I want to know why the following is true: Every semifinite (ie has trivial type III part) von Neumann algebra $R$ is the direct sum of a finite von Neumann algebra and a properly infinite von Neumann algebra. This is clearly true if $R$ is type II. If $R$ is type I and infinite is it necessarily properly infinite (ie has no finite nontrivial central projections)? I cannot find anything that implies this, even for type I$_{\infty}$.
Edit: $R$ is assumed to have a faithful normal state.
9.2.19. LEMMA. If $\omega$ is a faithful normal state of a semi-finite von Neumann algebra $\mathscr{R}$, there is a normal semi-finite faithful tracial weight $\rho$ on $\mathscr{R}$, and a positive element $K$ in the unit ball of $\mathscr{R}$, such that $I-K \in F_{p}$ and $$ \rho((I-K) A)=\omega(K A)=\omega(A K) \quad(A \in \mathscr{R}) . $$ Moreover, both $K$ and $I-K$ are one-to-one mappings. Proof. It suffices to consider separately the two cases in which $R$ (with a faithful normal state $\omega$ ) is either finite or properly infinite but semi-finite; for the general semi-finite $\mathscr{R}$ is a direct sum of two algebras, one of each of these kinds.