# Is $F: \mathbb T \times \mathbb R^d \times \Omega \to \mathbb R^d$ (constructed from Itô integral) Borel measurable in the product $\sigma$-algebra?

$$\newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\NN}{\mathbb{N}} \newcommand{\PP}{\mathbb{P}} \newcommand{\EE}{\mathbb{E}} \newcommand{\FF}{\mathbb{F}} \newcommand{\PPP}{\mathscr{P}} \newcommand{\KKK}{\mathscr{K}} \newcommand{\eps}{\varepsilon} \newcommand{\diff}{\mathop{}\!\mathrm{d}}$$

We fix $$T \in (0, \infty)$$ and let $$\TT$$ be the interval $$[0, T]$$. Let $$\sigma : \TT \times \RR^d \to \RR^d \otimes \RR^m$$ be measurable and $$a := \sigma \sigma^\top$$. We assume there exists a constant $$C>0$$ such that for $$t \in \TT$$ and $$x, y \in \RR^d$$: \begin{align} \frac{1}{C} |y|^2 \le \langle a (t, x) y, y \rangle & \le C |y|^2. \end{align}

Let $$(B_t, t \ge 0)$$ be a $$m$$-dimensional Brownian motion and $$\FF := (\mathcal F_t, t \ge 0)$$ an admissible filtration on a probability space $$(\Omega, \mathcal A, \PP)$$. We assume $$(\Omega, \mathcal A, \FF, \PP)$$ satisfies the usual conditions. We define a map $$F: \TT \times \RR^d \times \Omega \to \RR^d$$ by $$F(t, x, \cdot) := x + \int_0^t \sigma(s, x) \diff B_s.$$

Is $$F$$ Borel measurable w.r.t. the product $$\sigma$$-algebra $$\mathcal B (\TT) \otimes \mathcal B (\RR^d) \otimes \mathcal A$$?