Understanding the proof of a continuity principle for potentials I am reading Ransford's Potential Theory in the Complex Plane and I am stuggling with the inequality very last part of this proof (of (a)). 
Define $p_ {\mu} (z):=\int \log|z-w| \, d \mu (w)$. 

I have tried writing it out:
\begin{eqnarray*}
\liminf_{z \to \zeta_0} p_{\mu} (z) &\geq& \liminf_{\zeta \to \zeta_0, \zeta \in K}(p _{\mu} (\zeta)-\varepsilon \log 2- \int _{K \backslash \Delta (\zeta_0,r)} \log \left| \frac{\zeta-w}{z-w} \right | \, d \mu(w)) \\
&\geq&\liminf_{\zeta \to \zeta_0, \zeta \in K}p \mu (\zeta)-\varepsilon \log 2- \limsup_{\zeta \to \zeta_0, \zeta \in K}\int _{K \backslash \Delta (\zeta_0,r)} \log \left| \frac{\zeta-w}{z-w} \right | \, d \mu(w)
\end{eqnarray*}
Is this correct? And if it is, why is the last part zero as it is stated in the proof? 
And another thing: what does this theorem actually say, that potentials are continuous on the boundary of the support? 
Any hint would really be appreciated.
 A: What you stated is not correct since you still have a dependence on $z$ on the right-hand side of your inequality. 
First of all, notice that we actually need to only prove that the upper limit that makes the problem is non-positive, since we are substracting it.
Secondly, the way $\zeta$ is defined, it is actually a function of $z$. So if we take limit in $z \rightarrow \zeta_0$ we get as in the proof
$$
 \liminf_{z \rightarrow \zeta_0} p_{\mu}(z) \geq \liminf_{z \rightarrow \zeta_0} \left(p_{\mu}(\zeta(z)) - \epsilon\log 2 - \int_{K\setminus \Delta}  \log\left|\frac{\zeta(z) - w}{z-w}\right|\,d\mu(w) \right) \geq \\
\liminf_{z \rightarrow \zeta_0}p_{\mu}(\zeta(z)) + \liminf_{z \rightarrow \zeta_0} \left(- \int_{K\setminus \Delta(\zeta_0,r)}  \log\left|\frac{\zeta(z) - w}{z-w}\right|\,d\mu(w) \right) - \epsilon\log 2 \geq \\
\liminf_{\zeta \rightarrow \zeta_0,\; \zeta \in K}p_{\mu}(\zeta) - \limsup_{z \rightarrow \zeta_0} \left( \int_{K\setminus \Delta(\zeta_0,r)}  \log\left|\frac{\zeta(z) - w}{z-w}\right|\,d\mu(w) \right) - \epsilon\log 2.
$$
Now let's deal with the integral. Remember the assumption - $\mu$ is finite Borel measure supported on a compact set, so $\mu(K) < \infty$. Suppose now that we came very close to $\zeta_0$, so $\delta(z) = |z-\zeta_0| \ll r$, then by definition of $\zeta(z)$, $|z-\zeta(z)| \leq |z-\zeta_0|$, thus we get
$$
\log\left|\frac{\zeta(z) - w}{z-w}\right| = \log\left|\frac{\zeta(z) -z +z - w}{z-w}\right| \leq \log\left( 1+\left|\frac{\zeta(z) -z}{z-w}\right|\right) \leq \\
\log\left( 1+\left|\frac{\delta(z)}{r-\delta(z)}\right|\right),
$$
since $w \in K\setminus\Delta(\zeta_0,r)$. The estimation for the integral is straightforward now:
$$
\limsup_{z \rightarrow \zeta_0} \int_{K\setminus \Delta(\zeta_0,r)}  \log\left|\frac{\zeta(z) - w}{z-w}\right|\,d\mu(w) \leq \limsup_{z \rightarrow \zeta_0} \;\log\left( 1+\left|\frac{\delta(z)}{r-\delta(z)}\right|\right) \mu(K) = 0.
$$
As for your second question, well, I'm not sure.
