Local martingale: how is $\sup_{s \in[0, t]}\left|M_s\right|$ measurable?

Let $$M_t$$ be a local martingale. In order to prove that it is a martingale it is sufficient to prove that $$M_t^{\tau_k} \rightarrow M_t$$ in $$L^1$$ (as $$k \rightarrow \infty$$) for every $$t$$, that is, $$\mathrm{E}\left|M_t^{\tau_k}-M_t\right| \rightarrow 0$$; here $$M_t^{\tau_k}=M_{t \wedge \tau_k}$$ is the stopped process. The given relation $$\tau_k \rightarrow \infty$$ implies that $$M_t^{\tau_k} \rightarrow M_t$$ almost surely. The dominated convergence theorem ensures the convergence in $$L^1$$ provided that

$$(*) \quad \operatorname{E} \sup_k\left|M_t^{\tau_k}\right|<\infty \quad$$ for every $$t$$.

Thus, Condition $$(*)$$ is sufficient for a local martingale $$M_t$$ being a martingale. A stronger condition

$$(**) \quad \operatorname{E} \sup_{s \in[0, t]}\left|M_s\right|<\infty$$ for every $$t$$.

is also sufficient.

My understanding It's possible that the supremum of an uncountable family of measurable functions is not measurable. Clearly, $$[0, t]$$ is uncountable.

How is $$\sup_{s \in[0, t]}\left|M_s\right|$$ measurable (so that its expectation is well-defined)?

• You cannot prove measurability. I think (**) is supposed to say that the supremum is measurable and its expectation is finite. Mar 18, 2023 at 11:43
• @geetha290krm Besides the condition that $M$ has continuous paths, are there weaker conditions that imply $\sup_{s \in[0, t]}\left|M_s\right|$ is measurable? Mar 18, 2023 at 12:09
• Right/left continuity of paths is a standard assumption. Mar 18, 2023 at 12:10