Complete blocks of projections and spectral theorem This question was asked in a slightly different way here, and then I realized that it could be of higher interest in the physics community.
Let $\mathcal{P}(\mathcal{H})$ be an (atomic) lattice of projections on a separable Hilbert space $\mathcal{H}$. Let $\mathcal{B}$ be a maximal Boolean subalgebra (block) of $\mathcal{P}(\mathcal{H})$. Then, by spectral theorem, there exists a measure space $(X,\mu)$ and unitary operator $U:\mathcal{H}\rightarrow L^2(X,\mu)$ such that for each $P\in\mathcal{B}$, it holds that $UPU^{-1}$ is a multiplication by a measurable function $f_P:X\rightarrow\mathbb{R}$. 
The interesting implication is that if $\mathcal{B}$ is a complete** block, then $\mathcal{B}\simeq\mathcal{B_0}$, where $\mathcal{B_0}$ is a measure algebra of $(X,\mu)$. Obviously, whenever $\mathcal{B}$ is an atomic BA, $\mathcal{B_0}$ is also atomic. (Details can be found e.g. in Takeuti's "Two applications of logic to mathematics" pg. 63.) 
I have several questions to that issue:


*

*minor one: to what extent $(X,\mu)$ is unique in the spectral theorem?

*major one 1: my intuition is that if $\mathcal{B}$ is a complete block generated (and maximized by Zorn's lemma) by spectral measure of a self-adjoint operator with a continuous spectrum, then $\mathcal{B}$ is atomless. On the other hand, if a complete $\mathcal{B}$ comes similarly from a self-adjoint operator with a pure point spectrum, then $\mathcal{B}$ is atomic (must $\mu$ be discrete then?) - is this correct?*

*major one 2: what can be said about atomicity of a complete block $\mathcal{B}$ generated by a self-adjoint operator with mixed (both continuous and point) spectrum?

*aside: could this be related to the presence of minimal projections in von Neumann algebras? I suspect the answer is no, since from the very beginning every block in $\mathcal{P}(\mathcal{H})$ being a BA is commutative, hence not factor.



*This would come from the fact that projections $P$ under the isomorphism are mapped to characteristic functions $\chi_P$; if a projection's range is a one-dimensional subspace of $\mathcal{H}$, then $\chi_P=\chi_{\{p\}}$ for some $p\in X$.
**Edit: I realized that the completeness is a subtle point here: a complete BA of projections is not only complete as a BA (i.e. containing all the sup's), but also in the following, stronger sense: if $P=\mathrm{sup}(P_a)$, then $P(\mathcal{H})$ (the range of $P$) is the closure of the linear space spanned by $\bigcup P_a(\mathcal{H})$.
 A: First of all I am not sure to understand well your terminology, but I think that maximality implies completeness (the $\sup$ of a set of projectors, viewed as the strong limit of a class of projectors, commutes with all the projectors of the block using the strong operator topology and so it belongs to the block as it is maximal). So your blocks are complete.
Regarding your first question actually I do not know, I think it is unique but I should check my conjecture and I do not have much time now. Sorry.
Regarding your second question. I assume that atomic means that for every $P\neq 0$ in the block there is an atom $Q$ with $Q \leq P$. If the spectrum has only continuous component the lattice not only is not atomic, but it does not contain atoms because the atoms are associated to the single eigenvalues. If the spectrum of $A$ is a pure point spectrum, then $P_{\sigma_p(A)}=I$ by definition (notice that $\sigma_c(A)\neq \emptyset$ is still possible, think of a self-adjoint compact operator whose eigenvectors are accumulated by $0$ which, in turn, is not an eigenvalue). By difference, $P_{\sigma_c(A)}=0$. Therefore, every Borel set $E\subset \sigma(A)$ either intersects $\sigma_p(A)$ or produces the zero projector. In the first case an atom $Q \leq P_E$ exists since it is associated to some eigenvalue of $A$ included in $E$.
Regarding your third question, as the atoms are all of the form $P_{\{\lambda\}}$ with $\lambda \in \sigma_p(A)$, then the lattice may be atomic or not. As I pointed out above, pure point spectrum does not mean that $\sigma_c(A)$ is empty, but only that $P_{\sigma_p(A)}=I$, in this case however the lattice is atomic. In more complicated cases where the continuous part of the spectrum is full open segment for instance and the point spectrum includes also some isolated points, the lattice is not atomic.
I do not understand well your last question. Think of a Hermitian matrix $A$ so that its commutative lattice is generated by the atoms $P_\lambda$, where $\lambda$ is every eigenvalue of $A$. The commutative von Neumann algebra generated by the lattice coincides to the algebra of all complex functions $f(A)$ spectrally defined. It is clear that, even if the algebra is not a factor, the projectors $P_\lambda$ are minimal projectors of that algebra.
