I am going through some spectral theory, and I have found two results under this name. I state this results:
(I) Spectral Theorem I: Let $\mathcal{H}$ be a separable Hilbert space. If $A\in L(\mathcal{H})$ is self adjoint. Then there exists a unique projection valued measure $\mu^A$ on the Borel-$\sigma$ algebra of the spectrum $\sigma(A)$ such that $$\int_{\sigma(A)} \text{Id}_{\sigma(A)} \; d\mu^A =A. $$
(II) Spectra Theorem II Let $\mathcal{H}$ be a separable Hilbert space. If $A\in L(\mathcal{H})$ is self adjoint. Then there exists a $\sigma-$finite measure $\mu$ on the spectrum $\sigma(A)$, a direct integral $$\Gamma(A)=\int_{\sigma(A)}^\oplus H_\lambda\; d\mu (\lambda) $$ and a unitary map $U:\mathcal{H}\to \Gamma(A)$ such that $$[UAU^{-1}(s)](\lambda)=\lambda s(\lambda) $$ for all sections $s$ in $\Gamma(A)$.
So, maybe a bit of terminology is required. Given a $\sigma-$finite measure space $(X,\Omega, \mu )$ and a indexed family $\{H_\lambda \}_{\lambda\in X}$ of Hilbert spaces, we can costruct the obvious vector bundle $$\pi: \xi=\bigsqcup_{\lambda\in X} H_\lambda \to X$$ $$\psi\mapsto \lambda \quad \text{if } \psi\in H_\lambda $$ A section $s$ of $\xi$ is defined as a map between $X$ and $\xi$ such that $\pi \circ s=\text{Id}_X$ as it usual, but we further impose a measurability condition. This can only make sense with a measure structure is $\xi$. This measure structure is given by an indexed family of sequences $\{ \{e_j^\lambda \}_{j=1}^\infty\}_{\lambda \in X}$ such that for any $\lambda \in X$ we have $\{e^\lambda_j \}_{j=1}^\infty \subseteq H_\lambda$, $$\langle e_j^\lambda,e_k^\lambda \rangle=0 \text{ for } j\neq k$$ and the norm of every $e_j^\lambda $ is either $1$ or $0$. We also ask the maps $$\lambda\mapsto \text{dim}(H_\lambda) \quad \text{and} \quad \lambda \mapsto \langle e^\lambda_j,e^\lambda_k \rangle \quad \forall j,k>0 $$ to be measurable. With this, we call a section $s$ measurable if $\lambda\mapsto \langle s(\lambda),e^\lambda_j\rangle $ is a measurable map. Now, the direct integral $$\int_{X}^\oplus H_\lambda\; d\mu (\lambda) $$ Is the set $\Gamma(\xi)/ \sim $ where $\Gamma(\xi)$ is the set of measurable sections $s$ in $\xi$ for which $$\Vert s \Vert^2=\int_{X} \Vert s(\lambda) \Vert^2 \; d\mu(\lambda)<\infty $$ and $s_1\sim s_2$ if they agree $\mu-$almost every where.
Now, it seems like we could recover the spectral theorem (I) from (II). Let's try, suppose (II) and for each $E$ in the borel $\sigma-$algebra of $\sigma (A)$, define $V_E\subseteq \Gamma(A)$ as the set of all section $s$ such that $\text{support}(s)\subseteq E$. Let $P_E $ be the orthogonal projection onto $V_E$, now we define a projection valued measure $\mu^A$ on $\sigma(A)$ as $$\mu^A(E)=U^{-1}P_E U $$ where $U$ is as in (II). The question is how could I prove that $$\int_{\sigma(A)} \lambda \;d\mu^A(\lambda)=A. $$ Also, does (I) implies (II)? I think it does not, but cannot figure out where does the equivalence fail.
Here's my attempt: So, I need to show that for any $\psi\in \mathcal{H}$ we have $$\Bigr\langle \left(\int_{\sigma(A)} \lambda \; d\mu^A (\lambda)\right) \psi, \psi \Bigr\rangle =\langle A\psi, \psi \rangle. $$
Let $\psi=U^{-1}s$ for some $s$ in the direct integral, then $$\langle A\psi,\psi \rangle=\langle AU^{-1}s , U^{-1}s \rangle=\langle UAU^{-1}s , s \rangle =\int_{\sigma(A)} \langle UAU^{-1}s(\lambda) , s(\lambda)\rangle \;d\mu(\lambda)= \int_{\sigma(A)} \langle \lambda s(\lambda) , s(\lambda)\rangle \;d\mu(\lambda) = \int_{\sigma(A)} \lambda\langle \ s(\lambda) , s(\lambda)\rangle \;d\mu(\lambda). $$ Hence, it suffices to show that $$ \int_{\sigma(A)} \lambda\langle \ s(\lambda) , s(\lambda)\rangle \;d\mu(\lambda)=\int_{\sigma(A)} \lambda \;d\mu^A_{\psi}(\lambda) $$where $\mu^A_\psi$ is the real valued measure on $\sigma(A)$ defined by $$\mu^A_\psi(E)=\langle \mu^A(E)\psi , \psi \rangle=\langle U^{-1}P_EU \psi , \psi \rangle= \langle P_E s,s \rangle=\int_{\sigma (A)} \langle P_E s(\lambda),s(\lambda) \rangle \; d\mu(\lambda). $$ Here is where I am stuck.
Other notes By support($s)\subseteq E$ I mean that the $s(\lambda)=0$ for $\mu-$almost every $\lambda\in E^c$. Id$_X$ denotes the identity in $X$ and the inner product is assumed to be linear in the first entry. Also, all inner products are missing the label for in which it is defined, it is just to avoid writing the labels too much, figuring out what this label is should not be hard.
From the attempt (See question) it suffice to show that $P_Es=\chi_Es$, where $\chi_E$ is the characteristic function of $E$. Which is the case, for we can write $$s=P_Es+P_{E^\perp}s $$ where $P_{E^\perp}$ is the orthogonal projection onto $V_E^\perp$, observe that if $t\in V_E^\perp$, we have $\langle t,r \rangle=0$ for all $r\in V_E$, that is, support$(r)\subseteq E$, thus we have $$0=\int_{\sigma(A)} \langle t (\lambda), r(\lambda) \rangle d\mu(\lambda)=\int_{E}\langle t (\lambda), r(\lambda) \rangle d\mu(\lambda) $$ If we choose $r$ to always have unit lenght, then we have $t=0$ $\mu-$almost everywhere in $E$, hence support$(t)\subseteq E^c$. The converse of this observation is trivially true, and this implies that $P_{E^\perp}$ is the projection onto the space of all sections whose support is $\mu-$almost everywhere contained in $E^c$. This clearly implies that $\chi{E}s=P_Es$. And now the proof is complete.