Let $H$ be a non separable Hilbert space and $\Omega$ be a measurable space. 
Naturally, we say that $f:\Omega\to B(H)$ is $w$-measurable if $f^{-1}(O)$ is measurable for any open set $O$ in the weak operator topology. 


Question: Let $f$ and $g$ be two $w$-measurable functions on $\Omega$. Is the multiplication $fg$ a $w$-measurable function as well?



Operator-valued measurable functions
The answer is no even for functions with values in $L\_\infty(\mu)$ (for the purpose of this question embedded in $B(H)$), as long as the cardinality of the space $L\_\infty(\mu)$ is larger than continuum. The following argument works for many reasonable topologies, including the w.o.t. in the question.


Let $\Omega$ be the set $L\_\infty(\mu)\times L\_\infty(\mu)$ with the smallest $\sigma$-algebra that makes both projections onto $L\_\infty(\mu)$ measurable. The diagonal $D=\{(x,x)\mid x\in L\_\infty(\mu)\}$ is not a measurable subset of $\Omega$: If it were then there would be a countable family of subsets of $L\_\infty(\mu)$ separating the points of $L\_\infty(\mu)$, so the cardinality of $L\_\infty(\mu)$ would be at most continuum.


Define $f(x\_0,x\_1)=e^{x\_0}$, $g(x\_0,x\_1)=e^{-x\_1}$ for $(x\_0,x\_1)\in\Omega$. The function $fg=e^{x\_0-x\_1}$ is not measurable because the set $\{I\}$ is closed and its preimage is $D$.


