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.
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.