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Let $D \subseteq A$ be an abelian C*-subalgebra of the C*-algebra $A$ where $A \subseteq B(H)$ for some separable Hilbert space $H$. Assume that the von Neumann Algebra generated by $D$ is a maximal abelian subalgebra (masa) of the von Neumann algebra generated by $A$.

Does this already imply that $D \subseteq A$ is maximal abelian?

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For instance, take $A=\ell^\infty(\mathbb N)$, and $D=c_0(\mathbb N)$.

For a different example, take $A=C[0,1]$ seen as multiplication operators on $L^2[0,1]$, and $$D=\{f\in A:\ f(0)=0\}.$$ Then $D''=A''=L^\infty[0,1]$.

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