[proofplan]
We approximate the stopping time $T$ from above by dyadic stopping times $T_n = 2^{-n}\lceil 2^n T \rceil$, reduce the strong Markov property at $T_n$ to the simple Markov property at finitely many deterministic times, and then pass to the [limit](/page/Limit) $T_n \downarrow T$ using path [continuity](/page/Continuity) and the [Dominated Convergence Theorem](/theorems/4).
[/proofplan]
[step:Approximate $T$ by dyadic stopping times and verify the Markov property at each $T_n$]
Define $T_n = 2^{-n}\lceil 2^n T \rceil$. Then $T_n$ takes values in $\{k 2^{-n} : k \geq 0\}$, $T_n \geq T$ for all $n$, and $T_n \downarrow T$ as $n \to \infty$. For each $k \geq 0$, the event $\{T_n = k 2^{-n}\} = \{(k-1)2^{-n} < T \leq k 2^{-n}\} \in \mathcal{F}_{k2^{-n}}^+$ (since $T$ is a stopping time for $(\mathcal{F}_t^+)$). The shifted process at the deterministic time $k 2^{-n}$ is
\begin{align*}
B_t^{(k)} = B_{t + k2^{-n}} - B_{k2^{-n}}, \qquad t \geq 0,
\end{align*}
which is a standard Brownian motion independent of $\mathcal{F}_{k2^{-n}}^+$ by the [Simple Markov Property](/theorems/1175)(iii) and [Independence from the Germ Field](/theorems/1177).
[guided]
The central idea is to reduce a random-time Markov property to a deterministic-time one by "rounding up" the stopping time. Since $T_n$ takes only countably many values $k2^{-n}$, we can condition on each value separately and apply the simple Markov property at the deterministic time $k2^{-n}$.
Why round up rather than down? Because $T_n \geq T$ ensures $\mathcal{F}_T^+ \subset \mathcal{F}_{T_n}^+$ (a smaller $\sigma$-algebra at the stopping time means more independence, and we need $E \in \mathcal{F}_T^+$ to also lie in $\mathcal{F}_{T_n}^+$ for the factorisation to apply). The event $\{T_n = k2^{-n}\}$ belongs to $\mathcal{F}_{k2^{-n}}^+$ because $\{T_n = k2^{-n}\} = \{T \leq k2^{-n}\} \setminus \{T \leq (k-1)2^{-n}\}$ and both constituent events are in $\mathcal{F}_{k2^{-n}}^+$ (since $T$ is a stopping time for the right-continuous filtration).
[/guided]
[/step]
[step:Establish independence at the dyadic stopping time $T_n$]
Let $A$ be a Borel subset of $C([0,\infty); \mathbb{R}^d)$ (the space of continuous paths) and let $E \in \mathcal{F}_{T_n}^+$. Define $B_*^{(n)}(t) = B_{t+T_n} - B_{T_n}$. We compute
\begin{align*}
\mathbb{P}\bigl(\{B_*^{(n)} \in A\} \cap E\bigr) &= \sum_{k=0}^\infty \mathbb{P}\bigl(\{B^{(k)} \in A\} \cap E \cap \{T_n = k 2^{-n}\}\bigr).
\end{align*}
On the event $\{T_n = k2^{-n}\}$, the shifted process $B_*^{(n)}$ coincides with $B^{(k)}$. Since $E \cap \{T_n = k2^{-n}\} \in \mathcal{F}_{k2^{-n}}^+$ and $B^{(k)}$ is independent of $\mathcal{F}_{k2^{-n}}^+$, each summand factors:
\begin{align*}
\mathbb{P}\bigl(\{B^{(k)} \in A\} \cap E \cap \{T_n = k 2^{-n}\}\bigr) = \mathbb{P}(B^{(k)} \in A) \cdot \mathbb{P}(E \cap \{T_n = k2^{-n}\}).
\end{align*}
Moreover, $\mathbb{P}(B^{(k)} \in A) = \mathbb{P}(B \in A)$ for all $k$ (each $B^{(k)}$ is a standard Brownian motion). Summing over $k$:
\begin{align*}
\mathbb{P}\bigl(\{B_*^{(n)} \in A\} \cap E\bigr) = \mathbb{P}(B \in A) \cdot \sum_{k=0}^\infty \mathbb{P}(E \cap \{T_n = k2^{-n}\}) = \mathbb{P}(B \in A) \cdot \mathbb{P}(E).
