$ \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 $$ F(t, x, \cdot) := x + \int_0^t \sigma(s, x) \diff B_s. $$

Inspired by this answer, I would like to ask:

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

Thank you so much for your elaboration!



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