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I know that non-decreasing cadlag functions (functions that are right continuous with left limits) on $[0,\infty)$ have at most a countable number of discontinuities. Does the same result hold for more general (not necessarily non-decreasing) cadlag functions?

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    $\begingroup$ Yes, it does. In fact, if $f:[0,\infty)\to\mathbb R$ has a left limit and a right limit at each point, then $f$ has at most countably many discontinuity. This is because on any compact interval $[0,A]$, the function $f$ is a uniform limit of a sequence of step functions. $\endgroup$ – Etienne Sep 23 '13 at 11:15
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It clearly suffices to prove that there are only countably many discontinuities on any interval $[0,n]$.

The existence of left- and right-hand limits means that for all $x$ and for all $N$, there exists an $\epsilon_{x,N}$ with:

$$\forall y,z \in B(x; \epsilon_{x,N}): (y-x)(z-x) > 0 \implies |f(y)-f(z)| <\frac1N$$

where $(y-x)(z-x) > 0$ ensures that $y$ and $z$ are on the same side of $x$.

By compactness of $[0,n]$, finitely many of the $B(x;\epsilon_{x,N})$ cover $[0,n]$.

That is, except for finitely many points (the centers $x$ for the covering balls), we can be sure that no discontinuity of size bigger than $\frac1N$ exists in $[0,n]$.

Thus, for each $N$, the set:

$$\left\{x \in [0,n]: \left|\lim_{\xi\to x^+}f(\xi) - \lim_{\xi\to x^-}f(\xi)\right| > \frac1N\right\}$$

is finite. It follows that the set of discontinuities on $[0,n]$:

$$\left\{x \in [0,n]: \lim_{\xi\to x^+}f(\xi) \ne \lim_{\xi\to x^-}f(\xi)\right\} = \bigcup_{N \in\Bbb N}\left\{x \in [0,n]: \left|\lim_{\xi\to x^+}f(\xi) - \lim_{\xi\to x^-}f(\xi)\right| > \frac1N\right\}$$

is countable, and hence so is:

$$\left\{x \in [0,\infty): \lim_{\xi\to x^+}f(\xi) \ne \lim_{\xi\to x^-}f(\xi)\right\} = \bigcup_{n \in \Bbb N}\left\{x \in [0,n]: \lim_{\xi\to x^+}f(\xi) \ne \lim_{\xi\to x^-}f(\xi)\right\}$$

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