Proof, [Theorem 26.3, Zhu93]. ($\vnI$): By Zorn’s lemma, there exists a maximal family $\seqi{P}\subset \text{Proj}(A)$ of centrally orthogonal abelian projections. Let $P = \sum_{i \in I}P_{i}$, then $P$ is abelian by (2) of Lemma 37.8.5.

Let $P_{\vnI}= Z(P)$, then $A_{\vnI}:= P_{\vnI}AP_{\vnI}$ is a von Neumann algebra with identity $P_{\vnI}$. Let $R \in \text{Proj}(Z(A_{\vnI})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnI}$, $RP \le R$ is a non-zero abelian projection. Therefore $A_{\vnI}$ is of type $\vnI$.

($\vnII$): Assume without loss of generality that $I \in A$. Since $\seqi{P}$ is maximal and $(I - P_{\vnI}) \in Z(A)$, $(I - P_{\vnI})A(I - P_{\vnI})$ has no non-zero abelian projections.

By Zorn’s lemma, there exists a maximal family $\seqj{Q}\subset \text{Proj}((I - P_{\vnI})A(I - P_{\vnI}))$ of centrally orthogonal finite projections. Let $Q = \sum_{j \in J}Q_{j}$, then $Q$ is finite by (3) of Lemma 37.8.5.

Let $P_{\vnII}= Z(Q)$ and $A_{\vnII}= P_{\vnII}AP_{\vnII}$, then $A_{\vnII}$ is a von Neumann algebra with identity $P_{\vnII}$. Let $R \in \text{Proj}(Z(A_{\vnII})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnII}$, $RQ \le R$ is a non-zero finite projection by (4) of Lemma 37.8.4.

($\vnIII$): Let $P_{\vnIII}= I - P_{\vnI}- P_{\vnII}$ and $A_{\vnIII}= P_{\vnIII}AP_{\vnIII}$. Since $P_{\vnIII}\in Z(A)$ and $\seqi{P}$, $\seqj{Q}$ are maximal, $A_{\vnIII}$ has no non-zero finite projections. Therefore $A_{\vnIII}$ is of type $\vnIII$, and $A = A_{\vnI}\oplus A_{\vnII}\oplus A_{\vnIII}$.

($\vnII_{1}$): By Zorn’s lemma, there exists a maximal family $\bracsn{R_k}_{k \in K}\subset \text{Proj}(A_{\vnII})$ of orthogonal central finite projections. Let $P_{\vnII_1}= \sum_{k \in K}R_{k}$, then $P_{\vnII_1}$ is a central finite projection by (3) of Lemma 37.8.5. Hence $A_{\vnII_1}= P_{\vnII_1}AP_{\vnII_1}$ is of type $\vnII_{1}$.

($\vnII_{\infty}$): Let $P_{\vnII_\infty}= P_{\vnII}- P_{\vnII_1}$ and $A_{\vnII_\infty}= P_{\vnII_\infty}AP_{\vnII_\infty}$, then $P_{\vnII_\infty}$ is a central projection. By maximality of $\bracsn{R_k}_{k \in K}$, $A_{\vnII_\infty}$ admits no non-zero finite central projections. Therefore $A_{\vnII_\infty}$ is of type $\vnII_{\infty}$, $A_{\vnII}= A_{\vnII_1}\oplus A_{\vnII_\infty}$, and

(Uniqueness): Let $A = A_{\vnI}' \oplus A_{\vnII_1}' \oplus A_{\vnII_{\infty}}' \oplus A_{\vnIII}'$ be a decomposition of $A$ into von Neumann algebras of type $\vnI$, $\vnII_{1}$, $\vnII_{\infty}$, and $\vnIII$, respectively.

Let $P_{\vnI}'$, $P_{\vnII_1}'$, $P_{\vnII_\infty}'$, and $P_{\vnIII}'$ be the identity elements of $A_{\vnI}'$, $A_{\vnII_1}'$, $A_{\vnII_{\infty}}'$, and $A_{\vnIII}'$, respectively, then

is an orthogonal direct sum, and

1. ($\vnI$)

   Let $P_{1} = P'_{\vnI}(I - P_{\vnI})$, then by construction of $P_{\vnI}$, there exists no non-zero abelian projection $R \in \text{Proj}(A)$ with $R \le P_{1}$. As both $P_{\vnI}'$ and $(I - P_{\vnI})$ are central, $P_{1} \in A_{\vnI}'$, so $P_{1} = 0$ because $A_{\vnI}'$ is of type $\vnI$. Thus $P_{\vnI}' \le P_{\vnI}$. By symmetry, $P_{\vnI}= P_{\vnI}'$ and $A_{\vnI}= A_{\vnI}'$.

2. ($\vnII$, $\vnIII$)

   Let $P'_{\vnII}= P_{\vnII_1}' \oplus P_{\vnII_\infty}'$, $A_{\vnII}' = A_{\vnII_1}' \oplus A_{\vnII_\infty}'$, and $P_{2} = P'_{\vnII}(I - P_{\vnI}- P_{\vnII})$. By construction of $P_{\vnII}$, there exists no non-zero finite projection $R \in \text{Proj}(A)$ with $R \le P_{2}$. Since $P_{2} \in A_{\vnII}'$ and $A_{\vnII}'$ is of type $\vnII$, $P_{2} = 0$ and $P_{\vnII}' \le P_{\vnII}$. By symmetry, $P_{\vnII}= P_{\vnII}'$. Thus $P_{\vnIII}= P_{\vnIII}'$, $A_{\vnII}= A_{\vnII}'$, and $A_{\vnIII}= A_{\vnIII}'$.

3. ($\vnII_{1}$, $\vnII_{\infty}$)

   Let $Q_{2} = P'_{\vnII_1}(P_{\vnII}- P_{\vnII_1})$, then there exists no non-zero finite central projection $R \in \text{Proj}(A)$ with $R \le Q_{2}$. However, since $A_{\vnII_1}'$ is of type $\vnII_{1}$, $P'_{\vnII_1}$ is itself a finite projection, and every subprojection of $P'_{\vnII_1}$ is finite by (4) of Lemma 37.8.4. Thus $Q_{2} = 0$ and $P_{\vnII_1}' \le P_{\vnII_1}$. By symmetry, $P_{\vnII_1}' = P_{\vnII_1}$. Therefore $P_{\vnII_\infty}' = P_{\vnII_\infty}$, $A_{\vnII_1}' = A_{\vnII_1}$, and $A_{\vnII_\infty}' = A_{\vnII_\infty}$.

$\square$