In the following, the section numbers I mention are from Rudin's Functional Analysis text, Chapter 12.
Let $T$ be a bounded normal operator in the (not necessarily separable) Hilbert space $\mathfrak{H}$. Let $E$ be the resolution of the identity for $T$ on the Borel subsets of the spectrum $\sigma(T)$. Let $f$ be a bounded measurable complex function on $\sigma(T)$. Suppose $\mathfrak{M}$ is a closed subspace of $\mathfrak{H}$ which is reducing for $T$. That is, $T\mathfrak{M}\subseteq\mathfrak{M}$ and $T^*\mathfrak{M}\subseteq\mathfrak{M}$, or equivalently, $T\mathfrak{M}\subseteq\mathfrak{M}$ and $T\mathfrak{M}^\perp\subseteq\mathfrak{M}^\perp$. I would like to show that $f(T)\mathfrak{M}\subseteq\mathfrak{M}$, that is, that $\mathfrak{M}$ is an invariant subspace with respect to $f(T)$.
If I could show that there is a borel subset $\omega\subseteq\sigma(T)$ such that the range $\mathscr{R}(E(\omega))=\mathfrak{M}$ (that is, $E(\omega)$ is a projection on $\mathfrak{M}$), then I would be done, because by the spectral theorem (12.23), every $E(\omega')$ commutes with $T$, and by the properties of the resolution of the identity (12.17(c)), $E(\omega)$ commutes with every $E(\omega')$, so by 12.21, $E(\omega)$ commutes with $f(T)$. I could then write for $x\in\mathfrak{M}=\mathscr{R}(E(\omega))$, say $x=E(\omega)y$, $$f(T)x=f(T)E(\omega)y=E(\omega)f(T)y\in\mathscr{R}(E(\omega))=\mathfrak{M}.$$
So, is it true that such an $\omega$ must exist and how do I show it? If not, is it still true that $\mathfrak{M}$ is $f(T)$-invariant, and how would I show it?
It is not true in general that the orthogonal projection $P$ onto a reducing subspace $M$ is of the form $P = 1_\omega (T)$ for some subset $\omega$. Indeed, if $T$ is the identity operator, then every subspace is reducing, but the only spectral projections are the trivial projections.
Nevertheless, you have $T P= PT$ (why?), so that Theorem 12.24 in Rudin's functional analysis shows $f(T)P = Pf(T)$, which implies that $f(T)M \subset M$.