Let $A$ be a C*-subalgebra of $B(H)$ for some Hilbert space $H$. Let's denote by $B$ the strong operator closure of $A$ in $B(H)$.
Question: Is $B$ a von Neumann algebra? [Since a von Neumann algebra is defined as a C*-subalgebra of $B(H)$ that is closed in the strong operator topology, this comes down to determining whether or not $B$ is a C*-algebra].
I know from the literature, that the answer is "yes", but how can we be certain, that $B$ is closed under operator multiplication, given that multiplication $B(H) \times B(H) \to B(H)$ is not strong operator continuous?