Let $\mu$ be a finite complex Borel measure on unit circle $\mathbb{T}$ , and let $E$ be a shift invariant subspace of $L^2(\mu)$ . Show that $zE\subseteq E\subseteq L^2(\mu)$ implies $zE=E$ iff the normalized Lebesgue measure on $\mathbb{T}$ , say , $m$ is not absolutely continuous w.r.t. $\mu$ .
I have used Lebesgue decomposition of $\mu$ as follows $$d\mu=wdm+d\mu_s$$ where $\mu_s$ is the singular part of $\mu$ and $wdm$ with corresponding weight $w$ is the absolute part of $\mu$ . If $zE=E$ then we have $E=\chi_{\sigma}L^2(\mu)$ on a measurable set $\sigma\subset\mathbb{T}$ by Wiener's theorem . Again by Beurling decomposition theorem $$E=\theta H^2\oplus\chi_{\sigma}L^2(\mu_s)$$ for some measurable function $\theta$ such that $|\theta|^2w=1$ where $H^2$ is the Hardy space with $m(\sigma)=0$ . So it is clear that absolute part of $\mu$ is zero and therefore $\mu=\mu_s$ is singular w.r.t. $m$ , hence $m$ is not absolutely continuous w.r.t. $\mu$ .
Is this proof correct ? And can the converse part be proved similarly ? I am a newbie in Hardy space theory and trying to learn by myself . So any kind of help or sources will be appreciated .