I've been studying Martingale theory from William's Probability With Martingales, and I have a question.
Let $X\in L^1(\Omega)$ be a real function, and let $\{\mathcal{F}_{n}\}_{n=1}^{\infty}$ be a filtration on the probability space $\Omega$. Define $M_{n} := \mathbf{E}[X\mid \mathcal{F}_{N}]$. Then the set $\{M_{n}\}_{n=1}^{\infty}$ forms a martingale with respect to the filtration.
Williams proves that there exists a limit $M_\infty = \lim_{n \to \infty} M_{n}$ in both $L^1$ and almost surely. My question is, are there easy-to-understand, or insightful, necessary and sufficient conditions for when $M_{\infty} = X$.
One answer is to define the limit $\sigma$-algebra $\mathcal{F}_{\infty}:= \bigcup_{n=1}^{\infty}\mathcal{F}$. The limit function $M_\infty$ is $\mathcal{F}_{\infty}$-measurable and $M_\infty = \mathbf{E}[X\mid \mathcal{F}_\infty]$. Thus, we have $M_\infty = X$ if and only if $\sigma(X)\subseteq \mathcal{F}_\infty$.
Is this the only statement which is true in general? Can we add any additional conditions on $X$ which yield different results? Is there a cleaner characterization of when the martingale converges to the function which creates it?