Show that eigen-vectors belong in range of projection valued measure. Let $A$ be some self-adjoint bounded operator in Hilbert space, with associated projection valued measure $P$ such that
$A=\int_\mathbb R \lambda dP(\lambda)$.
I want to show that if $f$ is an eigenvector of $A$ with eigenvalue $\lambda$ (e.g. $Af-\lambda f=0$), then $f$ belongs in the range of $P(\lambda)$.
How to show this? Note that:
$$AP(\lambda)f=\lambda P(\lambda)f$$
I have
$$(P(\lambda)f,f)=\frac{1}{\lambda}(AP(\lambda)f,f)=\frac{1}{\lambda} \int_\mathbb R \lambda d\mu_{P(\lambda)f,f}(\lambda),$$
where $$\mu_{f,g}(A)=(P(A)f,g).$$
How to go forward here?
 A: If $Ax=\mu x$, then
\begin{align}
       0=\|Ax-\mu x\|^2= \left\|\int \lambda dP(\lambda)x-\int \mu dP(\lambda)x\right\|^2 \\
  = \left\|\int(\lambda-\mu)dP(\lambda)x\right\|^2 \\
  = \int |\lambda-\mu|^2d\|P(\lambda)x\|^2
\end{align}
This implies that $d\|P(\lambda)x\|^2$ is a measure that is concentrated at $\mu$. Therefore $x=P(\mathbb{R})x=P(\{\mu\})x$.
A: Using B.C. Hall's Quantum Theory for Mathematicians, I have found a chain of exercises that will lead you to a proof.
(1) Let $A$ a self-adjoint bounded operator on $\mathcal{H}$ and let $V \subseteq \mathcal{H}$ an $A$-invariant closed vector subspace, i.e. $A V \subseteq V$. Denote the spectrum of an operator $B$ by $\sigma(B)$. Then $\sigma(A|_V) \subseteq \sigma(A)$.
Hint: You can use the following theorem (which you could prove if you want or take): $\lambda \in \sigma(A)$ iff there exists a sequence $\psi_n$ of nonzero vectors in $\mathcal{H}$ s.t. $$\lim_{n\to\infty} \frac{\|A\psi_n - \lambda \psi_n \|}{\|\psi_n\|} = 0.$$
(2) If $f$ is a bounded measurable function on $\sigma(A)$, then $f(A|_V) = f(A)|_V$ where we use the functional calculus for $A|_V$ and $A$ resp.
(3) Suppose $\psi \in \mathcal{H}$ and $A \psi = \lambda \psi$ for some $\lambda \in \mathbb{C}$. Then for any bounded measurable function on $\sigma(A)$ $f$, $f(A)\psi = f(\lambda)\psi$. (Use (2))
From (3) your answer follows easily.
Edit: Here are the proofs.
Proof of (1): Let $\lambda \in \sigma(A|_V)$. Then there is some sequence $\psi_n$ of nonzero vectors in $V$ s.t.\ $\frac{\|A|_V\psi_n - \lambda \psi_n\|}{\|\psi_n\|} \to 0$. But then the sequence is also in $\mathcal{H}$, so $\lambda \in \sigma(A)$.
Proof of (2): Note that the two sides clearly agree on polynomials. By taking uniform limits we get that the two sides agree on continuous functions. Denote $\mathcal{F} := \{f \text{ bounded, measurable functions on } \sigma(A) \mid f(A|_V) = f(A)|_V\}$. Note that if $f, g \in \mathcal{F}$ and $c \in \mathbb{C}$, then $(f+cg)(A|_V) = f(A|_V) + cg(A|_V) = f(A)|_V + cg(A)|_V = (f+cg)(A)|_V$. So $\mathcal{F}$ is a complex vector space. By a standard result in measure theory, if a set $\mathcal{F}$ of bounded measurable functions on a compact metric space is (1) a complex metric space, (2) contains the continuous $\mathbb{R}$-valued functions, and (3) is closed under pointwise limits of uniformly bounded (i.e. dominated by a constant in the since of dominated convergence) sequences. Then $\mathcal{F}$ is the set of bounded measurable functions. (E.g. B.C. Hall, Quantum Theory for Mathematicians, Ex. 8.3). Thus it is sufficient to show that $\mathcal{F}$ is closed under uniformly bounded pointwise limits. But this follows directly: let $f_n$ a uniformly bounded sequence s.t. $f_n \to f$ pointwise. Then $f(A|_V)$ is the unique operator s.t. 
$$\forall \psi \in V: ( \int f d\mu^{A|_V} \psi, \psi) = \int f d\mu^{A|_V}_\psi$$
where $\mu^{A|_V}_\psi(E) = (\mu^{A|_V}(E) \psi,\psi)$. Then since $\mu^{A|_V}_\psi$ is finite, constants are integrable, so dominated convergence gives that 
$$(f(A|_V)\psi, \psi) = \int f d\mu^{A|_V}_\psi = \lim \int f_n d\mu^{A|_V}_\psi = \lim (\int f_n d\mu^{A|_V} \psi, \psi) = \lim (f_n(A|_V)\psi,\psi)$$
$$= \lim (f_n(A)|_V \psi, \psi) = \lim (f_n(A) \psi, \psi) = \lim \int f_n d\mu^A_\psi = \int f d\mu^A_\psi = (f(A)|_V \psi, \psi).$$
Proof of (3): Let $V:= \mathbb{C}\psi$. Note that $V$ is a closed vector subspace and that $A(c \psi) = c A \psi = \lambda c \psi \in V$, so $V$ is $A$ invariant and also note that $A|_V = \lambda I$, where $I$ is the identity operator. So we can apply (2), giving $f(A) \psi = f(A)|_V \psi = f(A|_V)\psi = f(\lambda I) \psi$. Now we just need to show that $f(\lambda I) = f(\lambda)I$ for any bounded measurable $f$. We will use the same proof technique as in the proof of (2). This clearly holds for polynomials and hence for continuous functions by taking uniform limits. Direct computation verifies the set of solutions form a complex vector space. So we just need to show closure under pointwise limits of uniformly bounded sequence. Let $f_n$ some such sequence. Then:
$$(f(\lambda I) \psi , \psi) = \int f d\mu^{\lambda I}_\psi = \lim \int f_n d\mu^{\lambda I}_\psi = \lim (f_n(\lambda I) \psi, \psi) = \lim (f_n(\lambda)I \psi, \psi)$$
$$= (\psi, \psi) \lim f_n(\lambda) = (\psi, \psi) f(\lambda) = (f(\lambda) I \psi, \psi).$$
So uniqueness gives the desired closure property.
(3) $\implies$ eigenvectors contained in spectral subspace: Let $\psi$ an eigenvector with eigenvalue $\lambda$. By (3), $1_{\{\lambda\}}(A)\psi = 1_{\{\lambda\}}(\lambda) \psi = \psi$. then since $1_{\{\lambda\}}(A)$ is orthogonal projection on the spectral subspace corresponding to $\{\lambda\}$, $V_{\{\lambda\}}$, $1_{\{\lambda\}}(A) \psi = \psi \implies \psi \in V_{\{\lambda\}}$.
