in its displayed formula. We take $\lambda^+=(1-\varepsilon)e^{i\theta}$, as required by that formula.
As $r\to1$, apply the <Sokhotski–Plemelj formula> to the kernel from part (b):
$$
\frac1{u+i0v}=\operatorname{PV}\frac1u-i\pi\operatorname{sgn}(v)\delta(u).
$$
The <principal value> contribution is $i\operatorname{sgn}(b_r)HRf/2$. The delta contribution is the same from either side and equals
$$
\frac12\int_{\mathbb R}\operatorname{sgn}(\tau-\tau')F(\tau',\rho,\theta)d\tau'
=\frac12Rf-\int_\tau^\infty F(\tau',\rho,\theta)d\tau'.
$$
Since $b_r<0$ for the inside limit, the resulting \b[boundary value] is
$$
\boxed{\mu(\lambda^+)=-P^-Rf(\rho,\theta)-\int_\tau^\infty F(\tau',\rho,\theta)d\tau'.}
$$
The outside limit similarly has $P^+Rf$ in place of $-P^-Rf$. Naming the signed convention explicitly prevents a sign error from treating $P^-$ as a positive projection.
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