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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!

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From what I just learned thanks to Jesse's link, the claim sounds not surprising and as if its proof should be straightforward though possibly messy in the technical details. Am I overlooking something? In other words: Where does one get stuck in a direct proof from the definitions? –  Hagen von Eitzen Jan 6 '13 at 20:57
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@ Davide: yes, you are right of course - it should read "metric" instead of "norm". sorry for the confusion. –  s_2 Jan 6 '13 at 21:38
    
@ Hagen: you might be right that a proof is straightforward, but messy. anyways, I could not figure out a solution via a "brute-force-attack". therefore I would appreciate a reference where a (detailed) proof is provided - or at least a similar question is answered. –  s_2 Jan 6 '13 at 21:41
    
Today I learned the words cadlag and Skorokhod: en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g –  Jesse Madnick Jan 8 '13 at 8:05

1 Answer 1

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.

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hi Alexander, many thanks for your answer! Very elegant. I just have one difficulty, namely verifying your last claim that $f^* \rightarrow f$ in L1, and why this means Riemann integrability (together with the result for $f_*$ I assume). maybe you can add a line or so? many thanks in advance!!! –  s_2 Jan 7 '13 at 9:55
    
@s_2: Added some details on Riemann integrability. –  Alexander Shamov Jan 7 '13 at 23:38
    
Thanks Alexander for your help, I really appreciate it. I think I get your point now - sorry for not getting it immediately. Just one more question (which is probably just cosmetic): how do I see (or show) that $f^*$ and $f_*$ are measurable (which I would need for Lebesgue's theorem? Maybe you know the answer - if not, never mind! Again, many thanks for your valueable time that you have invested in answering my questions! –  s_2 Jan 8 '13 at 18:18
    
In the case of cadlag functions it equals sup or inf over a dense countable set, so $f^\ast$ is obviously Borel. –  Alexander Shamov Jan 9 '13 at 0:53
    
And.. everything that is "explicitly constructed in a unique way from Borel functions" is at least universally measurable, so in most cases measurability is not an issue, but this topic is irrelevant to your question. –  Alexander Shamov Jan 9 '13 at 1:04

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