I'm reading this question, i.e.,
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 $\mathbb{E}\left[\underset{t\in[0,T]}{\sup}|M_t|\right]<+\infty$.
Given that $M$ is square integrable how to show that $t \mapsto \mathbb{E}\left[M_t^2\right]$ is continuous.
and its answer, i.e.,
- $M_t$ 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 $\sup_{t \in [0,1]} M_{t}^2$ is in $L_1$. It follows from Jensen's inequality for conditional expectations. Indeed, we have $\mathbb{E} (M_T | \mathcal{F}_t) = M_t$. Hence
$$\color{blue}{\eta = \mathbb{E} ( M_T^2 | \mathcal{F}_t)} \ge \Bigl(\mathbb{E} ( M_T | \mathcal{F}_t) \Bigr)^2 = M_t^2 \Longrightarrow \sup_{t \in [0,T]} M_{t}^2 \le \eta$$ and $\mathbb{E} \eta = \mathbb{E} M_T^2< \infty$, because our martingale is in $L_2$.
Could you explain how $\eta$ in part (2.) of the answer is defined?
If $\eta := \mathbb{E} ( M_T^2 | \mathcal{F}_t)$ then $\eta$ certainly depends on $\mathcal F_t$ and thus on $t$. However, the proof would not work with my interpretation.
