Let $M$ be a right-continuous local martingale such that $M^*_t \in L^1(P)$ for all $t \in \mathbb{R}_+$. Here \begin{align*} M^*_t(\omega) = \sup_{0 \leq s \leq t} |M_s(\omega)|. \end{align*}

Now, I want to prove that $M$ is a martingale.

Since $M^*_t \in L^1(P)$ we know that $\sup_{0 \leq s \leq t} |M_s(\omega)| < \infty$.

So there is a localizing sequence $(\tau_k)_{k \geq0}$ such that $P(\tau_k \uparrow \infty) = 1$ and $\forall k, M^{\tau_k} = \{M_{t \wedge \tau_k } : t \in \mathbb{R}_+\}$ is a martingale.

Since, \begin{align*} \mathbb{E}[M_{t \wedge \tau_k } \mid \mathcal{F}_s] &= M_{s \wedge \tau_k }\\ \mathbb{E}[M_{t \wedge \tau_k }] &= \mathbb{E}[M_{s \wedge \tau_k }] \end{align*} here $s<t$.

So for $ A \in \mathcal{F}_s$, letting $k \to \infty$ in the equation \begin{align*} \mathbb{E}[ \mathbb{1}_A M_{t \wedge \tau_k }] = \mathbb{E}[ \mathbb{1}_A M_{s \wedge \tau_k }], \end{align*} should help me further. Unfortunately, I do not see why. How to introduce the supremum property here to find for example that $\mathbb{E}[M_t]=\mathbb{E}[M_0]$? The latter implies that $M$ is indeed a martingale.


1 Answer 1


Use $\lim\limits_k M_t^{\tau_k} = M_t$, the dominated convergence theorem and your last equation to show that $E [1_A M_t] = E [1_A M_s]$ (note that $1_A|M_t^{\tau_k}|$ ist boundet by $M_t^*$ which you need here)

With $E[1_A E[M_t | \mathcal{F}_s]] = E [1_A M_t]]$ (for $A \in \mathcal{F}_s$) you get that $E[M_t | \mathcal{F}_s] = M_s$ since both sides are $\mathcal{F}_s$-measurable


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