Martingales in continuous time

Let $\left\{ N_{t}: 0 \leq t \leq 1\right\}$ be a family of negligible sets, for Lebesgue measure on $\mathcal{B}[0,1]$, such that $\cup_{t} N_{t} = [0,1]$. For $\omega \in \Omega : = [0,1]$, define $X_{t}(\omega ): = \left\{ \omega \in N_{t}\right\}$. Show that $\left\{ X_{t}: 0 \leq t \leq 1\right\}$ is martingale with respect to the filtration $\left\{ \mathcal{B}[0, t]: 0 \leq t \leq 1\right\}$. Need it have cadlag sample paths?

Show that an adapted process $\left\{ X_{t}: t \in \mathbb {R}^{+}\right\}$ with right-continuous sample paths must be progressively measurable.

Let $\left\{ X_{t}: t \in \mathbb {R}^{+}\right\}$ be a process with continuous sample paths, adapted to a filtration $\left\{ \mathcal{F}_{t}: t \in \mathbb {R}^{+}\right\}$. Define $\tau (\omega ): = \inf \left\{ t \in \mathbb {R}^{+}: X_{t}(\omega ) \in F\right\}$, for some closed set $F$. Show that $\left\{ \tau \leq t\right\} = \left\{ \omega : \inf_{s \in Q_{t}} d\left(X_{s}(\omega ), F\right) = 0\right\}$, where $Q_{t}$ denotes the set of rational numbers in $[0, t]$. Deduce that $\tau$ is a stopping time for the filtration.