When is $\lim\limits_{t \rightarrow \infty} \mathbb{E}[X|\mathcal{F}_{t}] =\mathbb{E}\left[X|\lim\limits_{t\rightarrow\infty}\mathcal{F}_{t}\right]$?
Is there a theorem like monotone convergence or dominated convergence for a problem of this sort?
One specific case of interest would be when $\{\mathcal{F}_{t}\}$ is a sequence of sub-sigma algebras such that $\forall s<t[\mathcal{F}_{s}\subseteq \mathcal{F}_{t}]$ (that is, it is non-decreasing).
You should state all the premises and define your notations. For example, let $(\Omega,\mathcal{F},P)$ be a probability space and let $\{\mathcal{F}_{t}\mid t\geq0\}$ be a filtration. Let $X:\Omega\rightarrow\mathbb{R}$ be an integrable random variable.
Define $\mathcal{F}_{\infty}=\sigma\left(\cup_{t}\mathcal{F}_{t}\right)$. For each $t\geq0$, note that $E\left[X\mid\mathcal{F}_{t}\right]$ is only determined a.e. What is the sense of convergence $\lim_{t\rightarrow\infty}E\left[X\mid\mathcal{F}_{t}\right]$ ? Pointwisely a.e. ?
If this is what you want, you need to be careful: For the case of sequence, if we want to talk about pointwise a.e. convergence $X_{n}\rightarrow X$, for each $n$, we may modify $X_{n}$ on a $P$-null set and it would not affect the conclusion. However, for limit process involving uncountably terms, like $X_{t}\rightarrow X$, we are not allow to "For each $t$, modify $X_{t}$ on a $P$-null set". Now, we immediately encounter a problem: What is $E[X\mid\mathcal{F}_{t}]$? It is not a concrete random variable, but it is only determined a.e.. It is true that $\{E[X\mid\mathcal{F}_{t}]\mid t\geq0\}$ is always a martingale. However, its sample paths are out of control. Note that, if the filtration is standard, we can always choose a cadlag modification for $\{E[X\mid\mathcal{F}_{t}]\mid t\geq0\}$ (a deep result due to Doob) then invoke Martingale Convergence Theorem (uniformly integrable version) and conclude that $\lim_{t\rightarrow\infty}E[X\mid\mathcal{F}_{t}]=E[X\mid\mathcal{F}_{\infty}]$ pointwisely a.e.