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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 $0\leq t \leq T$ ?

2 Answers 2

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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.

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    Could you please write the proof from the first equality to your second equality and detail the proof from the second equality to the last one (density, passage to the limit, ...) ? – 2017-01-12
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    @jeff_0 There really isn't all that much of details to write. The first thing is just a routine fact from measure theory; you could show it by looking at $X_n=\max \{ 1/n,X \}$ if you really want. The second thing is just a routine fact from analysis: if a continuous function is zero on a dense subset of its domain then it is identically zero. This follows pretty much immediately from the definitions of continuity and density. – 2017-01-13
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    Could you show the two facts rigorously so that the answer is complete and i can accept it ? Thanks. – 2017-01-13
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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.