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It is stated in a proof that there are only finitely many $s$ such that $|\Delta X_s|\geq\frac{1}{2}$ on each compact interval where $X$ is a càdlàg process.

I thought of a process with sample path recursvly defined as $X_i(\omega)=1$ for $i\in[0,1/2)$ and $X_i(\omega)=X_j(\omega)+1$ for $j\in[\sum_{i=1}^{n-1}\frac{1}{2^{i}},\sum_{i=1}^n\frac{1}{2^{i}})$ and $i\in[\sum_{i=1}^{n}\frac{1}{2^{i}},\sum_{i=1}^{n+1}\frac{1}{2^{i}})$ for $n\geq1$ and all $\omega\in\Omega$ (thus X is deterministic). Then we would have $\Delta X_s>\frac{1}{2}$ for infinitely many s on the compact interval $[0,1]$ and X should be càdlàg.

peer
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    If I'm understanding your construction correctly, you will fail to have right continuity at zero regardless of how you choose the actual value at zero. (In particular, your condition $X_i(\omega)=1$ when $i \in [0,1/2)$ is not consistent with the following conditions...) – Ian Sep 04 '16 at 17:21
  • Why? X is defined to be constant to 1 at the time interval $[0,1/2)$, thus right continuous, or not? – peer Sep 04 '16 at 17:24
  • Is it? Surely $1/3 \in [1/4,1/2)$, which should instantiate the recursive condition, no? (There is at least one typo which is obfuscating the situation; for things to make sense you should have $\frac{1}{2^{n+1}}$ in the left position and $\frac{1}{2^{n-1}}$ in the right position.) – Ian Sep 04 '16 at 17:24
  • Oh, you are right. I will fix it. – peer Sep 04 '16 at 17:28
  • But if that is the only problem then you have defined $X_{1/3}(\omega)$ more than once. And this will get worse and worse in a neighborhood of zero until you have $X_0(\omega)=1$ but $X_s(\omega)$ is unbounded in a neighborhood of zero. – Ian Sep 04 '16 at 17:28
  • Oh, now you've moved the problem to $1$, where right-continuity is more or less automatic. Are you sure you don't intend to have $|\Delta X_s| \geq 1/2$ at at most finitely many points in any interval compactly contained in the time domain? – Ian Sep 04 '16 at 17:38
  • @Ian Well, the book says: "Since X has càdlàg paths, there are only a finite number of s such that $|\Delta X_s|\geq\frac{1}{2}$ on each compact interval (fixed $\omega$)" But isn't the process X defined as above a counterexample to this statement? – peer Sep 04 '16 at 17:43
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    My point here is that your $X_1(\omega)$ is $+\infty$. Do you see why this either contradicts your statement or at least contradicts its "spirit" (so that the statement can be slightly revised to get a correct statement)? – Ian Sep 04 '16 at 18:38
  • @Ian $X_1$ is not defined yet. By definining $X_i(\omega)=1$ for $i\geq 1$ we have a process defined on the time interval $[0,\infty)$. Clearly, $lim_{t−>1}=\infty$. I see. The leftsided limes for t=1 does not exist. So the process is not càdlàg. Any hints for a "proof" of the desired statement? – peer Sep 04 '16 at 19:07

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Let's take the compact interval to be $[0,1]$. Suppose there are infinitely many times $s\in[0,1]$ with $|\Delta X_s|\ge 1/2$. By compactness there is a convergent sequence of such times, and extracting a subsequence if necessary we can assume that the sequence is monotone, say monotone increasing. Thus we have $0\le s_1<s_2<\cdots\le 1$ with $s:=\lim_ns_n$, and $|\Delta X_{s_n}|\ge 1/2$ for each $n$. Because $X$ has left limits, for each $n\ge 2$ there is $t_n\in(s_{n-1},s_n)$ with $|X_{s_n}-X_{t_n}|\ge 1/4$. But this is absurd because $$ \lim_n X_{s_n}=\lim_n X_{t_n}=X_{s-}. $$

John Dawkins
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  • Thanks, but can you change x to s and $t_s$ to $t_n$? – peer Sep 06 '16 at 14:20
  • How do we know that $|X_{s_n}-X_{t_n}|\ge 1/4$ why can't it be smaller than 1/4 or even zero? – scholar Jan 29 '23 at 01:35
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    You can choose $t_n\in(s_{n-1},s_n)$ so close to $s_n$ that $|X_{t_n}-X_{s_n-}|\le 1/4$. Then $$ 1/2\le|X_{s_n}-X_{s_n-}|\le|X_{s_n}-X_{t_n}|+|X_{t_n}-X_{s_n-}|\le |X_{s_n}-X_{t_n}|+1/4, $$ so $|X_{s_n}-X_{t_n}|\ge 1/4$. – John Dawkins Jan 30 '23 at 23:56