Stochastic differential equations

Question

In Oksendal's book, in the proof of uniqueness of the solution of an SDE, at the end of the proof (p.70 in latest edition), we have:

$v(t)=E(|X_t-\hat{X}_t|^2)=0, \space \space \space 0\leq t\leq T$

and then the author says that

$P(|X_t-\hat{X}_t|=0 \space \space \space \text{for all} \space \space \space t \in \mathbb{Q}\space \cap \space [0,T])=1.$

My question: Why do we need to first write the equality above for rationals, why is it not directly true for all $0\leq t \leq T$ ?

Answer 1

I don't have your book so I cannot check fully. However, this is usually done for the technical reason of countability. [0,T] is uncountable, however it's intersection with the rationals is countable. If you're taking unions or intersections of your set you could not do it over an uncountable interval.

Answer 2

For each fixed $t \in [0,T]$, the first equality implies that $P(\{ \omega : X_t(\omega)=\hat{X}_t(\omega) \})=1$. (This is just saying that a nonnegative r.v. $X$ has mean zero only if it is $0$ a.s., which is a standard fact from measure theory.) From this it follows that $P \left ( \cap_{t \in A} \left \{ \omega : X_t(\omega)=\hat{X}_t(\omega) \right \} \right )=1$ whenever $A$ is a countable subset of $[0,T]$; this is essentially just continuity of a finite measure from above. This does not immediately go through for an uncountable subset of $[0,T]$, such as $[0,T]$ itself.

However, if you have continuity of sample paths, then you can choose $A$ to be $\mathbb{Q} \cap [0,T]$ to get what you would really like, which is for $P \left ( \cap_{t \in [0,T]} \left \{ \omega : X_t(\omega)=\hat{X}_t(\omega) \right \} \right )=1$. This works because a continuous function is determined by its values on a dense subset of its domain, so if $X_t-\hat{X}_t=0$ on a dense set of values of $t$ then $X_t-\hat{X}_t=0$ for all $t \in [0,T]$.

The classic example of this kind of thing going awry is the stochastic process $X_t(\omega)=\begin{cases} 0 & t \neq \omega \\ 1 & t=\omega \end{cases}$ where $\omega$ is uniformly distributed on $[0,1]$. Then for each fixed $t$ we have $P(X_t(\omega)=0)=1$, but $P(X_t(\omega)=0 \text{ for all } t \in [0,1])=0$. This example is mentioned in Oksendal.