# What is the significance of Blumenthal and Getoor's result on the boundedness of paths of a standard Markov process?

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

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.

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 $$\infty$$).
• Thank you for your answer, but I need to disagree. Your example is not a valid counterexample, since your $x$ is not real-valued. Jul 18, 2022 at 13:28
• 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 $(k_l)_{l\in\mathbb N}\subseteq\mathbb N$. Jul 18, 2022 at 13:33
• 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. Jul 18, 2022 at 13:33
• 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 $x(b) = 0$. Jul 18, 2022 at 13:42