$ \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 bounded and measurable. 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 $F: \RR^d \times \Omega \to \RR^d$ by $$ F(x, \cdot) := x + \int_0^T \sigma(s, x) \diff B_s. $$

Let $X_0 : \Omega \to \RR^d$ be a $\mathcal F_0$-measurable random variable. Inspired by this answer, I would like to ask:

Is it true that $F(X_0, \cdot) = X_0 + \int_0^T \sigma(s, X_0) \diff B_s$ a.s.?

Thank you so much for your elaboration!

**My attempt:** For $n \in \NN^*$ and $k\in \{0, 1, \cdots, n\}$, let $t^n_k := \frac{kT}{n}$. We define random variables
\begin{align}
A^n (x) &:= \sum_{k=0}^{n-1} \sigma (t^n_k,x) (B_{t^n_{k+1}} - B_{t^n_{k}}), \\
B^n &:= \sum_{k=0}^{n-1} \sigma (t^n_k, X_0) (B_{t^n_{k+1}} - B_{t^n_{k}}).
\end{align}

By properties of Itô stochastic integral, for each $x \in \RR^d$ there is a subsequence $n \mapsto\varphi_n^x$ such that $A^{\varphi_n^x}(x) \xrightarrow{n \to \infty} F(x, \cdot)$ a.s. Similarly, there is a subsequence $n \mapsto\psi_n$ such that $B^{\psi_n} \xrightarrow{n \to \infty} X_0 + \int_0^T \sigma(s, X_0) \diff B_s$ a.s.