right continuity of the filtration generated by coordinate mappings Suppose $E$ is an interval in the real line. Let $(E^{\mathbb{R}_+}, \mathscr{F})$ be the canonical probability space. Suppose $\{X_t\}_{t \geq 0}$ is the collection of coordinate mappings and let $\{\mathscr{F}_t\}$ be the filtration generated by $\{X_t\}$. Is $\{\mathscr{F}_t\}$ right continuous, i.e., is $$\mathscr{F}_t = \cap_{s > t} \mathscr{F}_s$$ true for all $t \geq 0$?
 A: No.  For a counter example, we have that $$A:=\left\{\lim_{\substack{t \downarrow 0 \\ t \in \mathbb{Q}}} X_t = 0\right\} \in \bigcap_{s > 0} \mathcal F_s$$ but $\mathcal F_0 = \sigma(X_0)$ so $A \not \in \mathcal F_0$.  I will show $$B:=\left\{\limsup_{\substack{t \downarrow 0 \\ t \in \mathbb{Q}}} X_t \le 0\right\} \in \bigcap_{s > 0} \mathcal F_s,$$ the arguments for $\liminf X_t \ge 0$ are similar.  We write \begin{align*}
\left\{\limsup_{\substack{t \downarrow 0 \\ t \in \mathbb{Q}}} X_t \le 0\right\} &= \bigcap_{n \in \mathbb{N}} \bigcup_{T \in \mathbb{Q}} \bigcap_{\substack{t \in \mathbb{Q} \\t < T}}\left\{X_t \le \frac 1n\right\}.
\end{align*}
Note that for any $s > 0$ and $n \in \mathbb{N}$ we have $$\bigcup_{T \in \mathbb{Q}} \bigcap_{\substack{t \in \mathbb{Q} \\t < T}}\left\{X_t \le \frac 1n\right\} = \bigcup_{\substack{T \in \mathbb{Q} \\ T \le s}} \bigcap_{\substack{t \in \mathbb{Q} \\t < T}}\left\{X_t \le \frac 1n\right\},$$ so $B \in \mathcal F_s$ for all $s > 0$.  Therefore we conclude $B \in \bigcap_{s > 0} \mathcal F_s$.
For a more concrete example, we could consider a random variable $Y$ with $\mathbb{P}(Y = 1) = \mathbb{P}(Y = -1) = \frac 12$ and define $X_t := Yt$.  Then $Y$ is measurable with respect to $\bigcap_{s > 0} \mathcal F_s$ but not with respect to $\mathcal F_0$.
