Does there exists a von Neumann algebra which is neither of type I$_n$, II$_1$,II$_\infty$, nor III? In the "Type decomposition" theorem 6.5.2 in Kadison-Ringrose, vol II, they proved that every von Neumann algebra can be decomposed as a direct sum of type $I_n$, type $II_1$, type $II_\infty$ and type $III$ parts (some of these summands may be zero). I understood this.
My question is:

Does there exists a von Neumann algebra which is neither of type $I_n, II_1,II_\infty, \text{ nor } III$?

For example, consider $\mathscr{R}:=L^\infty(X, \mu)$ acting on $\mathcal{H}:=L^2(X, \mu)$. Then since $\mathscr{R}$ is abelian so every non-zero projection in it is abelian projection. Hence $\mathscr{R}$ cannot be of type $II$ nor $III$. Can I say from there that $L^\infty(X, \mu)$ is a type $I$ von Neumann algebra (without even finding a projeciton with central carrier $I$. In deed here $\chi_X$ is an abelian projection with central carrier $I=\chi_X$ proving $L^\infty$ or any abelian algebra is of type $I$.)?
 A: In any abelian von Neumann algebra, each projection is its own central carrier. The only projection with central carrier I is I; that's enough to make it type I.
A von Neumann algebra can have all the types as central summands. For instance take $M$ and $N$ of your favourite (different types), and form $M\oplus N$.
If you look at the proof of Theorem 6.5.2 in KRII, the central projection $P_d$ is constructed in such a way that $(I-P_d)R$ has no abelian projections.  That makes it unique.
In a similar way, $P_{c_1}$ is constructed as the largest finite central projection in  $(I-P_d)R$. And so on.
The point is that you don't get to choose. A von Neumann algebra $R$ is a direct sum of von Neumann algebras of fixed types.
That a direct sum of algebras of distinct types is of neither type is trivial to check. For instance if $R=M\oplus N$ with $M$ type I and $N$ type II$_1$, then there are not abelian projections under $P_N$, so $R$ is not type I. And there are abelian projections under $P_M$, so $R$ is not type II.
