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This question was asked in a slightly different way and answered here.

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".)

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 blocks in $\mathcal{P}(\mathcal{H})$ are commutative, hence not factors.

*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 $\bigcup P_a(\mathcal{H})$.

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