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Let us consider the standard Brownian motion and the natural filtration $(\mathcal{F}_t^B)$. It is known that $(\mathcal{F}_t^B)$ is not right-continuous at $t=0$. But what about $t>0$? Is it true that $(\mathcal{F}_t^B)$ is not right-continuous at $t>0$? If so, could you please explain why it is not right-continuous?

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I think so, surely it just follows from Markov property. Take any time $s>0$, $W_t=B_{t+s}-B_s$ is just again a standard Brownian Motion. The filtration generated by $W_t$ is not right-continuous at 0.

$\mathcal{F}_{t+s}^B=\sigma(\mathcal{F^B_t},\mathcal{F^W_{s}}) $

In particular $\mathcal{F}_t^B=\sigma(\mathcal{F^B_t},\mathcal{F^W_0}) $ and $\mathcal{F}_{t+}^B=\bigcap\limits_{s>0}\sigma(\mathcal{F^B_t},\mathcal{F^W_s})=\sigma(\bigcap\limits_{s>0}\mathcal{F^B_t},\mathcal{F^W_s})$

Now $\sigma(\mathcal{F^B_t},\mathcal{F^W_{s}})$ and $\sigma(\bigcap\limits_{s>0}\mathcal{F^B_t},\mathcal{F^W_s})$ are clearly different, because $\mathcal{F^B_t}$ and $\mathcal{F^W_{s}}$ are indepdnent, also $\bigcap\limits_{s>0}\mathcal{F^B_t}$ and $\mathcal{F^W_s}$ are independent plus the fact $\mathcal{F^B_t}$ and $\bigcap\limits_{s>0}\mathcal{F^B_t}$ generate different $\sigma$-algebras.

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