The elementary Markov Property as I learned it reads:
For a positive real number $a$ the stochastic processes $(B_t)_{0 \leq t \leq a}$ and $(W_t)_{t \geq 0}:=(B_{t+a}-B_a)_{t \geq 0}$ are independent, i.e. $\sigma(B_t:0 \leq t \leq a) \perp \!\!\! \perp \sigma(W_t:t\geq 0)$.
How can I conclude the following statement?
For any measurable function $F:C_0 \rightarrow \mathbb{R}$, $$\mathbb{E}\big[F((W_t)_{t\geq 0})|\sigma(B_s:s \leq a)\big]=\mathbb{E}\big[F((B_t)_{t\geq0})\big].$$
Hints: it can be shown that the Borel sigma algebra on $C_0$ is the smallest sigma algebra which makes the evaluation maps $f \to f(t)$ ($t \in \mathbb R$) measurable. From this it follows immediately that $\omega \to (W_t(\omega))_t\geq 0$ is a measurable function from $\Omega$ with the sigma algebra $\sigma((W_t): \geq 0)$ into $C_0$. Hence $F(W_t(\omega))_t)$ is measurable w.r.t. $\sigma((W_t):t \geq 0)$ from which the result follows by independence and the fact that $(W_t: t\geq 0)$ is also a standard Brownian motion. [ So it induces the same measure on $C_0$ as $(B_t: t \geq 0)$.