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Let $f$ be defined on $[0, \infty)$ have all its left limits and be right continuous. Then is the set of discontinuity points of $f$ which are all jumps necessarily countable?

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    $\begingroup$ The set of discontinuity points is a borel set. It would be enough for me to see that it has lebesgue measure $0$. $\endgroup$ – Jeff Jul 12 '13 at 1:50
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By compactness and the cadlag property if $\Delta f(s) = f(s) - f(s^-)$ then $D_\epsilon = \{ s : |\Delta f(s)| > \epsilon \}$ intersects with any closed bounded interval at at most finitely many points for any $\epsilon > 0$. Otherwise, the set of such $s$ has an accumulation point and clearly there cannot be a limit at that accumulation point. Now take a union over the sets $[m,m+1]\cap D_{\frac{1}{n}}$

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