Let $E$ be a locally compact separable$^1$ metric space, $D([0,\infty),E)$ denote the Skorohod space, $C_0(E)$ denote the space of continuous functions $E\to\mathbb R$ vanishing at infinity and $E^\ast:=E\uplus\left\{\infty\right\}$ denote the Alexandroff one-point compactification.

Moreover, let $\mathcal M_1(D([0,\infty),E))$ denote the space of probability measures on $\mathcal B(D([0,\infty),E))$ equipped with the topology of weak convergence$^1$.

Now let $X^n$ be a $D([0,\infty),E)$-valued random variable on a probability space $(\Omega,\mathcal A,\operatorname P)$ for $n\in\mathbb N$.

I want to show that if $\left(\left(f\circ X^n\right)_\ast\operatorname P\right)_{n\in\mathbb N}\subseteq\mathcal M_1(D([0,\infty),\mathbb R))$ is relatively compact for all $f\in C_0(E)$, then $\left(X^n_\ast\operatorname P\right)_{n\in\mathbb N}\subseteq\mathcal M_1(D([0,\infty),E^\ast))$ is relatively compact.

This is claimed and proved in Corollary 9.3 of Chapter 3 in the book of Ethier and Kurtz. However, I don't understand their proof and hope there is an easier proof available which doesn't rely on the more general Theorem 9.1 (of the same chapter).

It's clear to me that if $f^\ast\in C(E^\ast)$, then $$f:=\left.f^\ast\right|_E-f^\ast(\infty)\in C_0(E)\tag1$$ and hence, by assumption, $\left(\left(f\circ X^n\right)_\ast\operatorname P\right)_{n\in\mathbb N}\subseteq\mathcal M_1(D([0,\infty),\mathbb R))$ is relatively compact. However, I don't understand how they conclude that $\left(\left(f\circ X^n\right)_\ast\operatorname P\right)_{n\in\mathbb N}\subseteq\mathcal M_1(D([0,\infty),\mathbb R))$ is actually relatively compact for all $f\in C(E^\ast)$ (not only those which arise by $(1)$).

And even when this has been shown, how do we conclude the claim? As I said before, Theorem 9.1 is quite general. For example, since $E^\ast$ is compact, the "compact containment condition" (9.1) should be trivially satisfied. So, how can we proceed to prove the claim?

$^1$ I guess it's crucial to assume that $E$ (and hence $D([0,\infty),E)$) is separable, since then $\mathcal M_1(D([0,\infty),E))$ is metrizable which in turn implies that relative sequential compactness and relative compactness are equivalent. Maybe someone can elaborate on that.

  • $\begingroup$ (1/2) I can't check Ethier and Kurtz until I go to my office tomorrow to see what Theorem $9.1$ says but it's no problem to get relative compactness for every $f \in C(E^*)$. Our discussion on another question may have confused you so let me try to be clearer here. $C_0(E)$ is naturally identified with $\{f \in C(E^*): f(\infty) = 0\}$ where $\infty$ is the distinguished point. $\endgroup$ – Rhys Steele Apr 12 at 20:28
  • $\begingroup$ (2/2) As a result, up to this identification $C(E^*)$ is the linear span of $C_0(E) \cup \{\text{constant functions}\}$. It's trivially true that if $f$ is a constant function then $(f \circ X^n)_\ast P$ is a relatively compact sequence since it's constant which gives the result. $\endgroup$ – Rhys Steele Apr 12 at 20:29
  • $\begingroup$ @RhysSteele Ah, sure. Thank you for your comments. Would be great if you could take a look at Theorem 9.1 today. $\endgroup$ – 0xbadf00d Apr 13 at 5:30
  • $\begingroup$ @RhysSteele It might be helpful to know what I'm finally after: math.stackexchange.com/q/3185888/47771 $\endgroup$ – 0xbadf00d Apr 13 at 6:16
  • $\begingroup$ In light of Theorem 9.1 of Ethier and Kurtz, it would be enough to establish that relative compactness in $D([0,\infty),E^*)$ with all sample paths actually lying in $E$ implies relative compactness in $D([0,\infty),E)$, no? $\endgroup$ – Rhys Steele Apr 13 at 13:48

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