Suppose $(\Omega, \mu)$ is measurable space with $\mu(\Omega)<\infty$, here $\Omega$ is a compact subset of $\mathbb{C}$. $f,g\in L^2(\Omega)$ with $\mu(\{x:f(x)=0\})=0$. I want to find polynomial $p$ s.t. $pf$ can approximate $g$ in $L^2$ norm. Concretely, for any $\varepsilon>0$, how to construct polynomial $p_{\varepsilon}$ s.t. $$ \|p_{\varepsilon}f-g\|_{L^2(\Omega)}<\varepsilon. $$ I tried to decompose space $\Omega$ as $\Omega=\cup_{n\geq1}E_n$, where $$ E_n=\{x:\frac1n<|f(x)|<n\}. $$ so $g/f\in L^2(E_n)$. Take polynomial $p_k$ s.t. $\|p_k-g/f\|_{L^2(E_n)}^2<\eta_{n, k}$, which is a small number ready to defined. And we have $$ \int_{\Omega}|p_kf-g|^2=\int_{E_n}|p_k-g/f|^2|f|^2+\int_{\Omega\setminus E_n}|p_kf-g|^2\\ \leq n^2\eta_{n,k}+\int_{\Omega\setminus E_n}|p_kf-g|^2. $$ Since $p_kf-g\in L^2(\Omega)$, for any $\varepsilon>0$ we can find $\delta_{k}$ s.t. for any measurable set $E$ and $\mu(E)<\delta_{k}$ we have $$ \int_E|p_kf-g|^2<\varepsilon. $$ If $\mu(\Omega\setminus E_n)<\delta_k$, we have $$ \int_{\Omega\setminus E_n}|p_kf-g|^2<\varepsilon. $$ this need that for $k$ fixed, $n$ must be large enough. but in order that $n^2\eta_{n,k}$ small enough, $k$ must be large enough for any fixed $n$.
I have no idea how to deal with it. But the original problem is
If $\mu$ is compactly supported measure on $\mathbb{C}$ and $f\in L^2(\mu)$, then $f$ is star-cyclic vector for $N_{\mu}$ if and only if $\mu(\{x:f(x)=0\})=0$.
Here $N_{\mu}$ is a normal operator defined on $L^2(\mu)$ as $(N_{\mu}f)z=zf(z)$ and $f$ is star-cyclic vector for $N_{\mu}$ means that $$ L^2(\mu)=\overline{\{p(N_{\mu},N_{\mu}^*)f:p(x,y)\text{ is polynomial on $\mathbb{C}$}\}}. $$ This problem is from A course in functional analysis written by Conway and it is Exercise 1 in Section 3 of Chapter 9.
We just need to show that $$ \overline{\{p(z)f(z)\ |\ p\text{ is polynomial}\}}=L^2(\Omega). $$ Since $L^2(\Omega)$ is a Hilbert space, so we can consider $g\in\overline{\{p(z)f(z)\ |\ p\text{ is polynomial}\}}^{\perp}$, that means $$ \int_{\Omega}p(z)f(z)\overline{g(z)}\mathrm{d}\mu=0.\quad\forall p\text{ polynomial}. $$ Since the subspace consisting of polynomials is dense in $L^2(\Omega)$, we get $f\bar{g}=0$, $\mu$-a.e. But $f\neq0$ $\mu$-a.e., impose $g=0$ $\mu$-a.e.