$ \newcommand{\bR}{\mathbb{R}} \newcommand{\bN}{\mathbb{N}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bE}{\mathbb{E}} \newcommand{\coloneq}{:=} \newcommand{\colon}{:} \newcommand{\rD}{D} \newcommand{\cF}{\mathcal{F}} \newcommand{\cA}{\mathcal{A}} \newcommand{\diff}{\mathop{}\!\mathrm{d}} \newcommand{\qtext}[1]{\quad\text{#1}} \newcommand{\textq}[1]{\text{#1}\quad} \newcommand{\qtextq}[1]{\quad\text{#1}\quad} $ Let $(B_t)$ be a one-dimensional Brownian motion on a completed filtered probability space $(\Omega, \cA, (\cF_t), \bP)$. We assume that the filtration $(\cF_t)$ is admissible for $(B_t)$. Let $\sigma : \bR \to \bR$ be Lipschitz. We fix a random variable $Y \in L^2 (\cF_0) \coloneq L^2 (\Omega, \cF_0, \bP; \bR)$. Let $\{ X^Y_t : t \ge 0 \}$ be the unique solution to the SDE $$ X^Y_t = Y + \int_0^t \sigma (X^Y_s) \diff B_s . $$
We define the random variable $\tilde X^Y_t$ by $\tilde X^Y_t (\omega) \coloneq X^{Y (\omega)}_t (\omega) \coloneq \{ X^{Y (\omega)}_t \} (\omega)$ for $\omega \in \Omega$.
I would like to ask if there are references containing the following theorem:
Theorem 1 Let $t \in (0, \infty)$. Then $X^Y_s = \tilde X^Y_s$ for all $s \in [0, t]$, $\bP$-a.s.
Thank you for your elaboration. Below is my proof for above theorem.
Let $C_1$ be the Lipschitz constant of $\sigma$. We need the following lemmas
Lemma 2 Assume $Y$ takes a finite number of values $y_1, \ldots, y_n \in \bR$. Let $t \in (0, \infty)$. Then $X^Y_s = \tilde X^Y_s$ for all $s \in [0, t]$, $\bP$-a.s.
Lemma 3 For $(T, p) \in (0, \infty) \times [2, \infty)$, there exists a constant $C_2 = C_2 (T, p, C_1) >0$ such that $$ \bE [ \sup_{s \in [0, t]} |X^Y_s - X^Z_s|^p ] \le C_2 \bE [ |Y-Z|^p ] \qtextq{for all} t \in [0, T] \qtextq{and} Y, Z \in L^p (\cF_0) . $$
Lemma 4 (Kolmogorov Continuity Theorem) Let $(E, \mathrm{d})$ be a Polish space and $S$ a subset of $\bR^d$. Let $\{U_x : x \in S \}$ be a collection of $E$-valued random variables. If $\alpha, \beta, C$ are positive constants satisfying $$ \bE [\mathrm d^\alpha (U_x, U_y) ] \le C |x-y|^{d+\beta} $$ for all $x, y \in S$, then $U_x$ has a continuous modification. Furthermore, with probability one, this modification is almost surely $\gamma$-Hölder continuous on all bounded sets for all $0<\gamma<\beta / \alpha$.
