Show that $\inf\{s\geq 0\mid |M_s|\geq d\}$ is a stopping time. 
Let $(M_t)$ be a martingale and $(\mathcal F_t)_t$ an adapted filtration. Show that $$T=\inf\{s\geq 0\mid |M_s|\geq d\}$$ is a stopping time.

So I must show that $\{T\leq t\}\in \mathcal F_t$. We have that $$\{\omega \mid T\leq t\}=\{\omega \mid \forall u\in [0,T], |M_u(\omega )|<d \},$$
but how can I continue ?
 A: Without any additional assumptions on the regularity of the sample paths (continuity, càdlàg,...) or the filtration (completeness), the assumption does, in general, not hold true.
Counterexample Denote by $V \subseteq [0,1]$ the Vitali set (this is an uncountable set which is not contained in the Borel-$\sigma$-algebra $\mathcal{B}(\mathbb{R})$ generated by the Euclidean metric). Define a stochastic process on $\Omega := [0,1]$ endowed with the Lebesgue measure by
$$M_t(\omega) := \begin{cases} 1_{\{t\}}(\omega), & t \in V, \\ 0, &t \notin V. \end{cases}$$
Then $(M_t)_{t \geq 0}$ is a martingale with respect to the canonical filtration $$\mathcal{F}_t := \sigma(M_s; s \leq t)$$ since $M_t = 0$ almost surely for any $t \geq 0$. Note that each $M_t$ is Borel-measurable for $t \geq 0$, and therefore $\mathcal{F}_t \subseteq \mathcal{B}(\mathbb{R})$ for all $t \geq 0$. However, if we define
$$T := \inf\{s \geq 0; |M_s| \geq 1\}$$
then
$$\{T \leq 1\} = V \notin \mathcal{B}(\mathbb{R})$$
hence, $\{T \leq 1\} \notin \mathcal{F}_1$. This means that $T$ is not a stopping time with respect to $(\mathcal{F}_t)_t$.
