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Let $f$ be defined and bounded on $[a,b]\subset\mathbb{R}$, let $D$ be its set of discontinuities. I want to prove if $D$ has Lebesgue measure $0$ then $f$ is Riemann-integrable.

My approach is that I want to prove $D$ is compact, which reduce to $D$ being closed. If $D$ compact then $D$ has Jordan measure $0$, which means $f$ is Riemann-integrable. I proved everything but I get stuck at $D$ is closed (or $[a,b]\backslash$ is open), I am wondering whether this statement is true to begin with, if so please help me a bit?

Thank you!

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Let your interval be $I = [0,1]$, and let $D = \mathbb{Q}\cap I$. Then $D$ has Lebesgue measure $0$, but $D$ is neither closed nor open, so $D$ is not compact. Your proof will have to use a different approach, it appears.

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  • $\begingroup$ you are definitely right. Do you know any method? I don't want to use oscillation criterion $\endgroup$
    – mez
    Feb 27, 2013 at 16:00
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    $\begingroup$ I'm not personally familiar with another method, but this question math.stackexchange.com/questions/244580/… has a response that shows a method without oscillations for a function valued in a Banach space. Perhaps you can model your proof on that? $\endgroup$
    – Stahl
    Feb 27, 2013 at 16:06
  • $\begingroup$ This is great. Thank you. $\endgroup$
    – mez
    Feb 27, 2013 at 16:09

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