Spectral Decomposition: $A\psi = \lambda \psi \implies f(A)\psi = f(\lambda)\psi$ I'm reading Simon & Reed's Functional Analysis, and attempting to write out the proof of the functional calculus form of the Spectral Theorem.
See this link for their construction of $f(A)$ where $f \in \mathbb{B}(\mathbb{R})$ a bounded Borel function on $\mathbb{R}$:
Continuity of the functional calculus form of the Spectral Theorem
Let $A \in L(H)$ be self-adjoint.
My question is this: 
How do I show that $A\psi = \lambda \psi \implies f(A)\psi = f(\lambda)\psi$?
This argument is easy when proving the same statement in the continuous calculus because when dealing with a continuous $f$ it may be approximated by a sequence $p_n$ that converges to it uniformly, and hence point-wise too.
To my knowledge, a bounded Borel function, $f$, has the property that $\exists \{f_n\} \subset C(\sigma(A))$ s.t $$\lim_n \int_{\sigma(A)}|f_n-f|du_\psi = 0.$$ Also I found that we have point-wise convergence a.e$[u_\psi]$.
Using the definition of $f(A)$ I can show that $$\exists ~~\lim_n f_n(\lambda) \in \mathbb{C}$$ but not that it necessarily equals $f(\lambda)$.
I wanted this to complete the attempt of:
$(\psi, f(A)\psi) = \int_{\sigma(A)} fdu_\psi = \lim_n \int_{\sigma(A)}f_ndu_\psi = \lim_n(\psi, f_n(A)\psi) = \lim_n(\psi, f_n(\lambda)\psi)$
 A: Suppose $\mu$ is a finite Borel measure on $\mathbb{R}$, and consider
$$
          F(\lambda) = \int_{\mathbb{R}}\frac{1}{t-\lambda}d\mu(t),\;\; \lambda\in\mathbb{C}\setminus\mathbb{R}.
$$
For a fixed $t_0$ and $\epsilon > 0$,
\begin{align}
      F(t_0+i\epsilon)-F(t_0-i\epsilon) &=\int_{\mathbb{R}}\frac{1}{(t-t_0)-i\epsilon}-\frac{1}{(t-t_0)+i\epsilon}d\mu(t) \\
    &=\int_{\mathbb{R}}\frac{2i\epsilon}{(t-t_0)^2+\epsilon^2}d\mu(t).
\end{align}
Therefore, by the bounded convergence theorem,
$$
    \lim_{\epsilon\downarrow 0}i\epsilon\{F(t_0+i\epsilon)-F(t_0-i\epsilon)\}
   = \lim_{\epsilon\downarrow 0}\int_{\mathbb{R}}\frac{-2\epsilon^2}{(t-t_0)^2+\epsilon^2}d\mu(t) = -2\mu\{t_0\}.
$$
Now you can combine this with the functional calculus to get what you want. To be a bit more explicit, suppose $A\psi = t_0\psi$. Then, for $\lambda\notin\mathbb{R}$,
$$
     (A-\lambda I)\psi = (t_0-\lambda)\psi \\
      \psi = (A-\lambda I)^{-1}(t_0-\lambda)\psi \\
     (A-\lambda I)^{-1}\psi = \frac{1}{t_0-\lambda}\psi.
$$
See if you can take it from here.
A: $\newcommand{\d}{\operatorname{d}}$
In fact, maybe a little surprise,$$\mu(\{\lambda\})=\mu\{\sigma(A)\}.(\mu:=\mu_\psi)$$
Proof. Let $$p_n(z)=( 1-\lvert\frac{z-\lambda}{R}\rvert^2)^n,$$ where $R>\lVert A\rVert$,
then  $p_n\to \chi_{\{\lambda\}}$ pointwise, thus $$\int p_n\d\mu \to \int\chi_{\{\lambda\}}\d\mu =\mu\{\lambda\}$$ by the Dominated Convergence Theorem. Since each $p_n$ can be represented as $p_n(z)=1+p(z)( z-\lambda) $ for some polynomial $p$ and $(z-\lambda)(A) \psi=A\psi-\lambda\psi=0$,
$$\int p_n(z)\d\mu=\int 1\d\mu+\int p(z)(z-\lambda)\d\mu=\int 1\d\mu+(\psi,p(A)(A-\lambda I)\psi)=\mu(\sigma(A)).$$
Hence
$\mu(\{\lambda\})=\mu\{\sigma(A)\}.$
