Let $T \in B(H)$ be a decomposable operator $\int_X T_x d\mu(x)$ where $T_x \in B(H_x)$ and $(X,\mu)$ is some measure space. Now, suppose that $S$ commutes with $T$. Can we conclude that $S$ is also decomposable? That is, there exists a measurable field of operators $S_x \in B(H_x)$ such that $S = \int_X S_x d\mu(x)$? If not, are there some extra conditions that we can place on $T$ so that it is true?
I believe if $T$ and $S$ are both normal then it will be true since they generate an abelian von Neumann algebra and so they will be simultaneously diagonablisable. However, I am not sure if the above is still true in the general case.
It would be appreciated if anyone could point me to any relevant references.
Take this answer as coming from someone who is deadly afraid of direct integrals.
In spirit, the operator $T$ is "block diagonal". The problem with commutants of diagonal matrices is multiplicity. In finite dimension you can have the operator $$ \begin{bmatrix} 1&0&0\\0&1&0\\0&0&2\end{bmatrix}. $$ The double multiplicity of $1$ makes the commutant of $T$ to be $M_2(\mathbb C)\oplus\mathbb C$. In a direct integral you may have the same problem. If there is a measurable set $E$ with $\mu(E)>0$ and $T_x=T_y$ for all $x,y\in E$, then the commutant of $T$ will include "full-matrices" for the entries indexed by $E$.