A uniformly bounded local martingale is a martingale

Solution 1:

Let $(M_t)_{t \geq 0}$ be a uniformly bounded local martingale. Then there exists a sequence of stopping times $(\tau_k)_k$ such that $\tau_k \uparrow \infty$ and $(M_{t \wedge \tau_k})_t$ is a martingale for each $k \in \mathbb{N}$, i.e.

$$\mathbb{E}(M_{t \wedge \tau_k} \mid \mathcal{F}_s) = M_{s \wedge \tau_k}$$

for all $ s\leq t$ and $k \in \mathbb{N}$. Since $M$ is uniformly bounded, it follows from the dominated convergence theorem that

$$\begin{align*} M_s &= \lim_{k \to \infty} M_{s \wedge \tau_k} = \lim_{k \to \infty} \mathbb{E}(M_{t \wedge \tau_k} \mid \mathcal{F}_s) = \mathbb{E}(M_t \mid \mathcal{F}_s). \end{align*}$$ This shows that $M$ is a martingale.