Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $G$ be a sub-$\sigma$-field of $\mathcal{F}$. Suppose that $(X_k)_{k \in \mathbb{N}}$ is a sequence of square-integrable real-valued random variables defined on $P$ which is "not eventually $G$-measurable". By this, I mean that there does not exist any point in the sequence such that, past that point, all further terms are $G$-measurable. Suppose also that the sequence is "var-Cauchy" meaning that for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $k,l > N$ implies $\operatorname{Var}(X_k - X_l) < \epsilon$.
I have been able to show that there exists a square-integrable random variable $X$, defined on $P$, such that $\operatorname{Var}(X_{k} - X) \to 0$ as $k \to \infty$. However, is it possible for $X$ to be $G$-measurable even though the sequence is not eventually $G$-measurable?
Yes.
Let $(B_t)$ be a Brownian motion which generates a Brownian filtration $\mathcal{F}_t$ and $\mathcal{F} = \sigma \left( \bigcup_{t \geq 0} \mathcal{F}_t \right)$. Let $t_0 > 0$ be fixed time, set $\mathcal{G} = \mathcal{F}_{t_0}$ and set $X_k = B_{t_0 + \frac{1}{k}}$. Then, none of the terms in the sequence $(X_k)$ is $\mathcal{G}$-measurable. However, the sequence is Cauchy with limit $X = B_{t_0}$, which is $\mathcal{G}$-measurable.