Suppose you have a sequence of martingales converging uniformly in probability to some process, $X$ say. The first theorem of the paper gives you a couple of conditions to check to ensure that the limit $X$ is a local martingale.
Understanding the theorem statement
- What does stationarily mean? This word describes a sequence which is eventually stationary. That is, there is some random variable $N$ such that, for $n\geq N$, $\eta_n=T$.
- What is the "right" definition of a local martingale indexed by a compact set of times? It seems as though it should be similar to the familiar definition, except that the reducing sequence $(\sigma_n)_{n\in\mathbb{N}}$ should converge stationarily to T, as this gives the useful property that $\lim_{n\to\infty}X^{\sigma_n}_T = X_T.$ This is a property that the familiar local martingales have, which doesn't necessarily hold if you drop the stationarity requirement, since $X$ isn't necessarily continuous at $T$.
Understanding the theorem proof
Here we define a sequence of stopping times $(\tau_n)_{n\in\mathbb{N}}$ is such a way that the system is well-bahaved up to time $\tau_n$, for each $n$, but so that we still have $\tau_n\uparrow T$ - note here that $X$ is cadlag and thus bounded on compacts.
I think the author passes to a subsequence here as then that sequence $(\tau_{n_k})_{k\in\mathbb{N}}$ converges stationarily to $T$, and thus, since $\eta$ converges stationarily to $T$, $\sigma$ does also, which would fit with the above definition of a local martingale.
$(\Delta X_{\tau_n})^- = (X_{\sigma_n}-X_{\sigma_n-})^-\leq n - \theta_n, $ since $\sigma_n\leq \tau_n$, and $\sigma_n\leq \eta_n$.
Now note that it's not necessarily true that $M^n_{t\wedge\sigma_m}\geq X_{t\wedge\sigma_m} -1.$
Why? Take, for example $t=T$, and suppose $X$ jumps far from $M^n$ at time $\sigma_m$. For this reason, we need to take into account a possible jump of $X$ at time $t\wedge \sigma_m$, which we've just shown is bounded by $m-\theta_m$. This gives the required inequality.
We can now apply Fatou's lemma as follows: $\liminf_n ~~\mathbb{E}[ M^n_{t\wedge\sigma_m} - (2\theta_m-m-1)]\geq \mathbb{E} [X_{t\wedge\sigma_m} - (2\theta_m-m-1)],$
and we can cancel $2\theta_m-m-1$ from both sides since it is integrable. Alternatively, as TheBridge mentions, we can learn this as an extension of Fatou's lemma.
I hope that helps!