I'm reading a book that proves the following statement regarding Doob's Lp inequality:
Let be $X_t \in L^p$ a continuous time martingale and $D$ some countable subset of $[0,\infty)$**. Then $$P\Big(\sup_{s\in D\cap [0,t]}|X_s| \geq r \Big) \leq \frac{1}{r^p} E[|X_t|^p]$$
The way it's used in the book is $X_t$ is Brownian Motion: $$P\Big(\sup_{s\in [0,t]}|X_s| \geq r \Big) \leq \frac{1}{r^p} E[|X_t|^p]$$ Which is to say, $\sup$ is on an uncountable set $[0,t]$. Here I note: $$\{s: \sup_{s\in D \cap [0,t]}|X_s| \geq r \} \subset \{s: \sup_{s\in [0,t]}|X_s| \geq r \}$$
So that: $$P\Big(\sup_{s\in D\cap [0,t]}|X_s| \geq r \Big) \leq P\Big(\sup_{s\in [0,t]}|X_s| \geq r \Big)$$
So is the highlighted statement still true for the way its used later in the book?
**Also, I think this is a typo, it should be $[0,t]$
The subset $D$ should be dense in $[0,\infty)$. Then by path-continuity the supremum over $[0,t]$ equals to the supremum over $D\cap [0,t]$. This also guarantees that the supremum under question is measurable.
Taking an arbitrary countable subset does not make sense. For example, $t=2$ and $D=\{n^{-1}:n\ge 1\}$. $D$ is countable but it is clear that two suprema are different in this case.