Let $B=\left\{ \left(x,y\right)\in\mathbb{R}^{2}:x^{2}+y^{2}<1\right\} $
be the unit ball in $\mathbb{R}^{2}.$


Can we construct a subharmonic
function $f:B\rightarrow\left[-\infty,0\right]$ such that
$$
0<\int\_{\widetilde{B}}\left(1-x^{2}-y^{2}\right)^{-2}dV<\infty,
$$
where $\widetilde{B}=\left\{ \left(x,y\right)\in B:-1<f\left(x,y\right)\right\} $? Here $dV$ is the standard Lebesgue measure on $\mathbb{R}^{2}$.


I thought that in order to answer this we need to control (understand) the growth
on sublevel sets of a subharmonic function.


The question was already asked on mathstackexchange and received no answers.



A subharmonic function with a growth property
The answer is "yes". Let $E$ be some Jordan region in the unit disk on which
$$\int\_E(1-x^2-y^2)^{-2}dxdy<\infty,$$
and such that $E$ contains $[0,1)$, and the closure of $E$ is contained in the open unit disk, except the point $1$. Let $\phi$ be a conformal map of $E$ onto the right half-plane, such that $\phi(1)=\infty$.


Let $u=\Re \phi$. Then $u$ is positive and harmonic n $E$, zero on $\partial E\backslash\{1\}$. Extend $u$ to the whole unit disk by setting is equal to $0$
outside $E$. Then take $f=u-2.$


As you see from this example, the weight $(1-|z|^2)^{-2}$ is not very relevant. You can replace it by anything you like.


