The relation between eigenvalues and spectral integral Consider a bounded and self-adjoint operator $A$ such that $mI\le A\le MI$. Let $\{E_\lambda\}$ be the spectral family of $A$, so $A=\int_{m-0} ^M \lambda dE _\lambda$.   
In my notes there is the following statement: every $a<m$ can be taken as a lower bound for the integral, and if $A$ has an eigenvalue $m$, them $a=m$ can not be taken as a lower bound.   
Can someone please explain that statement? Why does the eigenvalues of $A$ matter? What is the relation between the eigenvalues and the spectral integral?
 A: The integral is the Riemann Stieljes integral. It is defined via partitions of intervals. Let $\tau=\{[a_i,a_{i+1}]\mid i\in\{0,..,n\}\}$ be a partition of $[a,b]$ into $n$ disjoint sub-intervals so that $\bigcup [a_i,a_{i+1}]=[a,b]$. Define
$$I_\tau(f, dg) :=\sum_{i=0}^{n-1} f(a_i) \big(g(a_{i+1}) -g(a_i)\big)$$
(I assume $f$ is continuous, othwerwise there are different definitions.) The set of partitons of $[a,b]$ can be given a partial ordering via refinement, and for any two partitions there exists a common refinement. So this is a directed set. The Riemann Stieljes integral is defined as the limit of this, provided it exists:
$$I_{[a,b]}(f,dg)=\lim_{\tau}\,I_\tau(f,dg)$$
A feature of this is that it will sum up the discontinuities of $g$ weighted with the values of $f$ at these discontinuities. Check for instance with $g(x)=\begin{cases} 0&x<c\\ \Delta & x≥c\end{cases}$ that $$I_{[a,b]}(f,dg)=\begin{cases}f(c)\Delta& x\in(a,b]\\ 0&x\notin(a,b]\end{cases}$$
The connection with the eigenvalues of $A$ is that if $A$ has an eigenvalue at $a$, then $E(\lambda)$ is discontinuous at $a$, jumping by the projection onto this eigenspace. From the way the Riemann Stieljes integral is defined, in order for you to see this discontinuity you need $a$ to be the in the interior of your domain of integration.
