Q:

Question about von Neumann algebra generated by a complete algebra of projections

Hi all, sorry if this is a dumb question, I don't know much about von Neumann algebras except the definition and a few relevant facts I've managed to prove by myself so I expect the answer will turn out to be well known. Anyway, let $\mathcal{H}$ be a Hilbert space, and suppose that $P$ is a commuting set of self-adjoint projections on $\mathcal{H}$, with the additional two properties:
1) $P$ is closed under complements, i.e. if $p \in P$ then so is $1 - p$.
2) $P$ is closed under suprema of arbitrary subsets, i.e. if $S \subseteq P$ then $\sup S \in P$ (here the projections on $\mathcal{H}$ are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$).
Now let $V$ denote the smallest von Neumann algebra containing $P$. Suppose that $p \in V$ is a self-adjoint projection. Is $p \in P$?
I know that $p$ is necessarily in the closure (relative to the weak operator topology) of the set of finite sums $\sum_i \lambda_i p_i$, where $p_i \in P$ and $\lambda_i \in \mathbb{R}$. It seems like it may be possible to derive a contradiction from the assumption that $q$ has a strictly smaller range than $p$, where $q \equiv \sup ${$ r \in P | r \leq p $}. But I don't know how to proceed.

A:

The answer "yes" follows from Theorem 2.8 of Bade's "On Boolean algebras of projections and algebras of operators," 1955, which is in the more general context of algebras of operators on a Banach space. 
Bade had previously proven a less general result that still covers your case, dealing with algebras of operators on reflexive spaces, in Theorem 3.4 of "Weak and strong limits of spectral operators," 1954.

