Invariant subspaces are reducing subspaces in $L^2(\mu)$; where $\mu$ is a singular measure w.r.t Lebesgue measure

Question

I have already posted this question on math.stackexchange but didn't get any answer. I hope this is the right place to ask this question.

Recently I was reading a book "Operator Function and system" written by Nikolski. There I found this statement.

Let $\mu$ be a finite positive measure on the unit circle in complex plane which is also singular with respect to Lebesgue measure on the unit circle. Now consider the Hilbert space $L^2(\mu)$ and consider the operator $M \colon L^2(\mu) \to L^2(\mu)$ defined by $(Mf)(z) = z f(z) $ for $z$ in the unit circle. Then it turns out that every invariant subspace for $M$ is also a reducing subspace for $M$.

Is there any simple, elementary way to prove the result?

If the measure $\mu$ is supported on finitely many point, then I have some idea how to prove the result. But in general, I have no idea.

Any answer or reference or hint will be very helpful.

Answer 1

I guess it depends on what counts as simple or elementary, but I think the standard argument goes like this: using the F. and M. Riesz Theorem, one can show that since $\mu$ is singular to Lebesgue measure, there is a sequence of analytic polynomials $p_n$ such that $p_n(z)\to \overline{z}$ weak-* in $L^\infty(\mu)$. It follows that the operators $p_n(M)$ converge to $M^*$ in the weak operator topology on $B(L^2(\mu))$, and thus any subspace invariant for $M$ will be invariant for $M^*$.

EDIT Actually, it occurs to me that there is another proof using Szeg\"{o}'s theorem which allows one to prove the sharp result: every $M$-invariant subspace is reducing if and only if $\int \log\left(\frac{d\mu}{dm}\right) dm=-\infty$.