Weak convergence in the Skorohod space $D([0, T])$ $\forall T$ implies Weak convergence in $D([0, \infty))$? Assume that $Z_n$ are random variables taking value in the Skorohod space $D([0, \infty),Y)$ (endowed with its usual Skorohod topology) of right-continuous functions $[0, \infty) \to Y$, where $Y$ is a real separable Banach space. Assume that $\forall T>0$, $Z_n$ converge weakly (in distribution) to $Z$ in the space $D([0, T],Y)$, where $Z$ is a random variable defined on $D([0, \infty),Y)$. Does it hold that $Z_n$ converge weakly to $Z$ in the space $D([0, \infty),Y)$? What about if an assumption is made on $Z$, for example $Z$ is a constant random variable or $Z$ is continuous, i.e. it belongs to $C([0, \infty),Y)$.
 A: We shall use Theorem 4.2. in Billingsley's book Convergence of probability measures (1968):

Theorem. Let $\left(S,d\right)$ be a separable metric space and for $u,n\in\mathbb N$, $Y_n,X_{u,n}$ and $X$ are $S$-valued random variables defined on a common probability space $\left(\Omega,\mathcal F,\mu\right)$. We assume that

*

*for all $u\in\mathbb N$, $X_{un}\to X_u$ in distribution as $n\to \infty$;

*$X_u\to X$ in distribution as $u\to \infty$, and

*for each $\varepsilon\gt 0$, $\lim_{u\to \infty}\limsup_{n\to \infty}\mu\left\{d\left(X_{u,n},Y_n\right)\gt\varepsilon\right\}=0$.

Then $Y_n\to X$ in distribution as $n\to \infty$.

The proof uses portmanteau theorem: we take $F\subset S$ a closed set, and we define $F_\varepsilon:=\{x,d(x,F)\leqslant\varepsilon\}$, where $d(x,F):=\inf\left\{d(x,y),y\in F\right\}$. Then
$$\mu\{Y_n\in F\}\leqslant \mu\{X_{u,n}\in F_\varepsilon\}+\mu\{d(X_{u,n},Y_n)\gt\varepsilon\},$$
hence taking the $\limsup_{n\to \infty}$,
$$\limsup_{n\to \infty}\mu\{Y_n\in F\}\leqslant \mu\{X_u\in F_\varepsilon\}+\limsup_{n\to \infty}\mu\{d(X_{u,n},Y_n)\gt\varepsilon\}.$$
Taking now $\limsup_{u\to \infty}$, we get in view of 3.
$$\limsup_{n\to \infty}\mu\{Y_n\in F\}\leqslant \limsup_{u\to \infty}\mu\{X_u\in F_\varepsilon\}\leqslant \mu\{X\in F_\varepsilon\},$$
and we conclude, using the fact that $\bigcap_{\varepsilon >0}F_\varepsilon=F$, $F$ begin closed.
Now, we shall use this result to prove that if $Z_n\to Z$ in distribution in $D[0,T]$ for all $T$, then actually the convergence takes place in $D[0,\infty)$. We define
$$X_{u,n}(t,\omega):=Z_n(t,\omega)\chi_{[0,u)}(t)$$
$$Y_n:=Z_n; \quad X:=Z.$$
We check the third condition. Notice that by definition of the metric on $D[0,\infty)$, we have that
$$d(X_{u,n},Y_n)\leqslant \int_u^{\infty}e^{-t}\mathrm dt,$$
because we can bound the distance restricting the infimum to the increasing Lipschitz maps $\gamma$ for which $\gamma(s)=s$ for $0\leqslant s\leqslant u$.
