I am reading the book "Introduction to Stochastic Integration (Second Edition)" by K.L. Chung and R.J. Williams. I have a question about the proof of Proposition 1.3 (on page 13) in that book. First, here are two definitions just so we are on the same page:
Definition. A collection $\{M_t, \mathcal{F}_t, t\in\mathbb{R}_+\}$ is called a martingale iff
- $M_t\in L^1$ for each $t$,
- $M_s = E(M_t\vert \mathcal{F}_s)$ for all $s<t$.
Definition. For $p\in [1,\infty)$, $M=\{M_t,\mathcal{F}_t, t\in\mathbb{R}_t\}$ is called an $L^p$ martingale iff it is a martingale and $M_t\in L^p$ for each $t$.
With these definitions in mind, the following is the statement and the proof of Proposition 1.3 (verbatim from the book):
Proposition 1.3. Let $p\in [1,\infty)$. Suppose $\{M_t^n, \mathcal{F}_t, t\in\mathbb{R}_+\}$ is an $L^p$-martingale for each $n\in\mathbb{N}$, and for each $t$, $M_t^n$ converges in $L^p$ to $M_t$ as $n\rightarrow\infty$. If $\mathcal{F}_0$ contains all of the $P$-null sets in $\mathcal{F}$, then $\{M_t, \mathcal{F}_t, t\in\mathbb{R}_+\}$ is an $L^p$-martingale.
Proof. It suffices to verify condition (2) in the definition of a martingale. Fix $s<t$ in $\mathbb{R}_+$. For each $n$, $$M_s^n = E(M_t^n\vert\mathcal{F}_s).$$ The left hand side above converges in $L^p$ to $M_s$, by hypothesis, and by Proposition 1.2, the right side converges to $E(M_t\vert \mathcal{F}_s)$ in $L^p$. Hence $$M_s = E(M_t\vert \mathcal{F}_s)\quad\text{a.s.}$$ If $\mathcal{F}_0$ contains all of the $P$-null sets in $\mathcal{F}$, it follows that $M_s\in\mathcal{F}_s$ and then the a.s. above may be removed. QED.
I understand the first sentence (condition (1) follows since $L^1\subseteq L^p$ for $1\le p\le \infty$ holds in probability spaces). The second sentence holds by definition of a martingale. The first half of the third sentence is clear, and the second half of the third sentence is true directly from Proposition 1.2 in the book which is the following:
Proposition 1.2. Suppose $\{X_n\}$ converges in $L^p$ to $X\in L^p$ for some $p\in [1,\infty)$. Then for any sub-$\sigma$-field $\mathcal{G}$ of $\mathcal{F}$, $\{E(X_n\vert\mathcal{G}\}$ converges in $L^p$ to $E(X\vert\mathcal{G})$.
The fourth sentence doesn't make perfect sense to me. In particular, I do not understand exactly why $$M_s^n\rightarrow M_s\text{ in } L^p\quad \text{and} \quad E(M_t^n\vert\mathcal{F}_s)\rightarrow E(M_t\vert\mathcal{F}_s)\text{ in } L^p\implies M_s = E(M_t\vert \mathcal{F}_s)\text{ a.s.}$$
I guess I do not really understand what the "a.s." is referring to or why we get that. Is it saying that for any representative of the equivalence class of $\mathcal{F}_s$-measurable functions satisfying the martingale property, it only equals $M_s$ almost surely? (The weird thing to me is that any two representatives of $E(M_t\vert\mathcal{F}_s)$ only equal each other almost surely, so is the a.s. in the proof "more off" in some sense?) It seems to suggest that $M_s$ is not necessarily measurable with respect to $\mathcal{F}_s$, but I do not understand why $M_s$ may not be. On a similar note, $M_s^n$ is a representative of $E(M_t^n\vert\mathcal{F}_s)$, but the limit of $E(M_t^n\vert\mathcal{F}_s)$ is $\mathcal{F}_s$-measurable while the limit of $M_s^n$ isn't? Finally, I do not really understand the fifth sentence. Why would completing the filtration fix this?
I apologize for rambling, and I am not sure if I articulated my question(s) well or not, but the point is I do not know exactly what is going in in the fourth and fifth sentences. I want to understand it down to the last $\omega$, so clarity and details, rather than brevity and terseness, would be very much appreciated.
Since $M_t^n\to M_t$ in $L^p$, there is a subsequence $(M_t^{n_k})$ s.t. $M_{t}^{n_k}\to M_t$ a.s. Since $\mathbb E[M_t^{n_k}\mid \mathcal F_s]\to \mathbb E[M_t\mid \mathcal F_s]$ in $L^p$, there is a subsequence $(\mathbb E[M_t^{n_{k_\ell}}\mid \mathcal F_s])$ s.t. $$\mathbb E[M_t^{n_{k_\ell}}\mid \mathcal F_s]\to \mathbb E[M_t\mid \mathcal F_s]\quad a.s.$$ Therefore $$M_s=\lim_{\ell\to \infty }M_{s}^{n_{k_\ell}}=\lim_{\ell\to \infty }\mathbb E[M_t^{n_{k_\ell}}\mid \mathcal F_s]=\mathbb E[M_t\mid \mathcal F_s]\quad a.s.$$ and thus $M_s=\mathbb E[M_t\mid \mathcal F_s]$ a.s.