I am learning the spectral theorem for unbounded operators by myself and the following question came up. I will first tell how I learned it from the books.
Let $A$ be a densely defined self-adjoint operator on some Hilbert space $\mathscr{H}$ and, for a Borel set $\Omega \subseteq \mathbb{R}$, consider the function $1_{\Omega}$ which is $1_{\Omega}(x) = 1$ if $x \in \Omega$ and zero otherwise. By the spectral theorem for bounded self-adjoint operators, one can define the operator $P_{\Omega} = 1_{\Omega}(A)$ for each $\Omega$ and the family $\{P_{\Omega}\}$ of operators is a projection-valued measure.
For each $x \in \mathscr{H}$, one can define: $$\mu_{x}(\Omega) = \langle x, P_{\Omega}x\rangle$$ and prove that $\mu_{x}$ is a Borel measure on $\mathbb{R}$ for each fixed $x$. Then, one can extend it to a complex measure $\mu_{x,y}$ by setting $\mu_{x,y}(\Omega) = \langle x, P_{\Omega}y\rangle$, where the latter is defined by polarization identity.
The way I know it, the spectral theorem says that given a continuous function $f$ on the spectrum of $A$, we can define $f(A)$ by: $$\langle x, f(A)y\rangle = \int f(\lambda) d\mu_{x,y}(\lambda).\tag{1}\label{1}$$ It standard to write $d\mu_{x,y}(\lambda) = d\langle \psi, P_{\lambda}\varphi\rangle$, so that one writes: $$f(A) = \int f(\lambda) dP_{\lambda}. \tag{2}\label{2}$$
This is the tricky part for me. In this formulation, one is encouraged to think of $dP_{\lambda}$ not as an operator itself, but as a shorthand notation for the measure. In other words, (\ref{2}) only makes sense as a quadratic form: $$\langle x, \int f(\lambda) dP_{\lambda} y\rangle = \int f(\lambda)\langle x, dP_{\lambda}y\rangle = \int f(\lambda)d\mu_{x,y}(\lambda),$$ so that one recovers (\ref{1}). However, I don't know the meaning of (\ref{2}) as an operator itself, that is, I don't how to compute: $$f(A)x = \int f(\lambda)dP_{\lambda}x. \tag{3}\label{3}$$ This is because I always think of $dP_{\lambda}$ only as a shorthand notation for the measure, not as an operator. I was fine with this, but then I saw some formulas on other books and on the internet which seems to indicate that (\ref{2}) is more than just a notation; actually $dP_{\lambda}$ is some sort of projection operator and (\ref{3}) can be made precise somehow.
In short, my question is: using the formulation I described in the post, is it possible to understand $dP_{\lambda}$ also as some sort of operator and give a precise meaning for (\ref{3})?
