Convergence of martingales is a martingale

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 ${M_t, mathcal{F}_t, tinmathbb{R}_+}$ is called a martingale iff

1. $M_tin L^1$ for each $t$,
2. $M_s = E(M_tvert mathcal{F}_s)$ for all $s<t$.

Definition. For $pin [1,infty)$, $M={M_t,mathcal{F}_t, tinmathbb{R}_t}$ is called an $L^p$ martingale iff it is a martingale and $M_tin L^p$ 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 $pin [1,infty)$. Suppose ${M_t^n, mathcal{F}_t, tinmathbb{R}_+}$ is an $L^p$-martingale for each $ninmathbb{N}$, and for each $t$, $M_t^n$ converges in $L^p$ to $M_t$ as $nrightarrowinfty$. If $mathcal{F}_0$ contains all of the $P$-null sets in $mathcal{F}$, then ${M_t, mathcal{F}_t, tinmathbb{R}_+}$ 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 $n$, $$M_s^n = E(M_t^nvertmathcal{F}_s).$$ The left hand side above converges in $L^p$ to $M_s$, by hypothesis, and by Proposition 1.2, the right side converges to $E(M_tvert mathcal{F}_s)$ in $L^p$. Hence $$M_s = E(M_tvert mathcal{F}_s)quadtext{a.s.}$$ If $mathcal{F}_0$ contains all of the $P$-null sets in $mathcal{F}$, it follows that $M_sinmathcal{F}_s$ and then the a.s. above may be removed. QED.

I understand the first sentence (condition (1) follows since $L^1subseteq L^p$ for $1le ple infty$ 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 $Xin L^p$ for some $pin [1,infty)$. Then for any sub-$sigma$-field $mathcal{G}$ of $mathcal{F}$, ${E(X_nvertmathcal{G}}$ converges in $L^p$ to $E(Xvertmathcal{G})$.

The fourth sentence doesn’t make perfect sense to me. In particular, I do not understand exactly why
$$M_s^nrightarrow M_stext{ in } L^pquad text{and} quad E(M_t^nvertmathcal{F}_s)rightarrow E(M_tvertmathcal{F}_s)text{ in } L^pimplies M_s = E(M_tvert mathcal{F}_s)text{ a.s.}$$

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 $E(M_tvertmathcal{F}_s)$ 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 $E(M_t^nvertmathcal{F}_s)$, but the limit of $E(M_t^nvertmathcal{F}_s)$ 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.