Is integration a continuous functional on the Skorohod space? I have read several times that integration is a continuous functional on the Skorohod space $D[0,1]$, i.e., the set of all cadlag functions on $[0,1]$ equipped with the Skorohod metric; in symbols, the statement says that the mapping $I:D[0,1] \rightarrow \mathrm{R}: f \mapsto \int_0^1f(x)\mathrm{d}x$ is continuous with respect to the Skorohod metric on $D[0,1]$ and the Euclidean norm on $\mathrm{R}$. Does anybody know a reference to a proof of that fact, or know how to prove it?
Many thanks for any help!
 A: Let $f \in D[0,1]$ be a cadlag function. Then by definition of the Skorokhod metric $d$, if $d(f,g) \le \varepsilon$, then there exists a time change $\tau$ such that
$\Vert \tau - t \Vert \le \varepsilon$
and
$f \circ \tau - \varepsilon \le g \le f \circ \tau + \varepsilon$.
Let's use the notation
$\displaystyle f^\ast(t, \varepsilon) := \sup_{|s-t| \le \varepsilon} f(s)$
$\displaystyle f_\ast(t, \varepsilon) := \inf_{|s-t| \le \varepsilon} f(s)$
From the inequalities on $f$ and $g$ it follows that $f_\ast - \varepsilon \le g \le f^\ast + \varepsilon$.
On the other hand, $f^\ast(\cdot, \varepsilon) \to f$ and $f_\ast(\cdot, \varepsilon) \to f$ in $L^1$ (actually this means Riemann integrability of $f$!). It follows that the inclusion $D \to L^1$ is continuous, from which continuity of integration follows immediately.
To relate $f^\ast \to f$ and $f_\ast \to f$ in $L^1$ to Riemann integrability, one may estimate $\intop (f^\ast - f_\ast) dt$ via differences between upper and lower Darboux sums, and vice versa. The idea is to introduce two partitions of $[0,1]$ into segments such that every segment of length $2\varepsilon$ lies inside one of the segments of the partitions. Then the differences between Darboux sums for those partitions together bound $\intop (f^\ast - f_\ast) dt$.
A one-line argument is that Riemann integrability is equivalent to boundedness and continuity almost everywhere, and clearly, $f^\ast$ and $f_\ast$ converge to $f$ at points of continuity, so $f^\ast - f_\ast \to 0$ in $L^1$ by Lebesgue theorem.
