How can I show that $\mathcal{F}_t^X$ is generated by sets of the form $F=\{(X_{t_1},\dots, X_{t_n}) \in \Gamma\}$ How can I show that $\mathcal{F}_t^X$ is generated by sets of the form $F={(X_{t_1},\dots, X_{t_n}) \in \Gamma}$ where $\Gamma \in\mathcal{B}(\mathbb{R}^n)$ and $0=t_1< \dots <t_n=t$.
Do I need some kind of continuity for this to hold. The question assumes that $X$ is left continuous.
I tried to use the definition of $\mathcal{F}_t^X=\sigma(X_s,0 \leq s \leq t)$ and tried to think of an application of the $\pi-\lambda$ theorem to prove this but I couldnt apply it. Any hints on how could I show this?
 A: By definition, $\mathcal{F}_t^X$ is the smallest $\sigma$-algebra for which $X_s$ is measurable for any $0\leqslant s\leqslant t$. 
Let 
$$
\mathcal C:=\left\{ \left\{(X_{t_1},\dots, X_{t_n})\in \Gamma\right\},\Gamma \in\mathcal{B}(\mathbb{R}^n),0=t_1< \dots <t_n=t,n\geqslant 1 \right\}.
$$
Then the $\sigma$-algebra generated by $\mathcal C$, denoted by $\sigma\left(\mathcal C\right)$, is such that $X_s^{-1}(B)$ is in $\sigma\left(\mathcal C\right)$ for all $B\in\mathcal B(\mathbb R)$. This shows that $\mathcal F_t^X\subseteq \sigma\left(\mathcal C\right)$. 
To see the other inclusion, let $\mathcal G$ be a $\sigma$-algebra for which $X_s$ is measurable for any $0\leqslant s\leqslant t$ and let us show that $\sigma\left(\mathcal C\right)\subseteq\mathcal G$. Since $\mathcal G$ is a $\sigma$-algebra, it suffices to establish that $\mathcal C\subseteq\mathcal G$. This follows from the fact that measurability of $X_s$ for all $s$ with respect to $\mathcal G$ is equivalent (using projections) to measurability of $(X_{t_1},\dots, X_{t_n})$ with respect to $\mathcal G$. Indeed, the mentioned measurability implies that for any Borel subset $B$ of $\mathbb R^n$, the set $\left\{\omega\mid (X_{t_1}(\omega),\dots, X_{t_n}(\omega))\in B\right\}$ is in $\mathcal G$.