Proof of Theorem 1 Let $(Y_n) \subset L^2$ be a sequence of simple functions such that $Y_n \to Y$ in $L^2$. By Lemma 2, $$ \sup_{s \in [0, t]} | X^{Y_n}_s - \tilde X^{Y_n}_s | = 0 \qtext{$\bP$-a.s.} $$
By Lemma 3, possibly passing to a subsequence, $$ \sup_{s \in [0, t]} | X^{Y_n}_s - X^{Y}_s | \to 0 \qtextq{$\bP$-a.s. as} n \to \infty. $$
By Lemma 4, there exists a modification of $\{ X^y_s : s \in [0, t] \}_{y \in \bR^d}$ such that $y \mapsto \{ X^y_s : s \in [0, t] \}$ is continuous $\bP$-a.s. w.r.t the supremum norm $\| \cdot \|_\infty$ on $C([0, t] ; \bR)$. We will also denote this modification by $\{ X^y_s : s \in [0, t] \}_{y \in \bR^d}$. Possibly passing to a subsequence, we have $Y_n \to Y$ $\bP$-a.s. It follows that $$ \sup_{s \in [0, t]} | \tilde X^{Y_n}_s (\omega) - \tilde X^{Y}_s (\omega) | = \sup_{s \in [0, t]} | X^{Y_n (\omega)}_s (\omega) - X^{Y (\omega)}_s (\omega) | \to 0 \qtext{$\bP$-a.s.} $$
This completes the proof. $$\tag*{$\blacksquare$}$$
Proof of Lemma 2 Let $A_i \coloneq \{\omega \in \Omega : Y(\omega) =y_i\}$. Then $1_{A_i}$ is $\cF_0$-measurable. We have \begin{align} X^Y_t 1_{A_i} & = 1_{A_i} \Big [ Y + \int_0^t \sigma (X^Y_s) \diff B_s \Big ] & = y_i 1_{A_i} + \int_0^t \sigma (X^Y_s) 1_{A_i} \diff B_s , \\ X^{y_i}_t 1_{A_i} & = 1_{A_i} \Big [ y_i + \int_0^t \sigma (X^{y_i}_s) \diff B_s \Big ] & = y_i 1_{A_i} + \int_0^t \sigma (X^{y_i}_s) 1_{A_i} \diff B_s . \end{align}
Step 1: We assume that $\sigma (0) = 0$, so $\sigma (X^Y_s) 1_{A_i} = \sigma (X^Y_s 1_{A_i})$ and $\sigma (X^{y_i}_s) 1_{A_i} = \sigma (X^{y_i}_s 1_{A_i})$. Then \begin{align} X^Y_t 1_{A_i} & = y_i 1_{A_i} + \int_0^t \sigma (X^Y_s 1_{A_i}) \diff B_s , \\ X^{y_i}_t 1_{A_i} & = y_i 1_{A_i} + \int_0^t \sigma (X^{y_i}_s 1_{A_i}) \diff B_s , \end{align} which implies $$ X^Y_t 1_{A_i} - X^{y_i}_t 1_{A_i} = \int_0^t [ \sigma (X^Y_s 1_{A_i}) - \sigma (X^{y_i}_s 1_{A_i}) ] \diff B_s . $$
Thus \begin{align} \bE [ \sup_{s \in [0, t]} |X^Y_s 1_{A_i} - X^{y_i}_s 1_{A_i}|^2 ] & \le \bE \Big [ \sup_{s \in [0, t]} \Big | \int_0^s [ \sigma (X^Y_r 1_{A_i}) - \sigma (X^{y_i}_r 1_{A_i}) ] \diff B_r \Big |^2 \Big ] \\ & \le C_3 \bE \Big [ \int_0^t | \sigma (X^Y_r 1_{A_i}) - \sigma (X^{y_i}_r 1_{A_i})|^2 \diff r \Big ] \\ & = C_3 \int_0^t \bE [ | \sigma (X^Y_r 1_{A_i}) - \sigma (X^{y_i}_r 1_{A_i})|^2 ] \diff r \\ & \le C_1 C_3 \int_0^t \bE [ | X^Y_r 1_{A_i} - X^{y_i}_r 1_{A_i}|^2 ] \diff r \\ & \le C_1 C_3 \int_0^t \bE [ \sup_{s \in [0, r]} | X^Y_s 1_{A_i} - X^{y_i}_s 1_{A_i}|^2 ] \diff r . \end{align}
The constant $C_3$ is due to Burkholder-Davis-Gundy inequality. Let $$ f_i (t) \coloneq \bE [ \sup_{s \in [0, t]} |X^Y_s 1_{A_i} - X^{y_i}_s 1_{A_i}|^2 ] \qtextq{for} t \ge 0. $$
Then $t \mapsto f_i (t)$ is continuous. By Gronwall's lemma $f_i (t) =0$ for all $t \ge 0$. Notice that $\tilde X^Y_t 1_{A_i} = X^{y_i}_t 1_{A_i}$. The claim then follows.
Step 2 We define $\tilde \sigma : \bR \to \bR$ by $\tilde \sigma (x) \coloneq \sigma (x) - \sigma (0)$. Then $\tilde \sigma$ is Lipschitz and $\tilde \sigma (0) = 0$. Also, \begin{align} X^Y_t 1_{A_i} & = y 1_{A_i} + \sigma (0) \int_0^t 1_{A_i} \diff B_s + \int_0^t \tilde \sigma (X^Y_s 1_{A_i}) \diff B_s , \\ X^{y_i}_t 1_{A_i} & = {y_i} 1_{A_i} + \sigma (0) \int_0^t 1_{A_i} \diff B_s + \int_0^t \tilde \sigma (X^{y_i}_s 1_{A_i}) \diff B_s . \end{align}
Repeating the procedure in Step 1 completes the proof. $$\tag*{$\blacksquare$}$$
