Conditional expectation wrt an increasing family of $\sigma$-algebras Let $M_t$ be a martingale in $L^2$ adapted to the filtration $\mathcal{F}_t$ and $(T_k)_{k\geq 0}$ a sequence of bounded stopping times such that $\lim_{k \to \infty} T_k = \infty$. I am trying to understand why the stopped martingale $(M^{T_k}_t)_{k\geq 0}$ converges in $L^2$ as $k \to \infty$. Clearly it can be written as $E(M_t|\mathcal{F}_{t \wedge T_k})$, but why does this converge in $L^2 $ to $M_t$?
 A: EDIT: My first answer was wrong and only proved $L^1$-convergence, so as elaborated in the comments, here the revised version:
$M^2_t$  is a submartingale by the conditional Jensen inequality and hence $E[M^2_{t∧T_k}]≤E[M^2_t]< \infty$, so the sequence $M_{t \wedge T_k}$ is an $L^2$-bounded martingale, with respect to the filtration $(\mathcal F_k)_{k \in \Bbb N}$ of $\sigma$-algebras generated by the $t \wedge T_k$-past, by the optional stopping theorem. Implicitly, I assume here that the stopping times are increasing.
By a consequence of the martingale convergence theorem and Doob's $L^p$-inequality, any $L^p$-bounded ($1<p<\infty$) discrete-time martingale $M_k$ converges $\Bbb P$-a.s. and in $L^p$ to some random variable $M_\infty$.
This means that the sequence converges in $L^2$ to some random variable, and given continuity in probability, this random variable must be $M_t$.
Without continuity assumptions, this does not work, because $L^2$-convergence implies convergence in probability, and we can exhibit a martingale that is not continuous in probability:
Are all local martingales continuous in probability?
