Measurable Operators

Question

Suppose I have an operator valued function, $\omega\mapsto A(\omega)$; for each $\omega$, $A(\omega):X\to Y$, is a bounded linear operator with $X$ and $Y$ real Hilbert separable Hilbert spaces. Suppose I know that for any $x,y$, $(A(\omega)x,y)$ is a measurable mapping from $(\Omega,\mathcal{F})$ to $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. Here $\mathcal{F}$ is some $\sigma$-algebra on $\Omega$.

Two questions:

Is it thus true that $\omega \mapsto L(\omega)$ is a measurable mapping from $(\Omega,\mathcal{F})$ to $(L(X,Y), \mathcal{B})$? I am giving $L(X,Y)$ the operator norm topology, though perhaps this is true in some other topology. I am also happy to make a separability assumption on $X$ and $Y$ if need be.

If it is true, how can I see this?

Answer 1

Let $\mathrm{B}$ be the ball of centre $f$ and radius $r > 0$ in the space of linear functions $X \to Y.$ You want to show $\{A \in \mathrm{B}\}$ is a measurable set (that is, belongs to $\mathscr{F}$). By definition, $A(\omega) \in \mathrm{B}$ is equivalent to $\|A(\omega) - f\| \leq r$ and this signifies $$(A(\omega) \cdot x \mid A(\omega) \cdot x) - 2(A(\omega) \cdot x \mid f(x)) \leq r(x|x) - (f(x)|f(x))$$ for all $x \in \mathrm{X}.$

If $\mathrm{X}$ were separable, and $\mathrm{S}$ were a separating set for $\mathrm{X},$ then the condition: $$(A(\omega) \cdot x \mid A(\omega) \cdot x) - 2(A(\omega) \cdot x \mid f(x)) \leq r(x|x) - (f(x)|f(x)) \quad (x \in \mathrm{X})$$ is equivalent to $$(A(\omega) \cdot x \mid A(\omega) \cdot x) - 2(A(\omega) \cdot x \mid f(x)) \leq r(x|x) - (f(x)|f(x)) \quad (x \in \mathrm{S}).$$ Let $\mathrm{T}_x$ the set of $\omega$ satisfaying the previous inequality for $x.$ The extended hypotheses (in my comments) imply $\mathrm{T}_x$ belongs to $\mathscr{F}.$ Then $$\mathrm{T} = \bigcap_{x \in \mathrm{S}} \mathrm{T}_x \in \mathscr{F}.$$ This allows deducing $A$ is measurable. For we just showed $A^{-1}(\mathrm{B})$ is measurable for any ball, hence, if $\mathscr{L}(\mathrm{X}, \mathrm{Y})$ were separable, $A^{-1}(\mathrm{U})$ would be measurable for any open set $\mathrm{U}$ in $\mathscr{L}(\mathrm{X}, \mathrm{Y})$ (for the separability of $\mathscr{L}(\mathrm{X}, \mathrm{Y})$ permits writting $\mathrm{U}$ as a countable union of balls). Having $A^{-1}(\mathrm{U})$ measurable for every open set $\mathrm{U},$ you can consider the set $\mathscr{Z}$ of Borel subsets $\mathrm{K}$ in $\mathscr{L}(\mathrm{X}, \mathrm{Y})$ such that $A^{-1}(\mathrm{K})$ belongs to $\mathscr{F}.$ It is easy to see $\mathscr{Z}$ is a sigma algebra and we showed, under restrictive hypotheses, it contains the open sets, hence $\mathscr{Z}$ is all Borel sets. Q.E.D.

Answer 2

I had another thought on this problem if I use a slightly different topology for $L(X,Y)$. Instead of the looking at the topology induced by the Borel sets with respect to the operator, norm, consider the strong operator topology induced by the sets $$ \{L \in L(X,Y): Lx \in B\}, \quad x \in X, \quad B \in \mathcal{B}(Y) $$ Furthermore, assume that $Y$ is separable, with sequence $\{y_n\}$ such that $$ \|y\|_Y = \sup_n |(y,y_n)_Y| $$ For any $x\in X$, and $n=1,2,\ldots$, define $$ a_{x,n}(\omega) = (A(\omega)x,y_n). $$ By the assumption, the $a_{x,n}$ are measurable. Now, if $x\in X$, $y\in Y$ and $r>0$, $$ \begin{split} A(\cdot)^{-1}(\{L \in L(X,Y): Lx \in B_r(y)\}) &= \{\omega: A(\omega)x \in B_r(y)\}\\ &=\{\omega: \|A(\omega)x - y\|_Y<r\}\\ &=\{\omega: \sup_{n}|(A(\omega)x - y,y_n)_Y|<r\}\\ &=\cap_{n=1}^\infty a_{x,n}^{-1}(B_r((y,y_n)))\in \mathcal{F} \end{split} $$ Since this holds for arbitrary balls $B_r(y)$, it will then apply to the case where $B\in \mathcal{B}(Y)$, and we have measurability in this weaker topology.

Thoughts?

I am still unsure if there is something to do I the operator norm topology. I began thinking about this while reading Da Prato & Zabcyzk who remark that $L(X,Y)$ will not be separable, even if $X$ and $Y$ both are, and they provide a counterexample for $X=Y=L^2(\mathbb{R})$.