Prove that $A$ has Lebesgue measure $0$. Suppose $G$ is a connected open set of $\mathbb{C}^n$. Prove that: 
(1). If $f \in$ PSH(G) and $f \not \equiv \infty$ then
$A=\{z \in G: f(z)=-\infty\}$ has Lebesgue measure $0$.
(2). If $f \in \mathcal{H}(G,F)=\{f: G \to F,~ \text{f is holomorphic function}\}$ and $f \not \equiv 0$ (not heterogeneous $0$) then $A=\{z \in G: f(z)=-\infty\}$ has Lebesgue measure $0$.
Any help (or hint or another solution) would be greatly appreciated. Thanks.
 A: For (2), you surely mean "$f(z) = 0$", not "$f(z) = -\infty$".
(1) This is true even if we just assume that $f$ is subharmonic. One way to see it is to argue like this. 
Theorem. If $G$ is connected and $u$ is subharmonic on $G$, then $u \in L^1_{loc}$ unless $u \equiv -\infty$.
Proof. Every subharmonic function is locally bounded above. Let $W$ be the set of points $x$ in $G$ such that $u$ is integrable in a neighbourhood of $x$. Then $W$ is open by definition, and $u > -\infty$ a.e. on $W$. Take $x \in \bar W$ and choose $a \in W$ such that $|a-x| < r = \frac12 \operatorname{dist}(x,\partial G)$ and $u(a) > -\infty$. Then, by the mean-value property for subharmonic functions,
$$
-\infty < u(a) \le \frac{1}{\operatorname{Vol}(B_r(a))} \int_{B_r(a)} u(y)\,dy
$$
which shows that $u$ is integrable in a neighboorhood of $x$, i.e. that $x \in W$. Hence $W$ is both open and closed, so $W = G$ or $W = \emptyset$.
As a consequence, $u^{-1}(-\infty)$ is either the whole of $G$ or has zero measure. (Sets that are $-\infty$-set of [pluri]subharmonic functions are called [pluri]polar, and are in many ways even smaller than of zero measure.)
(2) Follows directly from (1), taking $u = \log|f|$.
