If $X^n$ is a càdlàg process in a locally compact space $E$, show that $\{X^n\}$ is relatively compact as processes in the compactification of $E$

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.

• (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. – Rhys Steele Apr 12 at 20:28
• (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. – Rhys Steele Apr 12 at 20:29
• @RhysSteele Ah, sure. Thank you for your comments. Would be great if you could take a look at Theorem 9.1 today. – 0xbadf00d Apr 13 at 5:30
• @RhysSteele It might be helpful to know what I'm finally after: math.stackexchange.com/q/3185888/47771 – 0xbadf00d Apr 13 at 6:16
• 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? – Rhys Steele Apr 13 at 13:48