Show that the operator associated to a spectral decomposition on a Hilbert space is self-adjoint

Let $$H$$ be a $$\mathbb R$$-Hilbert space and $$(H_\lambda)_{\lambda\ge0}$$ be a spectral decomposition of $$H$$ (see below). Now, let $$\mathcal D\left(A_\varphi\right):=\left\{x\in H:\int_0^\infty\varphi(\lambda)\:{\rm d}\langle\pi_\lambda x,x\rangle_H<\infty\right\}$$ and $$\langle A_\varphi x,y\rangle_H:=\int_0^\infty\varphi(\lambda)\:{\rm d}\langle\pi_\lambda x,y\rangle_H\;\;\;\text{for all }x\in\mathcal D\left(A_\varphi\right)\text{ and }y\in H\tag1$$ (the integral has to be understood as a Lebesgue-Stieltjes integral) for Borel measurable $$\varphi:[0,\infty)\to\mathbb R$$.

How can we show that $$A_\varphi$$ is self-adjoint for any Borel measurable $$\varphi:[0,\infty)\to\mathbb R$$ and that $$A_{\varphi\psi}=A_\varphi A_\psi$$ for all bounded Borel measurable $$\varphi,\psi:[0,\infty)\to\mathbb R$$?

By the way: If I got it right, the spectral theorem states that if $$(\mathcal D(A),A)$$ is a nonnegative self-adjoint operator on $$H$$, $$(H_\lambda)_{t\ge0}$$ can be chosen such that $$A_\varphi=A$$, where $$\varphi(\lambda):=\lambda$$ for $$\lambda\in[0,\infty)$$. Does that mean that the domains $$\mathcal D(A_\varphi)$$ and $$\mathcal D(A)$$ coincide as well?

Definitions:

$$(H_\lambda)_{\lambda\ge0}$$ is called spectral decomposition of $$H$$ if

1. $$H_\lambda$$ is a closed subspace of $$H$$ for all $$\lambda\ge0$$;
2. $$(H_\lambda)_{\lambda\ge0}$$ is nondecreasing and right-continuous, i.e. $$\bigcap_{\mu>\lambda}H_\mu=H_\lambda\;\;\;\text{for all }\lambda\ge0;$$ and
3. $$\bigcup_{\lambda\ge0}H_\lambda$$ is dense.

Let $$\pi_\lambda$$ denote the orthogonal projection of $$H$$ onto $$H_\lambda$$ for $$\lambda\ge0$$. It can be shown that

1. $$[0,\infty)\ni\lambda\mapsto\pi_\lambda$$ is nondecreasing, i.e. $$\langle\pi_\lambda x,x\rangle_H\le\langle\pi_\mu x,x\rangle_H\;\;\;\text{for all }x\in H,$$ and right-continuous (with respect to the strong operator topology)

So,

1. $$[0,\infty)\ni\lambda\mapsto\langle\pi_\lambda x,x\rangle_H=\left\|\pi_\lambda x\right\|_H^2$$ is bounded (by $$\left\|x\right\|_H^2)$$, nondecreasing and right-continuous for all $$x\in H$$
2. $$[0,\infty)\ni\lambda\mapsto\langle\pi_\lambda x,y\rangle_H=2^{-1}\left(\langle\pi_\lambda(x+y),x+y\rangle_H-\langle\pi_\lambda x,x\rangle_H-\langle\pi_\lambda y,y\rangle_H\right)$$ is right-continuous and of bounded variation for all $$x,y\in H$$