Convergence and continuity of a martingale I need some help with the following:
$1$. I have a $M$ continuous martingale, how can I show that if $t_n \underset{n}{\rightarrow}t$ then:
$$
M_{t_n}\overset{L^1}{\underset{n}{\rightarrow}}M_t
$$
assuming that $\mathbb{E}\left[\underset{t\in[0,T]}{\sup}|M_t|\right]<+\infty$
$2$. Given that $M$ is square integrable how to show that
$t \mapsto \mathbb{E}\left[M_t^2\right]$ is continuous.
thanks
 A: *

*$M_t$ is continuous, $M_{t_n}(\omega) \to M(t)(\omega)$, hence we have almost sure convergence.

Further, $|M_{t_n}| \le \sup_{t \in [0,1]} M_t = \eta$.  As $\mathbb{E} \eta < \infty$ we have $M_{t_n} \to  M_t$ in $L_1$ by dominated convergence theorem.


*In general case it's not true - so I think you meant that $M$ is countinious in the second problem also.

As in previous case, it's sufficient to prove that $$\sup_{t \in [0,1]} M_{t}^2 \in L_1. \quad (*)$$
Previous version of proof of (*) (it's not correct, the correct version is above)
It follows from Jensen's inequality for conditional expectations. Indeed, we have
$\mathbb{E} (M_T | \mathcal{F}_t) = M_t$. Hence
$$ \eta =  \mathbb{E} ( M_T^2 | \mathcal{F}_t) \ge \Bigl(\mathbb{E} ( M_T | \mathcal{F}_t) \Bigr)^2 = M_t^2 \Longrightarrow \sup_{t \in [0,T]} M_{t}^2 \le \eta$$
and
$\mathbb{E} \eta = \mathbb{E} M_T^2< \infty$, because our martingale is in $L_2$.
New version of proof of (*). $X_t$ is martingale and hence $X_t^2$ is submartingale and $X_t^2 \ge 0$. Hence by Doob's maximal inequality  (see https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality or
https://fabricebaudoin.wordpress.com/2012/04/10/lecture-11-doobs-martingale-maximal-inequalities/) we get
$$\mathbf{E}\left[\sup_{0 \leq s \leq T} |X_s|^p\right] \leq \left(\frac{p}{p-1}\right)^p\mathbf{E}[|X_T|^p]$$ and if $p=2$ we have
$\mathbf{E}\left[\sup_{0 \leq s \leq T} |X_s|^2\right] < \infty$.
