There is a passage in a book that is not very clear to me:
A is a C*Algebra and $a$ is selfadjoint.
Then
"Indeed identifying A with an algebra of operators on a Hilbert space $\mathcal{H}$, by the spectral theorem there is a projection-valued compactly supported measure $E(\cdot \space; a)$ so that for every continuous function,
$f(a)=\int f(t)dE((-\infty,t],a)$"
I know that the spectral theorem says that if a is selfadjoint then it can be viewed as a multiplication operator . I also know what a spectral measure is.
It's not clear to me why in the dE there are intervals in the form $(-\infty,t]$. I've always seen only dE (without specifications). Maybe is only a matter of notation.
and in addition, $f(a)$ should be itself an operator in A (obtained by the taylorization of f) so why is equal a complex nuber?!
Thank you for your help :-)
Spectral theorem for a self-adjoint operator is often formulated as $a=\int\lambda dE_\lambda,$ where $E_\lambda:\mathbb R\to B(H)$ is the spectral resolution of $a=a^*\in B(H).$ In your book $E(\cdot\ ;a)$ is the spectral measure (function of Borel subsets $\mathbb R$) associated with $a.$ The mapping $\lambda\mapsto E_\lambda=E((-\infty,\lambda];a)$ is now the spectral resolution of $a$ and the integral is the projection-valued Riemann-Stieltjes integral.
See also What is the relationship between spectral resolution and spectral measure?