I've been working through parts of Schrödinger Operators by Cycon, Froese, Kirsch, and Simon, and I'm a bit stuck on one of the results they use without proving. Let $(\Omega,\mathcal{M})$ be a measurable space, and $\mathcal{H}$ some Hilbert space. We say an operator valued map $A:\Omega \rightarrow \mathcal{B}(\mathcal{H})$ is weakly measurable if for any $\varphi, \psi \in \mathcal{H}$, the function $\omega \mapsto \langle \varphi, A(\omega)\psi\rangle$ is measurable in the usual sense, i.e. pre-images of open sets in $\mathbb{C}$ are in $\mathcal{M}$.
The text says it's clear that the product of weakly measurable functions is measurable, i.e. if $A,B: \Omega \rightarrow \mathcal{B}(\mathcal{H})$ are both weakly measurable, then operator valued function $\omega \mapsto A(\omega)B(\omega)$ should also be weakly measurable. The only thing I can think of is to consider pre-images, working with open balls for simplicity, so my current plan of attack is to show: $$\forall \psi,\varphi \in \mathcal{H}, z \in \mathbb{C}, r >0 \qquad\{\omega \in \Omega\,:\, \lvert \langle \varphi, A(\omega)B(\omega)\psi \rangle - z \rvert < r\} \in \mathcal{M}$$
While this is more or less exactly what it would mean for the product to be measurable, I'm not sure such a direct approach will be fruitful, and the way the book puts it suggests to me that there's some obvious trick I'm missing that makes this a short proof.
(Edited to make notation consistent.)
Most likely these authors only consider separable Hilbert spaces $\mathcal{H}$. In this case pick an orthonormal basis $(e_n)_{n=0}^{\infty}$ and write $$ \langle\varphi,A(\omega)B(\omega) \psi\rangle= \lim_{N\rightarrow \infty}\sum_{n=0}^{N}\langle \varphi,A(\omega)e_n\rangle\ \langle e_n, B(\omega)\psi\rangle\ . $$ Sums and products of measurable functions are measurable so the expression is measurable for finite $N$. Finally, since pointwise limits of sequences of measurable functions are measurable, this takes care of the $N\rightarrow\infty$ limit.