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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.

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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

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  • $\begingroup$ Thank you very much. $\endgroup$ – Wenguang Zhao Oct 13 '19 at 6:52
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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.

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  • $\begingroup$ 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 $D([0,T];R^d)\times[0,T]$? $\endgroup$ – Wenguang Zhao Oct 12 '19 at 12:14
  • $\begingroup$ @Wenguang Zhao: You are right, that is a problem and I think that $\tilde f$ is not continuous, $\endgroup$ – Dieter Kadelka Oct 12 '19 at 12:35

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