Discrete Ville inequality and anytime threshold validity

For any $\gamma \in (0,1)$, set ${\tau }_{\gamma }=\inf \{t\ge 0:M_t\ge 1\mathrm{/}\gamma \}$ and $A_N=\{{\tau }_{\gamma }\le N\}=\{{\max }_{0\le t\le N}{M}_t\ge 1\mathrm{/}\gamma \}$. The events $A_N$ increase and their union is $\{{\tau }_{\gamma }<\mathrm{\infty }\}=\{\mathrm{\exists }t\ge 0:M_t\ge 1\mathrm{/}\gamma \}$.

By [F1], $P(A_N)\le \gamma$ for each $N$; [F2] gives $P({\tau }_{\gamma }<\mathrm{\infty })={\lim }_{N}P(A_N)\le \gamma$.

Put ${\gamma }_n=\alpha +(1-\alpha )\mathrm{/}(n+1)$ for $n\ge 1$. Then ${\gamma }_n\downarrow \alpha$ and $1\mathrm{/}{\gamma }_n<1\mathrm{/}\alpha$. The countable-supremum event $H=\{{\sup }_{t\ge 0}{M}_t\ge 1\mathrm{/}\alpha \}$ equals ${\bigcap }_{n\ge 1}\{{\tau }_{{\gamma }_n}<\mathrm{\infty }\}$: if the supremum is at least $1\mathrm{/}\alpha$, each strictly lower threshold $1\mathrm{/}{\gamma }_n$ is exceeded by some process value; conversely, crossing every such threshold forces the supremum to be at least their limit $1\mathrm{/}\alpha$. Thus $H$ is measurable and, by step 2.1, $P(H)\le {\gamma }_n$ for every $n$.

Letting $n\to \mathrm{\infty }$ in step 3.1 gives $P(H)\le \alpha$. Taking $\gamma =\alpha$ in step 2.1 also gives $P({\tau }_{\alpha }<\mathrm{\infty })\le \alpha$.

For a discrete stopping time $\sigma$, $\{\sigma <\mathrm{\infty },M_{\sigma }\ge 1\mathrm{/}\alpha \}={\bigcup }_{t\ge 0}(\{\sigma =t\}\cap \{M_t\ge 1\mathrm{/}\alpha \})\subseteq \{{\tau }_{\alpha }<\mathrm{\infty }\}$. The event is measurable by [F3]–[F4], and step 4.1 bounds its probability by $\alpha$.