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Let $f: [a,b] \to \mathbb{R}$ and then let $g : [a,b] \to \mathbb{R}$ given by, $$ g(x)= \begin{cases} f(x) &,\, x \in[a,b]\setminus E \\ K &, \, x \in [a,b]\cap E\end{cases}$$ where $K$ is any real number and $E$ is a countable (finite or countably infinite) subset of $[a,b]$.

Now if $f \in R[a,b]$ then is it the case that $g \in R[a,b]$ with $ \int\limits_{[a,b]}g=\int\limits_{[a,b]} f$?

The notation $f \in R[a,b]$ seems to have caused confusion. Apologies for that, by that notation, I meant $f$ is Riemann integrable on $[a,b]$

My question is motivated by this result. According to it, $f \in R[a,b]$ then is it the case that $g \in R[a,b]$

Greg Martin
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William
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  • @SouravGhosh I'm assuming as long as measure of $E$ is $0$? Well that's nice to know but I'm happy with this being true for countable as of now. If this is a well known result, could you point me to a proof of it? – William Aug 27 '22 at 10:41
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    If you think about it purely in terms of measure theory, the deletion of a null set will not change the integral since the integral of any (measurable) function over a null set is zero. – FShrike Aug 27 '22 at 11:21
  • No, the result at that link does not show what you say it shows! You need to assume the closure $\overline E$ has measure zero... – David C. Ullrich Aug 27 '22 at 11:58
  • @FShrike Wow, now I'm genuinely confused. Some people here seem to think it doesn't work. Unfortunately, I'm not good with Lebesgue (measure or integration) so I can't tell... – William Aug 27 '22 at 17:44
  • @Lelouch see the link in my first comment. Countable discontinuities are alright. – William Aug 27 '22 at 18:13

1 Answers1

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The answer is simply no. People have been saying curious things here, perhaps thinking of the Lebesgue integral. And the result at the link requires stronger hypotheses.

Standard counterexample: if $f(x)=1$, $K=0$, $E=[a,b]\cap\Bbb Q$ then $f$ is integrable but $g$ is not.

David C. Ullrich
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  • Let me write down the math, you're saying $f(x) =1, , \forall x \in [a,b]$ and then we define $$g(x)= \begin{cases} 1, \ x \in [a,b] \setminus ([a,b]\cap \mathbb{Q}) = [a,b]\setminus \mathbb{Q} \ 0 , , x \in [a,b] \cap ([a,b]\cap\mathbb{Q})= [a,b] \cap \mathbb{Q} \end{cases}$$ So far so good? But now $g$ is discontinuous at countably infinite points, don't we have this result that suggests $f$ is Riemann integrable if set of discontinuities is countable? (Countable sets have measure $0$) – William Aug 27 '22 at 17:35
  • @William Yes the issue with Riemann integrability. My comment only looked at Lebesgue’s take on the matter ... I hadn’t read you wanted to know if $g$ was Riemann integrable. – FShrike Aug 27 '22 at 17:48
  • $g$ will be Lebesgue integrable, however, with Lebesgue integral $\int g=\int f$. David’s example is a classic example of a Lebesgue-but-not-Riemann integrable function – FShrike Aug 27 '22 at 17:49
  • @FShrike I have edited my question to clear the notation ambiguity. Also how is it not Riemann integrable? (See my comment below the answer.) – William Aug 27 '22 at 17:51
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    @William I believe that $g$ (better known as the indicator $1_{\Bbb Q}$, or “Dirichlet’s function” - I think) is in fact discontinuous everywhere. If you want an example, similar to $g$, with only countably infinite discontinuities, you can check out Thomae’s function – FShrike Aug 27 '22 at 17:53
  • @FShrike Oh looks I've blundered big time. I was thinking "Oh look $g(x)$ is $1$ at every point except rationals." So I thought the set of discontinuities is all the rationals in $[a,b]$, never even considered irrationals. HA HA HA, guess I might as well quit math, move to countryside and raise chickens at this point :') Thank you both. – William Aug 27 '22 at 18:11
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    You can do math and raise chickens :D – Greg Martin Aug 27 '22 at 18:17
  • @DavidC.Ullrich so the conclusion is changing function values at finite number of points doesn't change value of Riemann integral but changing at infinite points does. Right? – William Aug 27 '22 at 18:23
  • right.......... – David C. Ullrich Aug 28 '22 at 12:00