I have a probability space $(\Omega, \mathcal{F}, P)$ and assume that $(M_n, \mathcal{F}_n)$ is a martingale on this space satisfying $M_n\geq0$ for all $n\geq1$.
I also have two stopping times $\sigma_1$ and $\sigma_2$, with $\sigma_1 \leq \sigma_2$. These stopping times have positive probability of being infinite.
The optional stopping theorem cannot be applied directly, but I would like to like to show that $$ E(M_{\sigma_2} \mid \mathcal{F}_{\sigma_1}) = M_{\sigma_1 } \qquad \text{a.s.} $$ What I know is that $(M_{\sigma_2 \wedge n}, \mathcal{F}_{\sigma_2 \wedge n})$ is also a martingale and that $(M_{\sigma_2 \wedge n})$ is uniformly integrable. I have applied to the optional sampling theorem to the martingale $(M_{\sigma_2 \wedge n}, \mathcal{F}_{\sigma_2 \wedge n})$ to conclude that $$ E(M_{\sigma_2 \wedge n} \mid \mathcal{F}_{\sigma_1 \wedge n}) = M_{\sigma_1 \wedge n} \qquad \text{a.s.} $$ I feel I am close to the result. How can I take the above expression "to the limit"? My problem is that I do not know how to tackle the minimum operator in the filtration.
What I also know is that $M_{\sigma_2}$ closes the martingale $(M_{\sigma_2 \wedge n}, \mathcal{F}_{\sigma_2 \wedge n})$, i.e. $E(M_{\sigma_2} \mid \mathcal{F}_n) = M_{\sigma_2 \wedge n}$ because $(M_{\sigma_2 \wedge n})$ is uniformly integrable... but I am not sure how to proceed. This question is part of a bigger assignment I am doing, but I think there are enough assumptions here to conclude the desired result.
By the optional stopping theorem, we have
$$\mathbb{E}(M_{\sigma_2 \wedge n} \mid \mathcal{F}_{\sigma_1 \wedge k}) = M_{\sigma_1 \wedge k}$$
for any $k \leq n$. Since $M_{\sigma_2 \wedge n}$ converges to $M_{\sigma_2}$ in $L^1$ as $n \to \infty$, this implies
$$\mathbb{E}(M_{\sigma_2} \mid \mathcal{F}_{\sigma_1 \wedge k}) = M_{\sigma_1 \wedge k}. \tag{1}$$
Now we can apply Lévy's convergence theorem to conclude
$$\begin{align*} M_{\sigma_1} = \lim_{k \to \infty} M_{\sigma_1 \wedge k} &\stackrel{(1)}{=} \lim_{k \to \infty} \mathbb{E}(M_{\sigma_2} \mid \mathcal{F}_{\sigma_1 \wedge k}) \stackrel{\text{Lévy}}{=} \mathbb{E}(M_{\sigma_2} \mid \mathcal{F}_{\sigma_1}). \end{align*}$$