Suppose $[a,b]$ is a compact interval of $\mathbb{R}$ and $f:[a,b]\to\mathbb{R}$ be integrable in the Riemann sense. Then, by Lebesgue's criterion, $f$ is bounded on $[a,b]$ and it's set of discontinuities has Lebesgue measure zero.

Now suppose $\tilde{f}:[a,b]\to\mathbb{R}$ is a new function built by changing the value of $f$ in a Lebesgue measure zero subset of $[a,b]$. Since $\tilde{f}$ remains bounded and it's set of discontinuities has still Lebesgue measure zero, we know that $\tilde{f}$ is Riemann integrable.

Is it true that $$\int_{a}^{b}f=\int_{a}^{b}\tilde{f}\qquad ?$$

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    $\begingroup$ One can change the value of a continuous function on a set of measure zero in such a way that the new function is continuous nowhere. $\endgroup$ – Angina Seng Nov 25 '18 at 11:42
  • $\begingroup$ So what I am asking Is always true when the Number of point in wich I change the value of $f$ Is finite. Right? $\endgroup$ – eleguitar Nov 25 '18 at 11:45
  • $\begingroup$ If you modify the zero function to take the value $1$ on each rational input, the Riemann upper and lower sum will always be $b-a$ and $0$, respectively - no convergence $\endgroup$ – Hagen von Eitzen Nov 25 '18 at 11:46
  • $\begingroup$ You could say that if $\int f$ and $\int \tilde f$ exist and $\tilde f$ differs from $f$ only on a Lebesgue zero set, then the intergrals are equal. $\endgroup$ – Hagen von Eitzen Nov 25 '18 at 11:48
  • $\begingroup$ There's no reason to think $\tilde f$ is bounded. $\endgroup$ – zhw. Nov 25 '18 at 19:19

Let $f=0$ and $g(x)=0$ for $x$ irrational, $1$ for $x$ rational. Then $f$ is Riemann integrable, $f=g$ almost everywhere but $g$ is discontinuous everywhere, is it is not Riemann integrable.

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  • $\begingroup$ Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right? $\endgroup$ – eleguitar Nov 25 '18 at 11:55
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    $\begingroup$ Yes, neither Riemann integrability nor the value of the integral changes if you change the function at a finite number of points. $\endgroup$ – Kavi Rama Murthy Nov 25 '18 at 11:59
  • $\begingroup$ Thanks sir. Could you please see if my attempt is correct here math.stackexchange.com/questions/3012683/… $\endgroup$ – eleguitar Nov 25 '18 at 12:06

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