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i'm seeking a very short (and self contained) proof that given a map $$f : t\in [0,T] \to f(t)\in \mathbb{R}$$ which is right continuous with left limits, setting $$\Delta f (t) = f(t)- f(t_{-})$$ the set $$\{t\in[0,T] : |\Delta f(t)| \neq 0 \}$$ is at most countable. Thank you.

guesklm
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  • See this: http://math.stackexchange.com/questions/65941/if-f-mathbbr-to-mathbbr-is-a-left-continuous-function-can-the-set-of-disc – Alex R. Feb 06 '15 at 20:29

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If not, there is a $q\in\Bbb Q$ such that $D=\{t\in[0,T]:f(t_-)<q<f(t)\}$ is uncountable. Right continuity implies that for each $x\in D$ there is an $\epsilon_x>0$ such that $f(u)>q$ for $u\in[x,x+\epsilon_x)$ and hence that $[x,x+\epsilon_x)\cap D=\varnothing$. But then $\{(x,x+\epsilon_x):x\in D\}$ is an uncountable family of pairwise disjoint non-empty open intervals, which is impossible.

Brian M. Scott
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  • I don't follow your first statement: Just because the support of $|\Delta f(t)|$ is uncountable, why does that imply that there exists some specific positive number $r$ such that the support of $\max(|\Delta f(t)|-r,0)$ is uncountable? That seems like a stronger statement than the statement one was asked to prove. – Mark Fischler Feb 06 '15 at 23:02
  • @Mark: What is the relationship of your $r$ to my answer? – Brian M. Scott Feb 06 '15 at 23:04
  • It is any positive numnber; if such an $r$ exists then you can produce a $q \in \Bbb{Q}$ meeting your first statement, simply by choosing any rational less than $r$. But if no such $r$ exists, then your first statement would be wrong. – Mark Fischler Feb 06 '15 at 23:15
  • @Mark: There is nothing wrong with my first statement. It follows immediately from the fact that every non-empty interval $(f(t_-),f(t))$ must contain a rational. There are only countably many rationals, so uncountably many of the intervals must contain the same rational. – Brian M. Scott Feb 06 '15 at 23:16
  • Very nice, now I see why your statement must be true. That justification, to me, is the key point of the proof. – Mark Fischler Feb 06 '15 at 23:42
  • Do you assume that f is increasing ? The problem is not exactly the same since it may only apply to functions of bounded variations ? – guesklm Feb 07 '15 at 11:23
  • @guesklm: The only assumptions that I made are those given in the question. – Brian M. Scott Feb 07 '15 at 11:57