I started reading this paper (Lamb, Charles W.. “Shorter Notes: A Short Proof of the Martingale Convergence Theorem”. Proceedings of the American Mathematical Society 38.1 (1973): 215–217) today. In paragraph 2 the author says
The martingale $(X_n,\mathscr{F}_n:n\geq{0})$ is called complete if there exists a random variable $X$ with $\mathbb{E}[X|\mathscr{F}_n] = X_n$ for $n\geq 0$. There is no loss of generality in assuming that $X$ is $\mathscr{F}_\infty$-measurable where $\mathscr{F}_\infty$ is the $\sigma$-field generated by $\bigcup \mathscr{F}_n$. It is an elementary exercise to show that a martingale is complete if and only if it is uniformly integrable. We remark only that the necessity is proved by defining a set function $\mu(A)= \lim \mathbb{E}[X_n: A]$ on the field $\bigcup \mathscr{F}_n$, proving from the uniform integrability that $\mu$ is a finite signed measure on $\bigcup \mathscr{F}_n$ [...]
I can see why $\mu$ is a well defined set function and why it is finitely additive on $\bigcup \mathscr{F}_n$, but I didn't manage to prove countable additivity. This is my approach: let $(A_k)$ be a collection of disjoint elements of $\bigcup \mathscr{F}_n$ such that $A\equiv\bigcup_k A_k \in \bigcup_n \mathscr{F}_n$, then I can show by the Monotone Convergence Theorem that for all $m\geq 0$, $\mu(A)=\lim_{m\to \infty} \sum_k \mathbb{E}[X_m:A_k]$, but I don't see how to exchange the limit and the infinite sum. I guess this has to do with the fact that $X$ is uniformly integrable, but I don't understand how.
The uniform integrability of $\{X_n\}$ implies that $\mu$ is absolutely continuous with respect to $\Bbb P$ on $\mathscr A:=\cup_n\mathscr F_n$, in the sense that for a given $\epsilon>0$ there exists $\delta>0$ such that if $B\in\mathscr A$ and $\Bbb P(B)<\delta$, then $|\mu(B)|<\epsilon$. Now let $\{A_k\}$ and $A$ be as you describe. Fix $\epsilon>0$, and then let $\delta$ correspond to $\epsilon$ as above. Because $\Bbb P$ is countable additive, there exists $N$ so large that if $n\ge N$ then $\Bbb P(\cup_{k>n}A_k)<\delta$. Observe that $\cup_{k>n}A_k = A\setminus \cup_{k=1}^n A_k$ is an element of $\mathscr A$. Thus $|\mu(\cup_{k>n}A_k)|<\epsilon$. Consequently, because $\mu(A)=\mu(\cup_{k=1}^nA_k)+\mu(\cup_{k>n}A_k)$ and $\mu$ is finitely additive on $\mathscr A$, $$ \left|\mu(A)-\sum_{k=1}^n\mu(A_k)\right|= \left|\mu\left(\cup_{k>n}A_k\right)\right|<\epsilon, $$ provided $n\ge N$. This shows that the series $\sum_{k=1}^\infty\mu(A_k)$ is convergent, and has sum $\mu(A)$.