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Weierstrass theorem says that any continuous function on a compact interval is bounded. Is this also true for functions continuous except countably many points (ex. monotone functions)? If not, how monotone functions (so continuous except countably many points) on compact intervals are always Riemann integrable?

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    $\begingroup$ This is not true for every a.e. continuous function; but its is true for monotonic functions (and very easy to check directly). $\endgroup$ – Etienne Sep 6 '14 at 9:42
  • $\begingroup$ Sorry, I mean continuous fucntions with countably many discontinuities. not a.e. $\endgroup$ – luka5z Sep 6 '14 at 9:43
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    $\begingroup$ This is not true either for functions with countably many discontinuities; even for functions with just one discontinuity point. $\endgroup$ – Etienne Sep 6 '14 at 9:44
  • $\begingroup$ But it is true for monotone functions (even with countably many discontinuities, yes)? $\endgroup$ – luka5z Sep 6 '14 at 9:45
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    $\begingroup$ Yes, indeed. Because if $f:[a,b]\to \mathbb R$ is monotonic, then $\vert f(x)\vert$ is always between $\vert f(a)\vert$ and $\vert f(b)\vert$. $\endgroup$ – Etienne Sep 6 '14 at 9:47
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Every monotone function on $[a,b]$ is bounded, because $f(x)$ is always between $f(a)$ and $f(b)$; hence $|f(x)|\le \max(|f(a)|,|f(b)|)$.

Boundedness is not enough for Riemann integrability, but monotonicity helps again: the same in-betweenness idea applies on subintervals, as in this answer by Robert Israel.

Without monotonicity, even one discontinuity point can make a function nonintegrable, like $f(x)=1/x^2$ on $[-1,1]$.

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