If $A \subseteq \mathcal B(\mathcal H)$ is an algebra of operators that is closed under adjoint, then its bicommutant $A''$ is a von Neumann algebra, and is the ultraweak closure of $A$; this is one version of von Neumann's bicommutant theorem. Does the theorem hold relative to an arbitrary von Neumann algebra $\mathcal M$? Concretely, what is the truth value of the following statement:

Let $\mathcal M$ be a W*-algebra, and $A\subseteq \mathcal M$ be a subalgebra closed under adjoint. Then the relative bicommutant $A'' = \{ m \in \mathcal M | \forall x \in \mathcal M. (\forall a \in A. ax = xa) \implies mx=xm \}$ is a W*-algebra, and is the ultraweak closure of $A$.

A W*-algebra is C*-algebra that is isomorphic to a von Neumann algebra. I use the term W*-algebra to emphasize that the bicommutant is being computed relative to $\mathcal M$ itself rather than relative to a Hilbert space on which $\mathcal M$ is represented.

**Edit:** As Matthew points out below, $A$ should contain the unit of the ambient algebra, i.e., of $\mathcal B(\mathcal H)$ in the bicommutant theorem, and of $\mathcal M$ in the statement in question.