Independence of a Stochastic Process at Distinct Time Suppose $X_t$ is a stochastic process of $t$ on $[0,\infty)$ with almost surely continuous sample path. Does $X_t$ have to be almost surely deterministic and almost surely continuous in $t$ so that $X_{t_1}$ and $X_{t_2}$ are independent of each other for arbitrary and distinct $t_1$ and $t_2$?
 A: Recall the following statement from probability theory:

Theorem: Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of pairwise independent random variables. Then $Y:=\limsup_{n \to \infty} Y_n$ is constant almost surely.

Fix $t \geq 0$ and a sequence $(t_n)_{n \in \mathbb{N}}$, $t_n \neq t$, such that $t_n \to t$. By the continuity of the process $(X_s)_{s \geq 0}$, we know that $$Y_n := X_{t_n} \to X_t =: Y \qquad \text{almost surely as $n \to \infty$.}$$ Appyling the above lemma, we get that $X_t$ is constant almost surely. Since this holds for all $t \geq 0$, we conclude from the continuity of the process that $X_t=f(t)$ almost surely for some continuous function $f:[0,\infty) \to \mathbb{R}$ (which does not depend on $\omega$).
Remark: Note that the situation is totally different if we assume that $(X_t)_{t \geq 0}$ has independent increments. This leads to so-called additive processes.

Proof of the theorem:  We use the following version of the Borel-Cantelli lemma (for a proof see e.g. Kai Lai Chung: A Course in Probability Theory, Theorem 4.2.5 + Corollary):

Let $(A_n)_{n \in \mathbb{N}}$ be a sequence of pairwise independent events. Then $$\mathbb{P} \left( \limsup_{n \to \infty} A_n \right) \in \{0,1\}. \tag{1}$$

Fix $c \in \mathbb{R}$ and set
$$A_n := \{Y_n \geq c\}.$$
Since the sequence $(Y_n)_{n \in \mathbb{N}}$ is pairwise independent, the sequence $(A_n)_{n \in \mathbb{N}}$ is also pairwise independent. Moreover,
$$\begin{align*} \limsup_{n \to \infty} A_n &= \bigcap_{n \in \mathbb{N}} \bigcup_{k \geq n} \{Y_k \geq c\} = \left\{ \limsup_{n \to \infty} Y_n \geq c \right\}. \end{align*}$$
From $(1)$ we conclude
$$\mathbb{P} \left( \limsup_{n \to \infty} Y_n \geq c \right) \in \{0,1\}.$$
This means that $\limsup_{n \to \infty} Y_n$ is almost surely constant.
