A formula for the potential part in Riesz decomposition of simple subharmonic functions? Consider a compact subset $E\subset\mathbb{C}$, holomorphic functions $f_j:V\to \mathbb{C}$, $1\leq j\leq k$, defined in a neighbourhood $V$ of $E$, and set $u:V\to\mathbb{R}\cup\{-\infty\}$, $u(z)=\max_{1\leq j\leq k} \log|f_j(z)|$. 
Is there an explicit formula for the potential part 
$$ \int \log |z-w|\,d\mu(w) $$
of the Riesz decomposition of the subharmonic function $u(z)$?
If not, is at least the support of $\mu$ contained in the set
$\Gamma=\Gamma_0\cup \cup_{i<j} \Gamma_{i,j}$, where
$\Gamma_0=\{z\in E: f_j(z)=0,\,\forall j\}$ and $\Gamma_{i,j}=\{z\in E: |f_i(z)|=|f_j(z)|=e^{u(z)}\}$?
And can we relate the density of $\mu$ with respect to Lebesgue on $\Gamma_{i,j}$ with $|f_i'(z)-f_j'(z)|$?
 A: The "formula" is this. 
 Let $j(z)$ be that $j$ for which $|f_j(z)|$ is maximal among all $|f_k(z)|$. (There can
be several $j$ with this property, choose one, or just remove from your set several
functions, to make this $j(z)$ unique). Then $j(z)$ is locally constant on some
open set, the boundary of this open set consists of analytic curves which can meet at
isolated points. If $m$ and $k$
are two values of $j(z)$ on the right and the left of such a curve, 
which means that $|f_n(z)|=|f_m(z)|$ on the curve, while $|f_m(z)|$ is the largest one
on the left, and $|f_n|$ on the right, then the density of the Riesz measure on
this curve (with respect to the arclength) is equal to
$$\frac{1}{2\pi}|(d/dn)\log|f_n(z)|-(d/dn)\log|f_m(z)||$$ where $d/dn$ is the derivative along the 
normal to the curve. The direction of the normal in this formula can be chosen arbitrarily, but the same in both summands. In other words, this is the jump
of the directional derivative of $\max_k\log|f_k|$, in the direction normal
to the curve on which this $\max$ is not smooth.
All this is easy to prove, and people frequently use this, see, for example, 
 Proc. Amer. Math. Soc., 140, (2012) 1397-1402, formula (7), but there is no convenient reference
where this fact is stated. People either prove this, in the situation where they need it,
or say that this is evident, or well-known. I've seen this formula as an exercise in some
Russian book, but I do not remember which book. (There was no proof anyway).
Of course, the gradient of $\log|f|$ is $|f'/f|$ but here we deal not with the gradient
but with the derivative in the normal direction of the curve. But of course you can produce from
the above formula a formula which is more convenient for your purpose.
