If $f\in C_0$ and $\lambda>0$, how can we show that $x\mapsto\int_0^\infty e^{-\lambda t}f(x(t))\:{\rm d}t$ is continuous wrt the Skorohod topology?

Question

Let $(E,d)$ be a locally compact separable metric space, $C_0(E)$ denote the space of continuous function from $E$ to $\mathbb R$ vanishing at infinity equipped with the supremum norm and $D([0,\infty),E)$ denote the space of càdlàg functions from $[0,\infty)$ to $E$ equipped with the Skorohod topology.

Let $f\in C_0(E)$ and $\lambda>0$. I want to show that $$D([0,\infty),E)\ni x\mapsto\int_0^\infty e^{-\lambda t}f(x(t))\:{\rm d}t\tag1$$ is bounded and continuous.

It's clear that $(1)$ is bounded, since $f$ is bounded and $$\int_0^\infty e^{-\lambda t}\:{\rm d}t=\frac1\lambda.\tag2$$ How can we show that it is continuous?

EDIT: Maybe we need to consider the integral over $[0,T]$ for a fixed $T>0$ in a first step. Let's take a look: Let $(x_n)_{n\in\mathbb N}\subseteq D([0,\infty),E)$ and $x\in D([0,\infty),E)$ with $$x_n\xrightarrow{n\to\infty}x\tag3.$$ Let $\varepsilon>0$. Since $f\in C_0(E)$, $f$ is uniformly continuous and hence there is a $\delta>0$ with $$|f(x)-f(y)|<\varepsilon\;\;\;\text{for all }x,y\in E\text{ with }d(x,y)<\delta\tag4.$$ By $(3)$, there is a sequence $(\lambda_n)_{n\in\mathbb N}$ of increasing functions mapping $[0,\infty)$ onto $[0,\infty)$ with $$\sup_{t\in[0,\:T]}|\lambda_n(t)-t|\xrightarrow{n\to\infty}0\tag5$$ and $$\sup_{t\in[0,\:T]}d(x_n(t),x(\lambda_n(t)))\xrightarrow{n\to\infty}0\tag6.$$ Thus, there is a $N\in\mathbb N$ with $$\sup_{t\in[0,\:T]}|\lambda_n(t)-t|+\sup_{t\in[0,\:T]}d(x_n(t),x(\lambda_n(t)))<\delta\;\;\;\text{for all }n\ge N.\tag7$$

Now we may be able to show that $x$ is "uniformly right-continuous" and write $|f(x_n(t))-f(x(t))|\le|f(x_n(t))-f(x(\lambda_n(t)))|+|f(x(\lambda_n(t)))-f(x(t))|$. The uniform right-continuity seems to be needed for the second term. After that we may write $$\left|\int_0^Te^{-\lambda t}f(x_n(t))\:{\rm d}t-\int_0^Te^{-\lambda t}f(x(t))\:{\rm d}t\right|\le\int_0^Te^{-\lambda t}|f(x_n(t))-f(x(t))|\:{\rm d}t\tag8.$$ How do we need to fill the gaps and proceed?

Answer 1

Note that in the Skorokhod topology, a necessary condition for convergence is that $x_n (t) \to x(t)$ for all $t$ that are continuity points of $x$.

To prove this assertion, suppose that $x_n \to x$ in the Skorokhod topology. Let $t$ be a continuity point of $x$, take a sequence of time changes $\lambda_n$ as in the definition of convergence in the Skorokhod topology and set $t_n = \lambda_n (t)$. Then $t_n \to t$ and we have

$$d(x_n (t),x(t)) \leq d(x_n(t),x(t_n)) + d(x(t_n),x(t)).$$ Clearly, the first term goes to $0$ by definition of Skorokhod convergence and the second one goes to $0$ because we assumed that $x$ is continuous at $t$.

In particular this means that $x_n \to x$ holds pointwise in a set of full Lebesgue measure, and your first observation on the boundedness allows us to conclude using the dominated convergence theorem.