Is any fixed element of $L^\infty(0, 1) \bar{\otimes} M$ contained in $L^\infty(0, 1) \bar{\otimes} M_0$ where $M_0$ is separable?

Question

Suppose $M$ is a von Neumann algebra (not necessarily separable). Given any element $a \in L^\infty(0, 1) \bar{\otimes} M$, does there exist a separable subalgebra $M_0 \subseteq M$ s.t. $a \in L^\infty(0, 1) \bar{\otimes} M_0$? More generally, if I replace $L^\infty(0, 1)$ by any countably decomposable abelian von Neumann algebra (or even by any countably decomposable von Neumann algebra), is the statement above true? (It is easy to see that the result is not true if $L^\infty(0, 1)$ is replaced by an algebra that is not countably decomposable.) It is clear the result holds if $M$ is countably decomposable itself (by utilizing the GNS construction associated with faithful normal states), but it's not clear whether it is true in the general case.

Answer 1

I've figured this out myself. The statement is true iff $M$ is countably decomposable. An counterexample when $M$ is not countably decomposable is given by exactly the operator defined in my first comment, though it is somewhat complicated to prove this.