Convergence in Skorokhod topology implies convergence in $L^1$.

Question

Suppose we have a sequence of càdlàg functions $(x_n)_{n\in\mathbb{N}}\subset \mathbb{D}[0,1]$ that converges in the Skorokhod topology to $x\in \mathbb{D}[0,1]$. Do we then know that $x_n \to x$ in $L^1[0,1]$? This seems like it shouldn't hold, but I wouldn't know how to construct a counter example.

As for proving the statement, I end up with for all $\epsilon > 0$ there is an $N$ such that for all $n\geq N$ there is a homeomorphism $\alpha_n: [0,1]\to[0,1]$ such that \begin{align*} \int_0^1 |x_n(t) - x(t)|dt &\leq \int_0^1 |x_n(t) - x(\alpha_n(t))| + |x(\alpha_n(t)) - y(t)| dt\\ &\leq \epsilon + \int_0^1 |x(\alpha_n(t)) - x(t)| dt. \end{align*} My intuition says that this last integral does not necessarily go to $0$ as $n \to \infty$. Is there an easy way to show that this does (not) hold?

Answer 1

It seems like it indeed does converge in $L^1$. Since $x$ is càdlàg, it is bounded. Since $\alpha_n(t) \to t$ as $n\to \infty$ for all $t\in [0,1]$, we have $x(\alpha_n(t)) \to x(t)$ almost anywhere on $[0,1]$ (the set of discontinuities of $x$ on $[0,1]$ is countable). Hence by the Dominated Convergence Theorem, we have $\int_0^1 |x(\alpha_n(t)) - x(t)| dt \to 0$, which proves the statement.