Let $X$ be some non-empty set and $\mathcal{H}$ be a Hilbert space over the complex numbers. Suppose that $A\in \mathcal{B}(H)$ is some fixed bounded operator and $B(t)\in\mathcal{B}(H),t\in\mathbb{R}$ is a family of bounded operators with a uniform bound, that is $\exists C\in\mathbb{R}_{\geq 0}: \sup_{t\in\mathbb{R}}||B(t)|| \leq C$. Let $f:X\to\mathbb{C}$ be a test function, in this case a continuous function with a compact support, and define $Q(f) = \int_{-\infty}^{\infty}f(s)B(s)ds$ as the unique bounded operator which satisfies $\left<Q(f)v, v\right> = \int_{-\infty}^\infty f(s)\left<B(s)v, v\right>ds$ for $v\in \mathcal{H}$. Given all this, how can use justify moving $A$ inside the integral defining $Q$ in the sense that
$$AQ(f)v = A\left(\int_{-\infty}^{\infty}f(s)B(s)ds\right)v = \int_{-\infty}^{\infty}f(s)AB(s)dsv$$
for $v\in\mathcal{H}$? I'm asking because I've been reading Brian Hall's book, "Quantum Theory for Mathematicians", where he essentially mentions this property of operator-valued integration without providing a proof or referring to one in Lemma 10.18 (page 211).
Prior to the lemma in question, I can't seem to find proofs or statements to such analogues of basic integration theory as "moving a constant quantity inside the integral" from Hall's or Rudin's books. I mean to say that while $A$ is not evolving w.r.t. $t$, and integration reduces to summing some bounded linear operators when one uses projection valued measures, $A$ is still not a scalar, but a mapping, and therefore I am hesitant to "just move it inside the damn thing". How can this equality be justified?
Reading the pdf version that I have Hall defines operator valued integration by the property that $$\langle \int_Xf(t)B(t)vdt,w\rangle=\int_Xf(t)\langle B(t)v,w\rangle dt ,\qquad\qquad v,w\in \mathcal H.$$ Which is different from your definition in the sense that $v$ may not equal $w$. With this definition the property follows by uniqueness of the operator. One can show that both of these operators satisfy the same uniqueness property. For this let $v\in\mathcal H$ then we have \begin{align*} \langle \int_Xf(t)AB(t)dt v,w\rangle&=\int_Xf(t)\langle B(t) v, A^\ast w\rangle dt\\ &=\langle \int_Xf(t)B(t)dt v,A^\ast w\rangle\\ &=\langle A\int_Xf(t)B(t)dt v,w\rangle. \end{align*} Showing that $\int_Xf(t)AB(t)dt$ and $A\int_Xf(t)B(t)dt$ both satisfy the defining property.