[proofplan] We approximate $s$ from above by a strictly decreasing [sequence](/page/Sequence) $s_n \downarrow s$, use path [continuity](/page/Continuity) to pass from the increment $(B_{t+s_n} - B_{s_n})$ to $(B_{t+s} - B_s)$, and apply the [Simple Markov Property](/theorems/1175)(iii) at each $s_n$ together with the [Dominated Convergence Theorem](/theorems/4) to obtain the independence factorisation at the right-continuous time $s$. [/proofplan]

[step:Approximate $s$ by a strictly decreasing sequence and invoke path continuity] Let $(s_n)_{n \geq 1}$ be a sequence with $s_n > s$ and $s_n \downarrow s$ strictly. By path continuity of $B$, for any fixed $t_1, \ldots, t_k \geq 0$, \begin{align*} (B_{t_1 + s_n} - B_{s_n}, \ldots, B_{t_k + s_n} - B_{s_n}) \to (B_{t_1 + s} - B_s, \ldots, B_{t_k + s} - B_s) \end{align*} almost surely as $n \to \infty$. [/step]

[step:Apply the simple Markov property at each $s_n$ and pass to the [limit](/page/Limit)] Fix $A \in \mathcal{F}_s^+$ and let $F : \mathbb{R}^{dk} \to \mathbb{R}$ be bounded and continuous. Since $\mathcal{F}_s^+ = \bigcap_{t > s} \mathcal{F}_t^B$ and $s_n > s$, we have $A \in \mathcal{F}_{s_n}^B$ for every $n$. By the [Simple Markov Property](/theorems/1175)(iii), the process $(B_{t + s_n} - B_{s_n})_{t \geq 0}$ is independent of $\mathcal{F}_{s_n}^B$, so \begin{align*} \mathbb{E}\bigl[F(B_{t_1 + s_n} - B_{s_n}, \ldots, B_{t_k + s_n} - B_{s_n}) \, \mathbb{1}_A\bigr] = \mathbb{E}\bigl[F(B_{t_1 + s_n} - B_{s_n}, \ldots, B_{t_k + s_n} - B_{s_n})\bigr] \cdot \mathbb{P}(A). \end{align*} The integrand on the left is bounded by $\|F\|_\infty$ and converges almost surely to $F(B_{t_1+s} - B_s, \ldots, B_{t_k+s} - B_s) \, \mathbb{1}_A$. By the [Dominated Convergence Theorem](/theorems/4), the left-hand side converges to $\mathbb{E}[F(B_{t_1+s} - B_s, \ldots, B_{t_k+s} - B_s) \, \mathbb{1}_A]$. The same argument (without $\mathbb{1}_A$) shows the first factor on the right converges to $\mathbb{E}[F(B_{t_1+s} - B_s, \ldots, B_{t_k+s} - B_s)]$. Therefore \begin{align*} \mathbb{E}\bigl[F(B_{t_1+s} - B_s, \ldots, B_{t_k+s} - B_s) \, \mathbb{1}_A\bigr] = \mathbb{E}\bigl[F(B_{t_1+s} - B_s, \ldots, B_{t_k+s} - B_s)\bigr] \cdot \mathbb{P}(A). \end{align*} Since this holds for every bounded continuous $F$ and every $A \in \mathcal{F}_s^+$, the process $(B_{t+s} - B_s)_{t \geq 0}$ is independent of $\mathcal{F}_s^+$. [/step]