# Weak convergence in Skorohod topology

Let $$D([0,T];R^d)$$ be the space of càdlàg functions endowed with the usual Skorohod topology. $$X_t(\omega):=\omega(t)$$ denotes the usual canonical process. Assume that a family of probability measures $$\mu^n$$ on $$D([0,T];R^d)$$ is tight with a weak limit $$\mu$$.

Then, is it true that for any bounded continuous function $$f$$, we have $$\lim_{n\to\infty}E^{\mu^n}\left(\int_0^Tf(X_r)dr\right)=E^{\mu}\left(\int_0^Tf(X_r)dr\right) ?$$ Or are there any references for this? Thanks a lot.

For bounded and continuous $$f$$, the map $$\omega\mapsto\int_0^T f(\omega(r))\,dr$$, from $$D([0,T]; \Bbb R^d)$$ to $$\Bbb R$$ is continuous and bounded. See, for example, https://math.stackexchange.com/questions/271738/is-integration-a-continuous-functional-on-the-skorohod-space
Since $$(X_t(\omega))_[0,T]$$ is a cadlag process it is progressively measurable, in particular $$(\omega,t) \to X_t(\omega)$$ is measurable. Let $$\tilde f(y,t) := f(y)$$, then $$\tilde f$$ is bounded and continuous again. W.l.o.g. assume $$T = 1$$. Then $$\mu_n \otimes \lambda \to \mu \otimes \lambda$$ weakly. It immediately follows that $$\lim_{n \to \infty} \int \tilde f d\mu_n \otimes \lambda = \int \tilde f d\mu \otimes \lambda.$$ Applying Fubini we get the result.
• Thanks for the answer. $\tilde f$ should be a function defined on space $D([0,T];R^d)$? Then, the problem is: $f$ is bounded continuous on $R^d$, is it true that $\tilde f(\omega,t):=f(\omega_t)=f(X_t(\omega))$ is bounded an continuous on $D([0,T];R^d)\times[0,T]$? – Wenguang Zhao Oct 12 '19 at 12:14
• @Wenguang Zhao: You are right, that is a problem and I think that $\tilde f$ is not continuous, – Dieter Kadelka Oct 12 '19 at 12:35