We fix some notation. Write $Y := F(X_0, \cdot)$, $Z := X_0 + \int_0^T \sigma(s, X_0) \, dB_s$, and consider the process $W_t := B_t - B_0$ together with its completed natural filtration $\mathcal W_t$. By construction $\mathcal W_t$ and $\mathcal F_0$ are independent, and further we have $\mathcal F_t = \mathcal W_t \vee \mathcal F_0$.

We will show that for all events $E \in \mathcal F_T$, we have

$$\mathbb E[\mathbf 1_E Y] = \mathbb E[\mathbf 1_E Z],$$ which will imply the desired conclusion that $Y = Z$ almost surely, since otherwise at least one of $Y > Z$ or $Z < Y$ happens with positive probability, and setting $E$ above to be either of these events we obtain a contradiction.

Since $\mathcal F_t = \mathcal W_t \vee \mathcal F_0$, by a standard but tedious monotone class argument it is enough to check the desired equality on events $E$ of the form $A \cap B$ for $A \in \mathcal W_T$ and $B \in \mathcal F_0$.

We write

$$\mathbb E[\mathbf 1_E Z] = \mathbb E[\mathbf 1_A \mathbf 1_B Z] = \mathbb E[\mathbf 1_B \mathbb E[\mathbf 1_A Z| \mathcal F_0]].$$

By considering the regular conditional probability with respect to $\mathcal F_0$ and applying the independence of $\mathcal W_T$ and $\mathcal F_0$ and the freezing lemma in turn, we have

$$\mathbb E[\mathbf 1_A Z| \mathcal F_0] = \mathbb E[\mathbf 1_A F(x, \cdot)]_{|x = X_0} = \mathbb E[1_A Y| \mathcal F_0]$$

and so

$$\mathbb E[\mathbf 1_E Z] = \mathbb E [\mathbf 1_B \mathbb E[\mathbf 1_A Y| \mathcal F_0 ]] = \mathbb E[\mathbf 1_E Y],$$
as desired.

Yes. We fix some notation. Write $Y := F(X_0, \cdot)$, $Z := X_0 + \int_0^T \sigma(s, X_0) \, dB_s$, and consider the process $W_t := B_t - B_0$ together with its completed natural filtration $\mathcal W_t$. By construction $\mathcal W_t$ and $\mathcal F_0$ are independent, and further we have $\mathcal F_t = \mathcal W_t \vee \mathcal F_0$.

We will show that for all events $E \in \mathcal F_T$, we have

$$\mathbb E[\mathbf 1_E Y] = \mathbb E[\mathbf 1_E Z],$$ which will imply the desired conclusion that $Y = Z$ almost surely, since otherwise at least one of $Y > Z$ or $Z < Y$ happens with positive probability, and setting $E$ above to be either of these events we obtain a contradiction.

Since $\mathcal F_t = \mathcal W_t \vee \mathcal F_0$, by a standard but tedious monotone class argument it is enough to check the desired equality on events $E$ of the form $A \cap B$ for $A \in \mathcal W_T$ and $B \in \mathcal F_0$.

We write

$$\mathbb E[\mathbf 1_E Z] = \mathbb E[\mathbf 1_A \mathbf 1_B Z] = \mathbb E[\mathbf 1_B \mathbb E[\mathbf 1_A Z| \mathcal F_0]].$$

By considering the regular conditional probability with respect to $\mathcal F_0$ and applying the freezing lemma with the independence of $\mathcal W_T$ and $\mathcal F_0$ (here we may have to assume some regularity on $\sigma$ to ensure the freezing lemma applies), we have

$$\mathbb E[\mathbf 1_A Z| \mathcal F_0] = \mathbb E[\mathbf 1_A F(x, \cdot)]_{|x = X_0} = \mathbb E[1_A Y| \mathcal F_0]$$

and so

$$\mathbb E[\mathbf 1_E Z] = \mathbb E [\mathbf 1_B \mathbb E[\mathbf 1_A Y| \mathcal F_0 ]] = \mathbb E[\mathbf 1_E Y],$$
as desired.

