# Prove that a bounded local martingale is a martingale

I would like to know if my proof of this fact is correct : A bounded local martingale is a martingale.

The sequence of stopping time has the following property : it is increasing and it converges to $$+\infty$$ such that for all $$n$$ $$1_{\tau_n}X_{t}^{\tau_n}$$ is a martingale.

Attempt : We notice that for all $$t\geq 0$$ we have $$X_{t}(\omega) = \lim_{n\to\infty}X_{t}^{\tau_n}(\omega)$$ almost surely. Moreover there exists $$M>0$$ such that $$\forall t\geq 0$$ we have $$X_t(\omega)\leq M$$ almost surely. Also $$lim_{n\to\infty}1_{\tau_n>0}=1$$ almost surely. Thus we have that $$1_{\tau_n>0} X_{t}^{\tau_n}$$ converges almost surely to $$X_t$$ and it is dominated, so we have the $$L_1$$ convergence also. Now by the continuity of the operator conditional expectation we get for $$s>t$$

$$\mathbb{E}(X_s | \mathcal{F}_t) = \mathbb{E}(\lim_{n\to\infty}1_{\tau_n>0} X_{s}^{\tau_n} | \mathcal{F}_t) = \lim_{n\to\infty}\mathbb{E}(1_{\tau_n>0} X_{s}^{\tau_n} | \mathcal{F}_t) = **\lim_{n\to\infty} 1_{\tau_n>0}\mathbb{E}( X_{s}^{\tau_n} | \mathcal{F}_t)**= \lim_{n\to\infty} 1_{\tau_n>0}X_{t}^{\tau_n} = X_t$$

Is this seems correct please ?

Thank you a lot !

The step with double stars is in fact false !

I would define what $$\tau_n$$ is (i.e. a localizing sequence for $$X$$), even though it can be inferred from the context. It's also not obvious to me why you take $$1_{\tau_n > 0}$$ out of the conditional expectation in the penultimate inequality. The definition of localizing sequence I'm familiar with is that $$1_{\tau_n > 0} X_t^{\tau_n}$$ is a martingale, so you can directly state $$\mathbb{E}[1_{\tau_n > 0} X_s^{\tau_n}|\mathcal F_t] = 1_{\tau_n > 0} X_t^{\tau_n}.$$