# Importance of the right-continuity of filtration in definition of strong Markov Property

Taking the definition from wikipedia,

With $$X = (X_t : t \geq 0)$$ as a stochastic process on a probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$ with natural filtration $$\{ \mathcal{F}(t) \}_{t \geq 0}$$.

$$\mathcal{F}^+(t) = \bigcap_{s>t} \mathcal{F}(s) = \bigcap_{\varepsilon>0} \mathcal{F}(s+\varepsilon)$$

$$\mathcal{F}^+(\tau)=\{A \in \mathcal{F}: \{\tau=t\} \cap A \in \mathcal{F}^+(t) ,\, \forall t \geq 0\ \}$$

$$X$$ is said to have the strong Markov property if, for each stopping time $$\tau$$, conditioned on the event $$\{ \tau < \infty \}$$, we have that for each $$t \geq 0$$, $$X_{\tau + t}$$ is independent of $$\mathcal{F}^+(\tau)$$ given $$X_\tau$$.

What is the importance of being independent of $$\mathcal{F}^+(\tau)$$ instead of $$\mathcal{F}(\tau)$$? I know that $$\mathcal{F}^+(t)$$ is right continuous, i.e.

$$\bigcap_{\varepsilon > 0} \mathcal{F}^+(t+\varepsilon) = \mathcal{F}^+(t)$$ I understand this as there being no additional information for each infinitesimally small step.

Am I interpretting $$\mathcal{F}^+(t)$$ correctly as a $$\sigma$$-algebra of events $$X \subseteq \mathcal{F}^+(t)$$ s.t. $$\{ \tau \leq t \}$$ is observable?

It would be most helpful if this was explained with Brownian motion as an example.