From continuous to bounded Borel functions I know that we can extend the functional calculus of bounded self-adjoint operators to bounded Borel functions. I want to do the same for unbounded self-adjouint operators.
Therefore assume that $T$ is a unbounded self-adjoint operator on a Hilbert space $H$. Assume that one know how to obtain $f(T)$ if $f$ is continuous on the spectrum of $T$. I want to extend this to Borel functions on the spectrum. Can this be done?
 A: A spectral measure $E$ comes out of the classical version of the Spectrum Theorem. The only difference between $E$ for a bounded selfadjoint operator and an unbounded one is that the support of $E$ for a bounded selfadjoint operator is bounded. For any Borel function $f$, you can define
$$
                f(T)x = \int_{-\infty}^{\infty} f(\lambda)dE(\lambda)x
$$
on the domain $\mathcal{D}(f(T))$ consisting of all $x\in X$ for which
$$
           \|f(T)x\|^{2} = \int_{-\infty}^{\infty} |f(\lambda)|^{2}d\|E(\lambda)x\|^{2} < \infty.
$$
That turns out to define a closed, densely-defined normal operator whose adjoint $f^{\star}(T)=f(T)^{\star}$ has the same domain at $f(T)$. If $f$ is bounded on $\sigma(T)$, then $f(T) \in \mathcal{L}(H)$.
If $f$ is Borel measurable and $g$ is a bounded Borel function, then
$$
   g(T) : \mathcal{D}(f(T))\subseteq \mathcal{D}(f(T)),
$$
and $g(T)f(T)=f(T)g(T)=(fg)(T)$ on $\mathcal{D}(f(T))$; even though the domains of $(fg)(T)$ and $f(T)$ may be different, clearly $\mathcal{D}(f(T))\subseteq\mathcal{D}((fg)(T))$ follows from the above integral characterization of the domain.
Of course, if $f$ and $g$ are bounded Borel functions, then $f(T)$, $g(T)$ and $(fg)(T)$ are defined everywhere and bounded; furthermore, $f(T)g(T)=g(T)f(T)=(fg)(T)$.
Construction of Extended Calculus: A fundamental identity for the functional calculus is the resolvent
$$
      (T-\mu I)^{-1} = \int_{-\infty}^{\infty}\frac{1}{\lambda-\mu}dE(\lambda),\;\;\; \mu \in \rho(T).
$$
This solidly (i.e., uniquely) connects $T$ with the spectral measure $E$. These bounded continuous functions $f_{\mu}(t)=(t-\mu)^{-1}$ applied to $T$ are really all that you need to construct the full Borel functional calculus using strong limits. These bounded continuous functions of $T$ give you the resolvent. First, you can construct the spectral measure on intervals:
$$
    \frac{1}{2}\{ E[a,b]+E(a,b)\}x=\lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{a}^{b}\{f_{u+i\epsilon}(T)-f_{u-i\epsilon}(T)\}x\,du.
$$
This is Stone's Formula. This gives you $E[a,b]x$ and $E(a,b)x$ separately as vector limits from outside and inside the interval, respectively. Alternatively, you can form compactly supported continuous functions of $T$ which approach characeristic functions of intervals.
Reference For Stone's Formula: Proving Stone's Formula for Constructively obtaining the Spectral Measure for $A=A^\star$
A: If $T$ actually is a multiplication operator on $L^2(M,\mu)$, say there is a real-valued function $m$ on $M$ so that
$$
T(h) = mh
$$
for $h \in L^2$, then the functional calculus tells you that $f(T)$ is the multiplication by $f(m)$.  The general self-adjoint operator is similar to such a multiplication operator.
