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In the book Markov processes and Potential Theory of Blumenthal and Getoor we can find the following result:

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I don't understand the significance of this result. If I don't misinterpret the assertion, the claim is that for allmost all $\omega\in\Omega$ and for all $t\in[0,\zeta(\omega))$, the set $\{X_s(\omega):s\in[0,t]\}$ is bounded.

However, it is a basic fact that every function $x:[a,b]\to E_\Delta$ which has left and right limits at every point is bounded. So, it seems like the assertion immediately follows.

It would clearly be a stronger statement if the claim would be that $\{X_s(\omega):s\in[0,\zeta(\omega))\text{ and }\omega\in\Omega\setminus N\}$ is bounded for some null set $N\subseteq\Omega$. But that doesn't seem to be claim and it doesn't seem to be the thing which is shown in the proof (since $n$ in the last paragraph depends on $\omega$).

What am I missing? It seems like a very basic fact is proven in a complicated way.

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1 Answer 1

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It's a simple fact, but not quite as simple as you claim. For example, the function $$ x: [a,b] \to \mathbb R\cup\{\pm\infty\} : t\mapsto \frac 1 {b-t} $$ is càdlàg but not bounded. The rough meaning of the proposition is that if $t\in[0,\infty)$ is strictly less than the first hitting time of $\Delta$ (that is, the first exit time of $E$), then the path of the process up to time $t$ is contained in a compact set. Put another way, the first hitting time of $\Delta$ is equal to $\lim_n T_n$, where the $T_n$ are as in the proof above. This relies on quasi-left continuity (on $[0,\zeta)$), and an example of a process that is not q-lc and doesn't satisfy the result is the deterministic motion $$ X_t = \tan(t \text{ (mod } \pi/2)), \qquad t\in[0,\infty). $$ Here the stopping time $T = \lim_n T_n$ is strictly less than the first hitting time of $\Delta$ (which is $\infty$).

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  • $\begingroup$ Thank you for your answer, but I need to disagree. Your example is not a valid counterexample, since your $x$ is not real-valued. $\endgroup$
    – 0xbadf00d
    Jul 18 at 13:28
  • $\begingroup$ I'm not sure what I'm missing, but assume that $(E,d)$ is a metric space and $x:[a,b]\to E$ is a function with left and right limits. Assume $x$ is unbounded. Then there is a $(t_n)_{n\in\mathbb N_0}\subseteq[a,b]$ with $d(x(t_0),x(t_n))\ge n$ for all $n\in\mathbb N$. Since $[a,b]$ is sequentially compact, $t_{n_k}\xrightarrow{k\to\infty}t$ for some increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$ and $t\in[a,b]$. $(t_{n_{k_l}})_{l\in\mathbb N}$ is monotonic for some increasing $(k_l)_{l\in\mathbb N}\subseteq\mathbb N$. $\endgroup$
    – 0xbadf00d
    Jul 18 at 13:33
  • $\begingroup$ Assume it is nondecreasing. Then, $d(x(t_{n_{k_l}}),x(t-))\xrightarrow{l\to\infty}0$ and hence $$\infty\xleftarrow{l\to\infty}n_{k_l}\le d(x(t_0),x(t_{n_{k_l}}))\le x(t_0),x(t-))+d(x(t_{n_{k_l}}),x(t-))\xrightarrow{l\to\infty}d(x(t_0),x(t-));$$ which is impossible. So, $x$ is bounded. $\endgroup$
    – 0xbadf00d
    Jul 18 at 13:33
  • $\begingroup$ My $x$ is valued in $E_\Delta$, where $E=\mathbb R$ and $\Delta$ is the infinity point $\pm\infty$ from the one-point compactification of $\mathbb R$. But in any case it is still a counter example if I take $x(t) = 1/(b-t)$ for $a\le t<b$, and $x(b) = 0$. $\endgroup$
    – user1118
    Jul 18 at 13:42
  • $\begingroup$ Well, it seems like you are right, but what is wrong in my proof above? $\endgroup$
    – 0xbadf00d
    Jul 18 at 13:44

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