If $M$ is a local martingale and $τ:=\inf\left\{t≥0:|M_t|≥ε\right\}$, then $\text P[[M]_∞≥δ]≤\text P[τ<∞]+\text P[[M]_τ≥δ]$

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge0}$ be a right-continuous filtration of $\mathcal A$
• $M$ be an almost surely continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $M_0=0$ almost surely
• $[M]$ denote the quadratic variation of $M$

Note that $[M]$ is increasing and hence convergent with $$[M]_t\xrightarrow{t\to\infty}[M]_\infty:=\sup_{n\in\mathbb N}[M]_n\;.\tag1$$ Now, let $$\tau:=\inf\left\{t\ge0:|M_t|\ge\varepsilon\right\}$$ for some $\varepsilon>0$. Note that $M^\tau$ is a bounded $\mathcal F$-martingale and hence $$N:=(M^2-[M])^\tau$$ is an uniformly integralbe $\mathcal F$-martingale with $N_0=0$ almost surely.

Let $\delta>0$. I want to that $$\operatorname P\left[[M]_\infty\ge\delta\right]\le\operatorname P\left[\tau<\infty\right]+\operatorname P\left[[M]_\tau\ge\delta\right]\;.\tag2$$

Clearly, $$\left\{[M]_\infty\ge\delta\right\}=\left\{[M]_\infty\ge\delta\text{ and }\tau<\infty\right\}\cup\left\{[M]_\infty\ge\delta\text{ and }\tau=\infty\right\}\;.\tag3$$ So, it seems like we need to show that $$\left\{[M]_\infty\ge\delta\text{ and }\tau=\infty\right\}\subseteq\left\{[M]_\tau\ge\delta\right\}\tag4\;.$$ How can we do that?

Answer 1

I'm not sure if I am missing something, but doesn't $$ \left[ M\right]_\infty \geq \delta \text{ and } \tau = \infty $$ directly imply $$ \left[ M\right]_\tau = \left[ M\right]_\infty \geq \delta $$ ?