\end{align*}
[guided]
The factorisation in each summand uses two ingredients: (i) independence of $B^{(k)}$ from $\mathcal{F}_{k2^{-n}}^+$, which is the [Simple Markov Property](/theorems/1175)(iii) strengthened by [Independence from the Germ Field](/theorems/1177), and (ii) the fact that $E \cap \{T_n = k2^{-n}\}$ belongs to $\mathcal{F}_{k2^{-n}}^+$. Condition (ii) holds because $E \in \mathcal{F}_{T_n}^+$ means $E \cap \{T_n \leq t\} \in \mathcal{F}_t^+$ for all $t$, so in particular $E \cap \{T_n = k2^{-n}\} = (E \cap \{T_n \leq k2^{-n}\}) \setminus (E \cap \{T_n \leq (k-1)2^{-n}\}) \in \mathcal{F}_{k2^{-n}}^+$.
The crucial point is that $\mathbb{P}(B^{(k)} \in A)$ does not depend on $k$. This follows from stationarity of increments: each $B^{(k)}$ has the same [distribution](/page/Distribution) as $B$ itself. Without this uniformity, the sum would not factor as a product.
[/guided]
[/step]
[step:Pass to the limit $T_n \downarrow T$ via dominated convergence]
By path continuity of $B$, as $T_n \downarrow T$ we have $B_{t+T_n} - B_{T_n} \to B_{t+T} - B_T$ almost surely for each fixed $t \geq 0$, and in fact the convergence is uniform on compact time intervals. Let $E \in \mathcal{F}_T^+$ and let $F : C([0,\infty); \mathbb{R}^d) \to \mathbb{R}$ be bounded and continuous (with respect to the topology of [uniform convergence](/page/Uniform%20Convergence) on compacts). Since $T_n \geq T$ implies $\mathcal{F}_T^+ \subset \mathcal{F}_{T_n}^+$, we have $E \in \mathcal{F}_{T_n}^+$ for every $n$, so the previous step gives
\begin{align*}
\mathbb{E}\bigl[\mathbb{1}_E \cdot F(B_*^{(n)})\bigr] = \mathbb{P}(E) \cdot \mathbb{E}[F(B)].
\end{align*}
The integrand $\mathbb{1}_E \cdot F(B_*^{(n)})$ is bounded by $\|F\|_\infty$ and converges almost surely to $\mathbb{1}_E \cdot F(B_{T+\cdot} - B_T)$. By the [Dominated Convergence Theorem](/theorems/4),
\begin{align*}
\mathbb{E}\bigl[\mathbb{1}_E \cdot F(B_{T+\cdot} - B_T)\bigr] = \mathbb{P}(E) \cdot \mathbb{E}[F(B)].
\end{align*}
Since this factorisation holds for every bounded continuous $F$ and every $E \in \mathcal{F}_T^+$, the process $(B_{T+t} - B_T)_{t \geq 0}$ is a standard Brownian motion independent of $\mathcal{F}_T^+$.
[guided]
The passage to the limit is where path continuity of Brownian motion is essential. Without continuity, the convergence $B_{t+T_n} \to B_{t+T}$ could fail on a set of positive probability, and the dominated convergence argument would break down.
Note also that the inclusion $\mathcal{F}_T^+ \subset \mathcal{F}_{T_n}^+$ is specific to the choice $T_n \geq T$. If we had approximated from below ($T_n \leq T$), the inclusion would reverse and we could not use the factorisation at $T_n$ to deduce the factorisation at $T$. This is the reason for rounding up.
To verify that the limit process $(B_{T+t} - B_T)_{t \geq 0}$ is indeed a Brownian motion (not just independent of $\mathcal{F}_T^+$): the factorisation $\mathbb{E}[F(B_{T+\cdot} - B_T)] = \mathbb{E}[F(B)]$ for all bounded continuous $F$ means $(B_{T+\cdot} - B_T)$ and $B$ have the same distribution on $C([0,\infty); \mathbb{R}^d)$, hence $(B_{T+t} - B_T)_{t \geq 0}$ is a standard Brownian motion.
[/guided]
[/step]
