Countable sum of orthogonal projections with orthogonal ranges; uniformly bounded monotone sequence of self-adjoint operators is strongly convergent Let $\mathcal{H}$ a Hilbert space. Let $P_j$ a sequence of orthogonal projections such that their ranges are mutually orthogonal. Then $\sum_1^\infty P_j \to P$ in the strong operator topology where $P$ is the orthogonal projection onto the closed span of the ranges of the $P_j$.
A version of this is used in B.C. Hall's Quantum Theory for Mathematicians and he claims that "it follows from an elementary argument". I can't figure that argument out, I can't even convince myself that $\sum_j P_j\psi$ converges in the norm topology for all $\psi \in \mathcal{H}$.
In your answer, you can suppose that I know that the analogous result holds for the finite sum case. 
Edit: JustDroppedIn provided enough info for me to understand the problem and develop an answer. The post they linked to gave the outline of a proof of a theorem necessary to proving the theorem I am trying to get in my question. Because this effectively answered my question, I accepted JustDroppedIn's answer. I couldn't find a uniform presentation of all the necessary results though (and I was unfamiliar with the linked theorem) so I wrote out proofs of every result, which I added in a separate answer, in case anyone might find this useful. Apologies for the bad formatting on my answer, I only know how to format things nicely on Latex.
 A: Step $1$: If a net/sequence of projections converges strongly, then the limit is also a projection. This is easy: let $(p_\lambda)$ be a net of projections with $p_\lambda\to p$. Then for any $x,y\in H$ it is $$\langle x,p^*y\rangle=\langle px,y\rangle=\lim_\lambda\langle p_\lambda x,y\rangle=\lim_\lambda\langle x,p_\lambda y\rangle=\langle x,py\rangle,$$ so $p=p^*$, i.e. $p$ is self-adjoint. Moreover,
$$\langle p^2x,y\rangle=\langle px,py\rangle=\lim_\lambda\langle p_\lambda x,p_\lambda y\rangle=\lim_\lambda\langle p_\lambda^2x,y\rangle=\lim_\lambda\langle p_\lambda x,y\rangle=\langle px,y\rangle.$$
So $p^2=p$
Step 2: If $(p_n)$ is a sequence of pairwise orthogonal projections, then any finite sum $\sum_{j=1}^Np_j$ is a projection. This is immediate.
Step 3: An increasing net/sequence of projections $(p_\lambda)$ converges strongly to the projection onto $M:=\overline{\bigcup_\lambda p_\lambda(H)}$.
Denote this projection by $p$. By a theorem of Vigier (This is not an easy theorem, but it is not an overkill. Its proof is mostly based on the Riesz' representation theorem), an increasing net of projections converges strongly. By step 1, the limit is a projection and we denote it by $q$, so $q$ is the strong limit of $(p_\lambda)$. Obviously, since for each $\lambda$ it is $p_\lambda(H)\subset M$, we will also have $q(H)\subset M$. But since $(p_\lambda)$ is increasing, we will have $p_\lambda\leq q$ for all $\lambda$, so $p_\lambda(H)\subset q(H)$ for all $\lambda$, so $M\subset q(H)$. Therefore $q(H)=M$ and $q=p$ as we wanted.
