Why is a completely additive isomorphism of Von-Neumann Algebras ultraweakly continuous?

Question

Suppose $M$ and $N$ are Von-Neumann algebras on say $B(H)$ and $B(K)$ and $\theta: M \rightarrow N$ is an isomorphism if $M$ and $N$ are considered $C^*$-algebras. If also $\theta(\sum\limits_{i\in I}p_i)=\sum\limits_{i\in I}\theta(p_i)$ for every mutually orthogonal family of projections $\{p_i : i\in I\}$ in $M$, I want to understand why $\theta$ is necessarily ultraweakly continuous.

Answer 1

For any normal state $\psi$ on $N$, you get that $\psi\circ\theta$ is a completely additive state of $M$. By the result for states, the state $\psi\circ\theta$ is normal (equivalently, ultraweakly continuous). And then $\theta$ is ultraweakly continuous (if $x_j\to x$ ultraweakly, then $\psi(\theta(x_j))\to\psi(\theta(x))$ for all normal $\psi$,a nd then $\theta(x_j)\to\theta(x)$.

So it is enough to prove the result (completely additive implies ultraweakly continuous) for states. The result appears in Theorem 7.1.12 of Kadison-Ringrose; the proof takes about two pages (Theorems 7.1.8 and 7.1.9), without counting a few references to previous results.