0
$\begingroup$

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

enter image description here enter image description here

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $\infty$).

$\endgroup$
15
  • $\begingroup$ 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, 2022 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, 2022 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, 2022 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, 2022 at 13:42
  • $\begingroup$ Well, it seems like you are right, but what is wrong in my proof above? $\endgroup$
    – 0xbadf00d
    Jul 18, 2022 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.