Continuity of the Second Moment of a Continuous Stochastic Process Given a continuous stochastic process with respect to time with finite variance at a given $t$. Does it necessarily imply the second moment and variance are continuous in time?
 A: Consider the probability space $([0,1],\mathcal{B}([0,1])$ endowed with the Lebesgue measure. Choose continuous functions $f_n: [0,\infty) \to [0,\infty)$ such that $\text{supp} \, f_n \subseteq (2^{-(n+1)},2^{-n})=: I_n$ and $f_n(I_n) = [0,1]$. We set
$$X_t(\omega) := \sum_{n \in \mathbb{N}} 2^{n/2} 1_{I_n}(\omega) f_n(t), \qquad \omega \in [0,1].$$
Then $t \mapsto X_t(\omega)$ is continuous for each $\omega \in [0,1]$ and $X_0(\omega)=0$. In particular, $\|X_0\|_2=0$. On the other hand, we can choose $t_n \in I_n$ such that $f_n(t_n)=1$ and therefore $$\|X_{t_n}\|_{L^2} = \|2^{n/2} 1_{I_n}\|_{L^2} = \frac{1}{\sqrt{2}}.$$ This shows $\|X_t\|_{L^2} \not \to \|X_0\|_{L^2}$ as $t \to 0$. Since $$\mathbb{E}(X_{t_n}) = \|2^{n/2} 1_{I_n}\|_{L^1} = 2^{n/2} 2^{-n+1} \to 0 \qquad \text{as} \, \, n \to \infty$$ we also get $\text{var}(X_{t}) \not \to \text{var}(X_0)=0$ as $t \to 0$.
Remark: It follows from Fatous lemma that the inequality $$\|X_t\|_{L^2} \leq \liminf_{n \to \infty} \|X_{t_n}\|_{L^2}$$ holds for any sequence $(t_n)_{n \in \mathbb{N}}$ such that $t_n \to t$.
