Dense subspace of square integrable functions on the complex disc

Question

Denote by $L^{2}(D,(1-|z|^{2})^{a-1}|z|^{2b-2}dx dy)$ the set of square integrable functions on the complex disc $D= \lbrace z \in C, \; |z| <1 \rbrace$ with respect to the measure $(1-|z|^{2})^{a-1}|z|^{2b-2}dx dy$, where $z=x+iy$ and $a > 0$ and $b > 0$. Now let $f \in L^{2}(D,(1-|z|^{2})^{a-1}|z|^{2b-2}dx dy)$, i.e $$\int_{D}|f(z)|^{2}(1-|z|^{2})^{a-1}|z|^{2b-2}dx dy < + \infty$$ My goal is to find out a dense subspace $\Xi$ of $L^{2}(D,(1-|z|^{2})^{a-1}|z|^{2b-2}dx dy)$, such that the element $g \in \Xi$, satisfies the convergence the following serie: $$\sum_{n=0}^{+ \infty} n^{2a+r} \int_{D}|g(z)|^{2}|z|^{2n+2b-2}(1-|z|^{2})^{a-1} dxdy < + \infty \mbox{ where } r > 0$$ I have tried many times to solve that problem but it didn't got results. I have took the following set: $$\Xi'=\lbrace g \; entire function/ \sup_{z \in D}|g(z)|(1-|z|^{2})^{-a} < + \infty \rbrace$$ I have found out on $\Xi'$ the convergence of the serie above for $a > r+1$, but i didn't found that $\Xi'$ is a dense subspace of $L^{2}(D,|z|^{2b-2}(1-|z|^{2})^{a-1}dxdy)$ I want other informations than the compact set of D where on outside of D g=0. Because also on that case i have found that the series is convergent. I want a construction of the set $\Xi$ just like to majorate g with some function which verifies the density without using the compact support on the disc