On the definition of spectral integrals in Conways "A course in functional analysis" I am trying to make sense of the spectral integral defined in Conways "A course in functional analysis" but I cant really settle on how to think about it. He does the following,
He just proved that $<E(\Delta)g,h>=E(\Delta)_{g,h}$ is a measure of bounded variation on the spectra or any $X$ with a sigma algebra if one prefers.
Now, the left hand side of the inequlities at the end seems to contain some kind of Riemann sum(which I cant define a integral for w.r.t a arbitrary measure) rather then simple functions as one would have in the case of a Lebgue integral.
So what integral concept is the sum< $\sum \phi(x_{k})E(\Delta_{k})g,h>$ associated to? Is he inventing his own concept here?
Edit!
It might be possible to consider $\phi(x_{k})$ a step function and do Lebague all the way, I am not sure tho.
 A: The sum you wrote approximates an actual integral:
\begin{align}
        \langle \sum_k \phi(\lambda_k)E(\Delta\lambda_k)x,y\rangle 
       & = \sum_k\phi(\lambda_k)\langle E(\Delta\lambda_k)x,y\rangle \\
       & \approx \int \phi(\lambda)d_{\lambda}\langle E(\lambda)x,y\rangle.
\end{align}
And the complex integral on the far right is sesquilinear form that can be used to define a unique operator $``\int \phi(\lambda)dE(\lambda)''$ as the unique operator satisfying
$$
     \int \phi(\lambda)d_{\lambda}\langle E(\lambda)x,y\rangle = \left\langle \int \phi(\lambda)dE(\lambda)x,y\right\rangle,\;\;\; x,y\in H.
$$
Such an operator exists because the left side is a bounded sesquilinear form on the Hilbert space. If $E$ is a spectral measure associated with a selfadjoint operator $A$, then $A=\int \lambda dE(\lambda)$, and $A^n = \int \lambda^n dE(\lambda)$ for $n=1,2,3,\cdots$. So the spectral integral represents a function of the operator $A$. For example $e^{A} = \int e^{\lambda}dE(\lambda)$. The rough idea is that $dE(\lambda)$ is the projection onto the part of the space associated with spectral component $\lambda$. That is, $AdE(\lambda)=\lambda dE(\lambda)$. For a selfadjoint with discrete spectrum, such as a matrix on a finite-dimensional space, the measure $E$ has discrete support that is equal to the set of eigenvalues of $A$, and $E(\{\lambda\})$ is the projection onto the eigenspace associated with eigenvalue $\lambda$ of $A$, i.e., $AE\{\lambda\}=\lambda E\{\lambda\}$ is exact. The general case of a selfadjoint operator is well-approximated by an operator with discrete spectrum, even if it has no actual eigenvalues, and this is how the Riemann-Stieltjes sum comes into play to approximate the general by the discrete case.
A: The expression $\langle \sum_k \phi(\lambda_k) E(\Delta_k) g,h \rangle$ can be connected to the Lebesgue integral with the measure $E_{g,h}$. Note that you have
$$
\langle \sum_k \phi(\lambda_k) E(\Delta_k) g,h \rangle = \int \sum_k  \phi(\lambda_k) \chi_{\Delta_k} dE_{g,h}.
$$
The function $\sum_k \phi(\lambda_k) \chi_{\Delta_k}$ is a simple function, that approaches $\phi$ as you decrease the $\varepsilon > 0$ mentioned in the statement of the proposition. It can also be proven of course that the above sum converges to the Lebesgue integral $\int \phi dE_{g,h}$ when $\varepsilon \to 0$. 