We fix some notation. Write $Y := F(X_0, \cdot)$, $Z := X_0 + \int_0^T \sigma(s, X_0) \, dB_s$, and consider the process $W_t := B_t - B_0$ together with its completed natural filtration $\mathcal W_t$. By construction $\mathcal W_t$ and $\mathcal F_0$ are independent, and further we have $\mathcal F_t = \mathcal W_t \vee \mathcal F_0$.

We will show that for all events $E \in \mathcal F_T$, we have

$$\mathbb E[\mathbf 1_E Y] = \mathbb E[\mathbf 1_E Z],$$ which will imply the desired conclusion that $Y = Z$ almost surely, since otherwise at least one of $Y > Z$ or $Z < Y$ happens with positive probability, and setting $E$ above to be either of these events we obtain a conclusion.

Since $\mathcal F_t = \mathcal W_t \vee \mathcal F_0$, by a standard but tedious monotone class argument it is enough to check the desired equality on events $E$ of the form $A \cap B$ for $A \in \mathcal W_t$ and $B \in \mathcal F_0$.

We compute

$$\mathbb E[\mathbf 1_E Z] = \mathbb E[\mathbf 1_A \mathbf 1_B Z] = \mathbb E[\mathbf 1_B \mathbb E[\mathbf 1_A Z| \mathcal F_0]].$$

By considering the regular conditional probability with respect to $\mathcal F_0$ and applying independence of $W_t$ and $\mathcal F_0$,

$$\mathbb E[\mathbf 1_A Z| \mathcal F_0] = \mathbb E[\mathbf 1_A F(x, \cdot)]_{|x = X_0} = \mathbb E[1_A Y| \mathcal F_0]$$

and so

$$\mathbb E[\mathbf 1_E Z] = \mathbb E [\mathbf 1_B \mathbb E[\mathbf 1_A Y| \mathcal F_0 ]] = \mathbb E[\mathbf 1_E Y],$$
as desired.

We fix some notation. Write $Y := F(X_0, \cdot)$, $Z := X_0 + \int_0^T \sigma(s, X_0) \, dB_s$, and consider the process $W_t := B_t - B_0$ together with its completed natural filtration $\mathcal W_t$. By construction $\mathcal W_t$ and $\mathcal F_0$ are independent, and further we have $\mathcal F_t = \mathcal W_t \vee \mathcal F_0$.

We will show that for all events $E \in \mathcal F_T$, we have

$$\mathbb E[\mathbf 1_E Y] = \mathbb E[\mathbf 1_E Z],$$ which will imply the desired conclusion that $Y = Z$ almost surely, since otherwise at least one of $Y > Z$ or $Z < Y$ happens with positive probability, and setting $E$ above to be either of these events we obtain a contradiction.

Since $\mathcal F_t = \mathcal W_t \vee \mathcal F_0$, by a standard but tedious monotone class argument it is enough to check the desired equality on events $E$ of the form $A \cap B$ for $A \in \mathcal W_T$ and $B \in \mathcal F_0$.

We write

$$\mathbb E[\mathbf 1_E Z] = \mathbb E[\mathbf 1_A \mathbf 1_B Z] = \mathbb E[\mathbf 1_B \mathbb E[\mathbf 1_A Z| \mathcal F_0]].$$

By considering the regular conditional probability with respect to $\mathcal F_0$ and applying the independence of $\mathcal W_T$ and $\mathcal F_0$ and the freezing lemma in turn, we have

$$\mathbb E[\mathbf 1_A Z| \mathcal F_0] = \mathbb E[\mathbf 1_A F(x, \cdot)]_{|x = X_0} = \mathbb E[1_A Y| \mathcal F_0]$$

and so

$$\mathbb E[\mathbf 1_E Z] = \mathbb E [\mathbf 1_B \mathbb E[\mathbf 1_A Y| \mathcal F_0 ]] = \mathbb E[\mathbf 1_E Y],$$
as desired.