Optional stopping with unbounded stopping times $\sigma\le \tau$ case

Question

Let $M_t$ be a càdlàg martingale process. Then it is evident, by the optional stopping theorem, that for $\mathcal F_t$-stopping times $\sigma, \tau$ (not necessarily bounded) where $\sigma\le \tau$ we have that $$\mathbb{E}[M_{\tau\wedge n}\mid\mathcal{F}_{\sigma\wedge n}]= M_{\sigma\wedge n},\quad \forall n\in\mathbb{N_0}$$ because $\tau\wedge n$ and $\sigma\wedge n$ are bounded stopping times for each $n$.

Now, is it somehow possible to show that $$\mathbb{E}[M_{\tau}\mid\mathcal{F}_{\sigma}]= M_{\sigma}$$ by some limiting argument, under some additional conditions?

Would the case when $(M_{\tau\wedge n})_{n=0}^\infty$ is a uniform integrable sequence be an enough condition for above to hold?

Are there any references for this?

Edited: Changed X to M. Changed | to \mid.

Answer 1

If $X_{\tau \wedge t}$ is uniformly integrable, this follows from the following lemma:

Lemma: If $E|X| < \infty$, $E|X_n-X| \to 0$ as $n \to \infty$, $\mathcal{M}_n$ is an increasing family of $\sigma$-algebras, and $\mathcal{M}$ is generated by the union of all $\mathcal{M}_n$, then $E[X_n|\mathcal{M}_n]$ converges to $E[X|\mathcal{M}]$ in $\mathcal{L}^1$.

Proof of the lemma: By Doob's theorem, the uniformly integrable martingale $E[X|\mathcal{M}_n]$ converges to $E[X|\mathcal{M}]$ in $\mathcal{L}^1$. Furthermore, by Jensen's inequality for conditional expectations, $E[|E[X_n|\mathcal{M}_n] - E[X|\mathcal{M}_n]] \leqslant E[E[|X_n - X| \, | \mathcal{M}_n]] = E|X_n - X| \to 0$. $\square$

It suffices to take $X_n = M_{\tau \wedge n}$ and $\mathcal{M}_n = \mathcal{F}_{\sigma \wedge n}$: then $X_n$ converges in $\mathcal{L}^1$ to $M_\tau$, while $E[X_n|\mathcal{M}_n]=M_{\sigma \wedge n}$ converges in $\mathcal{L}^1$ to $M_\sigma$.

I guess the lemma is taken from Chung and Walsch, Markov Processes, Brownian Motion, and Time Symmetry, but it seems to be quite standard.