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Let $(E,d)$ be a metric space, $x:[0,\infty)\to E$ be càdlàg and $$x^\pm(t):=x(t\pm):=\lim_{s\to t\pm}x(s)\;\;\;\text{for }t\ge0,$$ where $x(0-):=x(0)$,.

How can we show that $x^-(t+)=x(t)$ for all $t\ge0$?

I really got trouble to wrap my head around this. Let $t>0$ and $\varepsilon>0$. Since $x$ is right-continuous at $t$, there is a $\delta_+>0$ with $$\forall u\in(t,t+\delta_+):d(x(t),x(u))<\varepsilon.\tag1$$ And since $x$ has a left-limit at $t$, there is a $\delta_-\in(0,t)$ with $$\forall s\in(t-\delta_-,t):d(x(s),x(t))<\varepsilon\tag2.$$ But what we need is a $\delta>0$ with $$\forall u\in(t,t+\delta):d(x^-(s),x(t))<\varepsilon\tag3.$$ How can we show this?

0xbadf00d
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  • What do you mean by $x^-(t+)$? This doesn't match any of your definitions – Snaw Dec 11 '21 at 19:18
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    What language is càdlàg and what does it mean? Is $E$ complete? – Ted Shifrin Dec 11 '21 at 19:20
  • @TedShifrin I have never seen it either. But voilà: https://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g – B. S. Thomson Dec 11 '21 at 19:26
  • @B.S.Thomson Holy cow! I am relatively fluent in French and have never seen this before. – Ted Shifrin Dec 11 '21 at 19:37
  • @TedShifrin They are simply regulated functions that are right-continuous. Who knew they deserved a new name. La Vache! indeed. – B. S. Thomson Dec 11 '21 at 19:46
  • @TedShifrin ... càdlàg is not a French term, it is a mathematics term. I imagine few French speakers who are not mathematicians have heard of it. And many mathematicians who speak no French have heard of it. – GEdgar Dec 11 '21 at 22:35
  • @GEdgar: I understand that. In my 50 years as a mathematician and more of speaking French, I've never encountered it. Clearly it shows up only in a very narrow band of mathematical discourse. – Ted Shifrin Dec 11 '21 at 22:38
  • @Snaw: Note that $x^-: [0,\infty)\rightarrow E$ is a function. Therefore, $x^-(t+)$ denotes the right-hand limit $\lim_{s\rightarrow t+} x^-(s)$. We need to show that it exists and equals to $x(t)$. – Danny Pak-Keung Chan Dec 11 '21 at 22:52
  • @DannyPak-KeungChan Thanks. – Snaw Dec 11 '21 at 23:18
  • @TedShifrin it is standard notation throughout probability and stochastic analysis. Anyone familiar with the basics of the general theory of Markov processes on reasonable (say Polish) spaces in continuous time will have seen the term. – Chris Janjigian Dec 11 '21 at 23:54
  • @ChrisJanjigian Thanks. :) – Ted Shifrin Dec 11 '21 at 23:58

1 Answers1

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In fact, the proof is elementary and just plays with definition.

Let $t_{0}\geq0$ be fixed. Denote $x(t_{0})=L$. Let $\varepsilon>0$ be arbitrary. Choose $\delta>0$ such that $d(x(t),L)<\varepsilon$ whenever $t\in[t_{0},t_{0}+\delta)$. This is possible because $x$ is right-continuous.

Let $t\in(t_{0},t_{0}+\delta)$ be arbitrary. By the definition of left-limit $x^{-}$, there exists $s\in(t_{0},t)$ such that $d\left(x^{-}(t),x(s)\right)<\varepsilon$. It follows that \begin{eqnarray*} & & d\left(x^{-}(t),L\right)\\ & \leq & d\left(x^{-}(t),x(s)\right)+d\left(x(s),L\right)\\ & < & 2\varepsilon \end{eqnarray*} by observing that $s\in(t_{0},t)\subseteq[t_{0},t_{0}+\delta)$. Therefore $\lim_{t\rightarrow t_{0}+}x^{-}(t)=x(t_{0})$.

Danny Pak-Keung Chan
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