# Changing value of a Riemann integrable function on a Lebesgue measure 0 set implies the new function has the same Riemann integral?

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 $$\int_{a}^{b}f=\int_{a}^{b}\tilde{f}\qquad ?$$

• 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. Nov 25, 2018 at 11:42
• So what I am asking Is always true when the Number of point in wich I change the value of $f$ Is finite. Right? Nov 25, 2018 at 11:45
• 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 Nov 25, 2018 at 11:46
• 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. Nov 25, 2018 at 11:48
• There's no reason to think $\tilde f$ is bounded.
– zhw.
Nov 25, 2018 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.
• Thanks. My question holds whenever the set of point at wich I change the value of $f$ Is finite. Right? Nov 25, 2018 at 11:55