I would like to prove the following statement (assuming that it is true)....
Let $\mathbb{F}:=\left\{\mathscr{F}_t\right\}_{t\in\mathbb{R}^+}$ be a filtration on a complete probability space $(\Omega,\mathscr{F},\mathbb{P})$ and define \begin{equation*} \mathscr{F}_\infty:= \sigma\left( \bigcup_{t\in\mathbb{R}^+} \mathscr{F}_t\right) \end{equation*} Then if $X:\mathbb{R}^+\longrightarrow \mathbb{R}$ is a stochastic process adapted to the filtration $\mathbb{F}$ and $X_t \longrightarrow X_\infty$ almost surely i.e. \begin{equation*} \lim_{t\rightarrow\infty} X_t(\omega) = X_\infty(\omega) \qquad \forall \omega\in \Omega\backslash N \end{equation*} where $N$ is some negligible set where $X_\infty:\Omega\longrightarrow \mathbb{R}$ is just some real valued function. Then $X_\infty$ is $\mathscr{F}_\infty$-measurable.
If I was to replace the time indexing set $\mathbb{R}^+$ into some finite time interval say $[0,T]$, the above statement would be one that says that $X_T$ is $\mathscr{F}_T$-measurable, and it would be easy to prove. However once the time indexing set is infinite/changed to $\mathbb{R}^+$ I am finding it difficult to prove (as the statement seems it would be intuitively correct). Any help would be greatly appreciated!