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I was reading Spectrum and dynamics where Paolo Facchi discusses projection-valued measures and integrals. The discussion and constructions are all based on the fact that one can define characteristic functions of self-adjoint operators, i.e. if $A$ is a self-adjoint operator on a Hilbert space $\mathscr{H}$ and $\chi_{\Omega}$ is the characteristic function associated with the Borel set $\Omega \subseteq \mathbb{R}$, then $\chi_{\Omega}(A)$ is a well-defined object. Again, this is used as the starting point for the discussion and, at the begining of the text, the author says that he'll defer the problem of explicitly defining it. However, I've read the entire text and got confused, since I did not find any such definition. So, my question is: how do we define $\chi_{\Omega}(A)$?

Remark: As mentioned in my other question Extension of functional calculus of continuous functions, I know there is this Borel functional calculus where one can define functions of self-adjoint operators. The problem is that if I use the Borel functional calculus to define $\chi_{\Omega}(A)$, the integral formalism discussed in the text (which is what I'm interested in at the moment) becomes kinda cyclic, since it becomes another way to define the Borel functional calculus. So, I was looking for a more practical definition of $\chi_{\Omega}(A)$ which can be used to give meaning to the linked text hypothesis.

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    $\begingroup$ Any standard textbook that covers operator theory contains a proof of that spectral theorem, but the text you cites is apparently an unfinished draft. $\endgroup$ Commented Sep 16, 2021 at 0:35
  • $\begingroup$ @NarutakaOZAWA thanks for your comment! I know the Borel functional calculus is standard. However, there are many different approaches and I got interested on this construction made in the linked text, i.e. to construct the functional calculus directly from these operator-valued integrals. The only missing point here is how to define $\chi_{\Omega}(A)$ in the first place. I don't know any reference with this exactly construction. $\endgroup$
    – MathMath
    Commented Sep 16, 2021 at 1:09
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    $\begingroup$ Your link to your other question instead pointed to a comment by @MaoWao. This seemed likely to be a typo, so I edited to point to the question. $\endgroup$
    – LSpice
    Commented Sep 16, 2021 at 1:12
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    $\begingroup$ @LSpice thank you! $\endgroup$
    – MathMath
    Commented Sep 16, 2021 at 1:14

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