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Let $H$ be a $mathbb R$-Hilbert space and $(H_lambda)_lambdage0$ be a spectral decomposition of $H$ (see below). Now, let $$mathcal Dleft(A_varphiright):=leftxin H:int_0^inftyvarphi(lambda):rm dlanglepi_lambda x,xrangle_H<inftyright$$ and $$langle A_varphi x,yrangle_H:=int_0^inftyvarphi(lambda):rm dlanglepi_lambda x,yrangle_H;;;textfor all xinmathcal Dleft(A_varphiright)text and yin Htag1$$ (the integral has to be understood as a Lebesgue-Stieltjes integral) for Borel measurable $varphi:[0,infty)tomathbb R$.

If I got it right, the spectral theorem states that if $(mathcal D(A),A)$ is a nonnegative self-adjoint operator on $H$, $(H_lambda)_tge0$ can be chosen such that $A_varphi=A$, where $varphi(lambda):=lambda$ for $lambdain[0,infty)$. However, I've always seen the identity $$A_1=operatornameid_H,tag2$$ but this seems to be wrong. By $(1)$ and the definition of the Lebesgue-Stieltjes measure, $$langle A_1x,yrangle_H=lim_lambdatoinftylanglepi_lambda x,yrangle_H-langlepi_0x,yrangle_Htag3.$$ Clearly $lim_lambdatoinftylanglepi_lambda x,yrangle_H=langle x,yrangle_H$, so it seems like $(1)$ should be replaced by $$langle A_varphi x,yrangle_H:=varphi(0)langlepi_0 x,yrangle_H+int_0^inftyvarphi(lambda):rm dlanglepi_lambda x,yrangle_H;;;textfor all xinmathcal Dleft(A_varphiright)text and yin Htag4.$$ On the other hand, we could also extend $H_lambda$ for $lambda<0$ by setting $H_lambda:=left0right$ (and hence $pi_lambda=0$) for all $lambda<0$. With this definition we could define $$langle A_varphi x,yrangle_H:=int_mathbb Rvarphi(lambda):rm dlanglepi_lambda x,yrangle_H;;;textfor all xinmathcal Dleft(A_varphiright)text and yin Htag5$$ by setting $varphi(lambda)=0$ for all $lambda<0$. With $(4)$ we would again obtain $(2)$.

So, I'm highly confused. Which is the usual definition when one is interested in the spectral measure for a nonnegative self-adjoint operator? (Please note that I'm only interested in Lebesgue-Stieltjes integration; not in Riemann-Stieltjes integration.)

Definitions:

$(H_lambda)_lambdage0$ is called spectral decomposition of $H$ if

1. 
   $H_lambda$ is a closed subspace of $H$ for all $lambdage0$;


2. 
   $(H_lambda)_lambdage0$ is nondecreasing and right-continuous, i.e. $$bigcap_mu>lambdaH_mu=H_lambda;;;textfor all lambdage0;$$ and


3. 
   $bigcup_lambdage0H_lambda$ is dense.

Let $pi_lambda$ denote the orthogonal projection of $H$ onto $H_lambda$ for $lambdage0$. It can be shown that

4. 
   $[0,infty)nilambdamapstopi_lambda$ is nondecreasing, i.e. $$langlepi_lambda x,xrangle_Hlelanglepi_mu x,xrangle_H;;;textfor all xin H,$$ and right-continuous (with respect to the strong operator topology)

So,

5. 
   $[0,infty)nilambdamapstolanglepi_lambda x,xrangle_H=left|pi_lambda xright|_H^2$ is bounded (by $left|xright|_H^2)$, nondecreasing and right-continuous for all $xin H$
   


6. 
   $[0,infty)nilambdamapstolanglepi_lambda x,yrangle_H=2^-1left(langlepi_lambda(x+y),x+yrangle_H-langlepi_lambda x,xrangle_H-langlepi_lambda y,yrangle_Hright)$ is right-continuous and of bounded variation for all $x,yin H$