If $E(|X_t-Y_t|^2)=0$, then $P\{\omega:|X_t(\omega)-Y_t(\omega)|=0,\forall t\in [0,T]\}=1$?

Question

Let $X_t, Y_t$ be stochastic processes with almost sure continuous paths defined on some probability space. Define $$v(t):=E(|X_t-Y_t|^2)$$ for $0\leq t\leq T$. Clearly $v(t)$ is continuous on $[0,T]$. Now suppose $v(t)=0$ for all $t\in [0,T]$. From this can we conclude that $$P\{\omega:|X_t(\omega)-Y_t(\omega)|=0,\forall t\in [0,T]\}=1$$

I am wondering how this can be proved. Any help is appreciated!

Answer 1

Hints:

1. Let $f:[0,T] \to \mathbb{R}$ be a continuous function. Show that $f$ is not identical zero, if and only if, there exist some $k \in \mathbb{N}$ and $q \in \mathbb{Q} \cap [0,T]$ such that $$|f(q)| > \frac{1}{k}.$$
2. Deduce from Step 1 and the a.s. continuity of the sample paths that $$\mathbb{P}\left(\exists t \in [0,T]: |X_t-Y_t| \neq 0 \right) \leq \sum_{q \in \mathbb{Q} \cap [0,T]} \sum_{k \in \mathbb{N}} \mathbb{P}(|X_q-Y_q|>1/k). \tag{1}$$
3. Use the fact that $\mathbb{E}(|X_q-Y_q|^2)=0$ to conclude that the right-hand side (and hence, the left-hand side) of $(1)$ equals zero.