Chapter 3, section 12 of Billingsley Convergence of Probability Measures contains the following theorem on p. 136:
Theorem 12.6. Suppose that $E \in \mathcal D$ and $T_0$ is a countable, dense
set in $[0,1]$. Suppose further that, if $x, x_n \in E$ and $x_n(t) \to x(t)$ for
$t \in T_0$, then $x_n \to x$ in the Skorohod topology. If $P_nE = PE = 1$ and
$P_n\pi_{t_1,\ldots,t_k}^{-1} \Rightarrow P\pi_{t_1,\ldots,t_k}^{-1}$ for all $k$-tuples an $T_0$, then $P_n \Rightarrow P$.
Here is Exercise 12.5 a couple pages later:
Exercise 12.5. Suppose that $\xi$ is uniformly distributed over $[\frac13,\frac23]$, and consider the random
functions $X = 2I_{[\xi,1]}$, $X^n = I_{[\xi-\frac1n,1]}+I_{[\xi+\frac1n,1]}$.
Show that $X_n \not\Rightarrow X$, even though $X^n_{t_1,\ldots,t_k} \Rightarrow X_{t_1,\ldots,t_k}$ for all $t_1,\ldots,t_k$. Why does Theorem 12.6 not apply?
Intuitively, it seems that we shouldn't have weak convergence with the Skorohod topology and metric since that only deals with tiny wiggles in space and time. Here, $X^n$ has sample paths that jump from 0 to 1 to 2, but $X$ has sample paths that jump directly from 0 to 2. So, even if sample path $x_n$ only takes the value 1 for a short time, it will always be Skorohod distance one from a path $x$ that jumps directly from 0 to 2.
I'll consider both processes to have path space $E$ the cadlag paths which only take values in $\{0,1,2\}$ and have finitely many discontinuities. This should be a closed subset of $D[0,1]$. To show $X_n \not\Rightarrow X$, I tried to find a continuous function $f$ and show that $E^n(f)\not\to E(f)$.
I tried $f(x)=\inf \{t\in[0,1]\mid x(t)=1\}$. If cadlag paths $x,y$ are close to each other by the Skorohod metric, then their jump times must be close. Hence, $f(x)$ and $f(y)$ must be close. So this $f$ seems like a continuous function (on this $E$ with the Skorohod metric/topology).
This $f$ fails to be continuous if we let the paths take values close to rather than exactly one though since that allows $x,y$ to be close in the Skorohod metric but $f(x)$ and $f(y)$ not close. Let $x$ jump to one at $t_0$, and let $y$ jump to $1+\epsilon$ at $t_0$ (both starting at zero). Then $f(y)=+\infty$ since it never takes the value one, but we can choose $\epsilon$ small to make these functions close in the Skorohod metric.
Back to the problem, we get $E^n(f)=E(\xi-\frac1n)=\frac12-\frac1n\to\frac12$, but $E(f)=+\infty$ since the random function $X$ never takes the value one. This shows that we do not get weak convergence. (Assuming my reasoning is correct…?)
Now, why Theorem 12.6 does not apply is where I am confused. Clearly $E$ satisfies $P_nE=PE=1$ here, since both processes (random functions) always have sample paths in the space of cadlag sample paths which only take values in $\{0,1,2\}$. Of course, most of those functions are not valid sample paths for either process. Really $X^n$ takes values in the set $E_{012}$ of functions that just jump $0$ to $1$ to $2$, with exactly 2 jumps, and $X$ takes values in $E_{02}$ the set of paths that have a single jump from $0$ to $2$. So I feel the language of Theorem 12.6 isn't precise enough, and should say something in the first sentence like the set $E$ not being able to be refined further and with $PE=1$.
Maybe it has to do with no countable $T_0$ can be found?
What am I missing?
Best Answer
I find the wording of the Theorem of Billingsley a bit awkward. I am certain it is meant that
The Theorem is not claiming that in general convergence of finite-dimensional projections implies convergence in the Skorokhod topology, as i am sure you are aware that such a statement is false. It is only claiming that if the processes $X^{(n)},X$ are concentrated on a special subset $E\subset D([0,1],\mathbb R)$ for which such an implication is true, then we have that convergence of finite-dimensional projections implies convergence in the Skorokhod topology.
Now to the specific example given in the Exercise. Clearly, the processes are concentrated on the set $$E:=\bigg\{x(t)=\mathbb 1\{u\leq t\leq 1\}+\mathbb 1\{v\leq t \leq 1\}: u\leq v \in [0,1]\bigg\}.$$ (We can narrow this down even further, but there is no need to do this.)
Question 1: Why does Theorem 12.6 not apply?
We can show that Theorem 12.6 is not applicable here by showing that $E$ is not a set for which the aforementioned implication holds true. To show this, let $s\in (0,1)$ and $$x(t)=2\cdot\mathbb 1\bigg\{s\leq t\leq 1\bigg\}\hspace{1cm}x^{(n)}(t)=\mathbb 1\bigg\{s-\frac{1}{n}\leq t\leq 1\bigg\}+\mathbb 1\bigg\{s +\frac{1}{n}\leq t\leq 1\bigg\}$$ Then clearly $x^{(n)}(t)\to x(t)$ for every $t\neq s$. It is not difficult to show that $x^{(n)}$ does not converge to $x$ in the Skorokhod topology. Your argument was essentially correct: Since $x$ is $0-2$ valued and $x^{(n)}$ always has value $1$ for some $t$, we have for any increasing homeomorphism $\lambda:[0,1]\rightarrow [0,1]$ that $\sup_{0\leq t\leq 1} |x^{(n)}(\lambda(t))-x(t)|\geq 1$.
Since $s$ was arbitrary, there can be no countable $T_0$ such that the implication is true.
Question 2: Why do the given $X^{(n)}$ not weakly converge to $X$ in $D([0,1],\mathbb R)$?
This question is essentially answered by repeating the arguments for Question 1. But to give another argument, consider the well-known criterion for weak convergence in $D$ given by Theorem 13.2 in Billingsley's "Convergence of Probability Measures":
While condition (i) surely is satisfied here, one may verify that for $\epsilon$ small enough we have $$\lim_{\delta\to 0}\limsup_{n\to\infty}\mathbb P[w'(X^{(n)},\delta)>\epsilon]=1$$
where $w'$ is defined by
\begin{aligned} &w_x^{\prime}(\delta)=w^{\prime}(x, \delta)=\inf _{\left\{t_i\right\}} \max _{1 \leq i \leq v} w_x\left[t_{i-1}, t_i\right),\\ &w_x(\delta)=w(x, \delta)=\sup _{0 \leq t \leq 1-\delta} w_x[t, t+\delta] \end{aligned}