The problem is to prove that
Show that $\limsup_n X_n$ is $\mathcal{T}_\infty$-measurable for independent sequence of r.v. $(X_n)_{n=1}^\infty$, where $X_n$ are r.v. defined on $(\Omega, \mathcal{F}, \mathbb{P})$ and $\mathcal{T}_\infty = \cap_{n=1}^\infty \sigma( X_n, X_{n+1}, .....)$.
I first show the following auxiliary result:
(*) Let $j \in \mathbb{N}$ be given. Then $\limsup_n X_n = \limsup_n X_{n+j}$.
$\textbf{Proof.}$ Let $\omega \in \Omega$. Then $$ \limsup_n X_n(\omega) = \lim_{n\to \infty} ( \sup_{k \geq n} X_k(\omega) ) = \lim_{n\to \infty} ( \sup_{k \geq n+j} X_k(\omega) )= \lim_{n\to \infty} ( \sup_{k \geq n} X_{k+j}(\omega) ) = \limsup_n X_{n+j}(\omega). $$ In the second equality, I've used the fact that monotone sequence always converges in extended real numbers, and that shifting the index doesn't affect the limit of a sequence of real numbers. Since the equality holds for any $\omega$, this proves (*).
Using the result above, since $X_{n+j}$ is $\sigma(X_j , X_{j+1}, ....)$-measurable for all $ n \geq 0$, the function $ \limsup_n X_n = \limsup_n X_{n+j}$ is also $\sigma(X_j , X_{j+1}, ....)$-measurable. Since $j$ is arbitrarily chosen, we conclude that $\limsup_n X_n$ is $\mathcal{T}_\infty$-measurable.
I would also like to ask other solution to this problem. Thanks.