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My question concerns the generalization of the Fundamental Theorem of Calculus. On the Wiki page it is stated that the equality holds at some point $x_0$ under the condition of $f(x)$ is continuous at $x_0$. I'm also aware that Lusin's theorem states that measurable functions are nearly continuous. Nevertheless, it seems that there is still a set of non-zero measure at which the function is discontinuous. Or can I actually apply, for Lebesgue spaces, the FTC almost everywhere?

Y N
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  • @Surb, I'm aware that for the AC functions the Newton-Leibnitz formula holds. But the formulation in my question is a weaker condition, and it certainly holds for all continuous (not just AC) functions, even for the Cantor function. – Y N Nov 06 '21 at 11:14
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    Immediate from Lebesgue's Theorem on differentiation. You can find it in this site or Wikipedia or Rudin's RCA. – Kavi Rama Murthy Nov 06 '21 at 11:28
  • You are interpreting Lusin's Theorem wrongly. An $L^{2}$ function can be dis-continuous at every point. – Kavi Rama Murthy Nov 06 '21 at 11:30
  • @KaviRamaMurthy, Thank you, the equality in my question indeed follows immediately from the Lebesgue's Theorem on differentiation. I'm still confused about how I misinterpreted Lusin's Theorem. Does it have to do with the fact that it says that it is restriction that is continuous, not the function itself? – Y N Nov 06 '21 at 11:46
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    Exactly. Only the restriction is continuous and the function itself need not be continuous at any point of that subset when looked at as a function on the whole space. – Kavi Rama Murthy Nov 06 '21 at 11:55
  • @KaviRamaMurthy Ok, great. If it matters, you could write an answer by copying your comments and I will accept it. – Y N Nov 06 '21 at 11:57

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