For any $x \neq 0$ let $P^x$ be the distribution of Brownian Motion started at $x$. Then $(P^x)$ is a collection of probability measures on the space of cadlag functions denoted by $\Omega$.
For any $t>0$ and $\omega \in \Omega$ we can define $X(t)(\omega) = \omega(t)$ and this gives a stochastic process.
Also, for any $T>0$ let $\theta_T(\omega)(t) = \omega(T+t)$ be the time-shifted path.
I am trying to understand why this process satisfies the following Markov property: For all $x \neq 0$, $T\ge 0$ and $f:\Omega \rightarrow \mathbb{R}$ bounded and measurable we have:
$$\mathbb{E}^{P^{x}}[f(\theta_T(\omega))|\sigma(X_t)_{t \le T}] = \mathbb{E}^{P^{X_T}}[f]$$
I do know the Markov property of Brownian Motion, i.e. that for any $T>0$ the process $B_{T+t} - B_{T}$ is a Brownian Motion independent of $\sigma(B_t)_{t \le T}$, under the measure $P^0$ on $\Omega$.
But this statement seems a lot more abstract and I am struggling to see how it is a consequence of the Markov Property of Brownian Motion.