This question is about a proof from Banach algebra Techniques in Operator theory by Ronald Douglas

The theorem is this: If $\mu$ is a positive regular Borel measure on the unit circle, then a closed subspace $\mathcal{M}$ of $L^2(\mu)$ satisfies $\chi_1 \mathcal{M}=\mathcal{M}$ if and only if there exist a Borel subset $E$ of the unit circle such that $\mathcal{M}=L_E^2(\mu)=\{f\in L^2(\mu):f(e^{it})=0 \text{ for } e^{it}\notin E\}$ where $\chi_1(e^{it})=e^{it}$.

In the proof, they show that the projection onto $\mathcal{M}$ is of the form $M_\phi$ for some $\phi\in L^\infty(\mu)$. Then they say that the result then follows. I am not seeing why the result then follows.

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$M_\phi$ is a projection, so $M_\phi = M_\phi^2$. Since $M_f M_g = M_{fg}$ it follows that $M_\phi^2 = M_{\phi^2}$. Therefore $M_\phi = M_{\phi^2}$, so that $\phi^2 = \phi$ a.e. and hence $\phi(x) \in \{0,1\}$ for a.e. $x$. This means that $\phi$ is a.e. equal to a characteristic function, say $\phi = \chi_E$.
Since for $f \in \mathcal{M}$ we have $f = M_\phi f = \chi_E f$, it follows that $f(x) = \chi_E(x) f(x) = 0$ for $x = e^{it} \notin E$. On the other hand, $f(e^{it}) = 0$ for $e^{it} \notin E$ implies that $\chi_E f = f$, so $\mathcal{M} = \{f\in L^2(\mu):f(e^{it})=0 \text{ for } e^{it}\notin E\}$.