Let $\Omega\subseteq\Bbb C$ open, $\varphi:\Omega\to[-\infty,+\infty[$ is subharmonic on $\Omega$ if
- $\varphi$ is uppersemicontinous on $\Omega$
- For all $K\Subset\Omega$ compact, and for all $h\in\mathcal C(K)\cap\mathcal H(\stackrel{\circ}{K})$, if $\varphi\le h$ on $\partial K$, then $\varphi\le h$ on the whole $K$
Now we have the following
Ok, I can't understand, why the author, proving the second condition, passes thru the use of that sequence of compact subsets, instead of arguing directly: $\varphi(z):=\inf_{\nu}\varphi_{\nu}(z)=\lim_{\nu}\varphi_{\nu}(z)\le h(z)$ on $\partial K$. So, given $\varepsilon>0$, it's clear that $\varphi(z)\le h(z)+\varepsilon$ on $\partial K$. Thus there exists $\bar{\nu}=\nu(\varepsilon)$ s.t. $\varphi_{\nu}(z)\le h(z)+\varepsilon$ on $\partial K$, for all $\nu\ge\bar{\nu}$. But the $\varphi_{\nu}$'s are subharmonics, thus they verify 2, so $\varphi_{\nu}(z)\le h(z)+\varepsilon$ on the whole $K$, for all $\nu\ge\bar{\nu}$ which is the same conclusion of the book.
Is there some error in my argument? I thought that if the author used compact sets there must be a reason, for which a direct argument, like mine, is wrong.
This doubt came to my mind reading "since the limit for $\nu\to+\infty$ is empty": the limit he refers to is $\cap_{\nu}K_{\nu}$, and in order to prove this is $\emptyset$, I reasoned as I wrote above; and reasoning I noticed that, my argument seemed to prove all. So the proof of the claim is all contained in the proof of a detail! This is suspect enough to make me write here.
As far as I know, you need compactness to show that, given $\epsilon>0$, there exists a $\bar{\nu}=\bar{\nu}(\epsilon)$ such that
$$\sup_{\nu\geq\bar{\nu}}\varphi_{\nu}(z)\leq h(z)+\epsilon,\quad\forall z\in\partial K$$
To show that for fixed $z\in\partial K$ such a $\bar{\nu}$ exists so that the pointwise inequality holds is a trivial consequence of the definition of limits. But to show that this $\bar{\nu}$ holds uniformly in $z\in\partial K$ requires some argument. Namely, the sets $K_{\nu}$ are closed subsets (by upper semicontinuity) of the compact set $\partial K$, hence compact. By the finite intersection property, there exists a $\bar{\nu}$ such that $K_{\nu}=\emptyset$ for $\nu\geq\bar{\nu}$.