Question on the spectral theorem for unitary operators. I am reading spectral theorem for unitary operators $U$ on $L^2(\mu)$ where $(X,\mathcal A,\mu)$ is some measure space. There I found that there always exists a unique projection valued measure $P$ on the Borel-$\sigma$-algebra of subsets of $[-\pi,\pi]$ such that $$U = \int_{-\pi}^{\pi} e^{it}\ dP(t).$$
Now is there any way to find out $P(\{0\})\ $? Our instructor said that it is nothing but the projection onto the space $\{f \ |\ Uf = f\}.$ But I failed to understand that. Could anybody please provide some hint?
Thanks a bunch.
 A: I will say some words, hoping that an answer can be extracted...
To have a fixed framework, i will work using the notations from Encyclopedia of Math :: wiki :: Spectral decomposition . Related, but rather confusing is the wiki link Borel functional calculus .
My choice for the letter for the spectral measure is $E$. And to have a better structural insight, it feels better to write the spectral integral over the spectrum $\sigma(U)$ of the given unitary operator $U$. This makes things simpler (to write and understand). There is no point to parametrize the unit circle $S^1$ as $S^1=\{\ e^{it}\ :\ t\in I\ \}$ for some interval $I$ of length $2\pi$, then move $E$ on $\sigma(U)\subseteq S^1$ to some $P$ on (some subset of... but then extend by zero... so on) $I$, so that $1\in\sigma(U)$ goes to $0\in I$. (And by chance, $0$ was taken to be the middle of $I$. People may get even more nervous when taking $I=[0,2\pi)$... although the measure theory makes no distinction, but psychologically...)
The confusion / question in the OP addresses exactly this translation of $z=1$ into $t=0$ via $z=e^{it}$, so that one may no longer recognize the eigenvalue $z=1$ of $U$ for the corresponding projection. The simplest way to now or a posteriori understand, or rather fix the claim of the instructor is as follows. The spectral decomposition is useful for one main thing, it is compatible with (step by step polynomial, analytic, smooth, continuous, and finally borelian) functional calculus in the following sense. I will profit now from the two ways for the choice of the letter for the spectral projection. (The $E$ versus $P$ discussion...)
Below, $E(A)$ is the spectral projector corresponding to a measurable $A\subset \sigma(U)$.
And $P(J)$ is the same projector $E(A)$, where $J\subseteq I$ is obtained by the above translation from $A\subseteq \sigma(U)$.

Given

*

*$E$ the spectral measure for the given (bounded) operator, here $U$,

*and a function $f$, on the spectrum $\sigma(U)$ of this operator - in the one or the other class - depending on the point where the lecturer claimed the property, so that the functional calculus $f(U)$ is defined...

*and the very special function $f=1_A$, the characteristic function of  $A=\{1\}$, which is a Borel measurable set in $\sigma(U)$
in parallel:
$$
\begin{aligned}
U &= \int_{\sigma(U)}z\; dE(z)\ ,\\
f(U) &= \int_{\sigma(U)}f(z)\; dE(z)\ ,\\
\color{blue}{(*)}\qquad 1_A(U) &= \int_{\sigma(U)}1_A(z)\; dE(z)=\int_A dE=E(A)\ .
\end{aligned}
$$
In the last line we have the spectral projection onto the $A$-eigenspace.

Now the question in the OP has something to do only with the translation of $A=\{1\}\subset\sigma(U)$ into $J=\{0\}\subset I$, for short
$$
P(\{0\})=P(J)=E(A)=E(\{1\})\ .
$$
Yes, we do not need the functional calculus above for the given "disguised" eigenvalue  $1\in S^1$, as the translated $0\in[-\pi,\pi)$, but this was the simplest way for me to write down something that binds some structure. (Instead of saying this is the definition...)
From the last relation in $\color{blue}{(*)}$ we need in the discussion only that right most $E(A)$. But the left most $1_A(U)$ is also useful to fix the ideas, since recalling the spectrum change due to functional calculus...
$$
\begin{aligned}
\sigma(f(U)) &=f(\sigma(U))\ ,&&\text{ so in our case}\\
\sigma(1_A(U)) &= 1_A(\sigma(U))\ ,&&\text{ which is $\{1\}$ or empty set.}
\end{aligned}
$$
(Again, the last line is a connection to the $1$-eigenspace.)
