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.

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    $\begingroup$ 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...) $\endgroup$ – Ian Sep 4 '16 at 17:21
  • $\begingroup$ Why? X is defined to be constant to 1 at the time interval $[0,1/2)$, thus right continuous, or not? $\endgroup$ – peer Sep 4 '16 at 17:24
  • $\begingroup$ 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.) $\endgroup$ – Ian Sep 4 '16 at 17:24
  • $\begingroup$ Oh, you are right. I will fix it. $\endgroup$ – peer Sep 4 '16 at 17:28
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    $\begingroup$ 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)? $\endgroup$ – Ian Sep 4 '16 at 18:38

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 $$ \lim_n X_{s_n}=\lim_n X_{t_n}=X_{s-}. $$

  • $\begingroup$ Thanks, but can you change x to s and $t_s$ to $t_n$? $\endgroup$ – peer Sep 6 '16 at 14:20

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