# Operator-valued measurable functions

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?

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 $D$.
• Thanks, It was really nice example. Could you please just give some more concerning why such a countable family of subsets exists which can separate the points of $L^{\infty}(\mu)$? If there are any references to find more examples, please inform me too. – Ali Bagheri Dec 31 '15 at 7:44
• That such a countable family exists follows from this easy general theorem: Let $\cal U$ be a set of subsets of a set $X$, and let $\Sigma$ be the smallest $\sigma$-algebra on $X\times X$ that contains all rectangles $A\times B$ where $A,B\in{\cal U}$. Then for every $E\in\Sigma$ there is a countable family ${\cal V}\subseteq{\cal U}$ such that $E$ is in the smallest $\sigma$-algebra that contains all rectangles $A\times B$ where $A,B\in{\cal V}$. – PassingThru Dec 31 '15 at 19:51
• The argument showing that $D$ is not measurable is standard: See e.g. 3.10.44 in "Measure Theory" by V.I. Bogachev. – PassingThru Jan 3 '16 at 15:08