Suppose $(X_n)_{n\in \mathbb{N}}=\mathbb{X}$ is a sequence of independent random variables with $X_n\to^{\mathbb{P}} X$, that is $X_n$ converges to $X$ in probability. For $\limsup_nX_n$ and $\liminf_n X_n$, we can use monotonicity to show both are $\mathcal{T}_\infty(\mathbb{X})$ measurable and are constant almost surely.
$X_n \to^{\mathbb{P}} X$ implies there exists a subsequence $(X_{n_k})_{n_k\in\mathbb{N}} \to X$ almost surely.
My question: does this imply $\limsup_{n_k}X_{n_k}=\liminf_{n_k}X_{n_k} =X$? and is $\limsup_{n_k}X_{n_k} \in \mathcal{T_\infty}(\mathbb{X})$?
For $\limsup_{n_k}X_{n_k}=\inf_{n_k}\sup_{m\geq n_k}X_{m}.$ $\sup_{m\geq n_k}X_{m}$ is $\mathcal{T}_\infty(\mathbb{X})$ measurable since all these $X_m$'s are and by monotonicity so is the infimum.
But does this $\limsup_{n_k}X_{n_k}=X?$
If $\{a_n\} \subset \mathbb R$ and $a_n \to a$ then $$\limsup a_n =\liminf a_n =a.$$However there is one point you have to note here. If $X$ is almost surely equal to a $Z$ where $Z$ is measurable w.r.t. some $\sigma$-algebra, say $\mathcal T_{\infty}$ you cannot say $X$ is measurable w.r.t. that sigma-algebra. To make things work you have to complete the $\sigma$-algebra. Once you do this everything works fine, $X$ is indeed almost surely constant and $\limsup X_{n_{k}} =\liminf X_{n_{k}}=X$ almost surely.