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I'm reading about local martingale from this Wikipedia page, i.e.,


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)?

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  • $\begingroup$ You cannot prove measurability. I think (**) is supposed to say that the supremum is measurable and its expectation is finite. $\endgroup$ Mar 18, 2023 at 11:43
  • $\begingroup$ @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? $\endgroup$
    – Akira
    Mar 18, 2023 at 12:09
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    $\begingroup$ Right/left continuity of paths is a standard assumption. $\endgroup$ Mar 18, 2023 at 12:10
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    $\begingroup$ @geetha290krm Could you post your previous comments as an answer? $\endgroup$
    – Akira
    Mar 18, 2023 at 12:19

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Measurability of the supremum cannot be proved. If the process has right(or left) continuous paths then the supremum is measurable. I think condition (**) is to be interpreted as saying that the supremum is measurable and has finite mean.

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