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In my textbook, the definition for Riemann Integrability is as follows:

DEFINITION. A function $f$ is said to be Riemann integrable or more simply integrable on a finite closed interval $[a,b]$ if the limit

$$\int_{a}^{b} f(x) \space \text{d}x = \lim_{\text{max}\Delta x_k\rightarrow 0}\sum_{k=1}^{n}f(x_k^*)\Delta x_k$$

exists and does not depend on choice of the partitions or on the points $x_k^*$ in the subintervals.

My teacher says that any function with a discontinuity is not integrable and as I understand it, a function must be continuous over the entire interval as such quoted:

does not depend on choice of the partitions or on the points $x_k^*$ in the subintervals.

Which means if there is a removable discontinuity, the function is not Riemann Integrable or simply Integrable because the value of $\Delta x_k^*$ now depends on the fact that $x_k^*$ does not land on the $x$ coordinate of the removable discontinuity. Should it land on such value of $x$ then $f(x_k^*)$ would be undefined.

However, I found this similar question Riemann integrability; discontinuity. It says that in order for a function to be Riemann integrable, it must have a Lebesgue Measure of $0$.(Some fancy measure a first year calculus student would not know.) However it later says that the number of discontinuities can be uncountable. Which I believe contradicts the definition because a uncountable number of discontinuities means your choices of partitions have been narrowed down if not to absolute none.

My interpretation of this definition could be very wrong, but my confusion lies here...
Is the definition for Riemann Integrability stated above correct at the most rigorous standards or just outright wrong? and how does Riemann Integrability differ from the implied General Integrability?

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  • $\begingroup$ The definition of Riemann Integral is the standard one, you are correct. Your teacher is incorrect or you've misheard him - a function is integrable ( on a interval...) if it is continuous except on a set of points which have measure zero, so one or two discontinuities is fine. Your question is not clear - "General Integrability" - I suspect you mean Lebesgue Integrability? The answer wouldn't make much sense until you've covered Lebesgue measure and integration, its not something we can explain in just a few lines. $\endgroup$
    – user247608
    May 18, 2016 at 9:52

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Should it land on such value of xx then $f(x^∗_k)$ would be undefined.

$f(x^∗_k)$ is defined by hypothesis. If the value of $f(x^∗_k)$ is "wrong", isn't a big deal because $\Delta x_k\to 0$.

Which I believe contradicts the definition because a uncountable number of discontinuities means your choices of partitions have been narrowed down if not to absolute none.

A uncountable set can be small in the sense of measure theory. Measure 0 means that you can cover the set with a countable collection of intervals with total length arbitrarily small.

Example: in the case of the characteristic function of the Cantor set, take a partition with intervals $[(k-1)/3^n,k/3^n]$.

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  • $\begingroup$ Okay. As we can see, if there’s countable number of discontinuities with measure 0 in an interval [a,b] we can integrate. But what happens in these examples: $\lim_{t \to 0}\int_t^x \ln(s) ds$ and $lim_{t \to 0}\int_t^x \frac{ds}{s}$ ?? The first is fine, there’s a discont and it converges After all. But what about the last one? $\endgroup$
    – Mr. N
    May 29, 2020 at 3:10
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    $\begingroup$ @Mr.N, your examples are improper integrals (limits of). No relation with the problem discussed here. Both integrands are continuous and bounded in each $[t,x]$, $t > 0$ but undefined at $t = 0$ and unbounded in $(0,x]$. $\endgroup$
    – Martín-Blas Pérez Pinilla
    May 29, 2020 at 4:52
  • $\begingroup$ Hmm I see, @Mart. But is there a way to know if such integrals will converge or not?Let's assume we do NOT know their anti-derivatives. $\endgroup$
    – Mr. N
    May 29, 2020 at 13:22
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    $\begingroup$ @Mr.N, 1st: the so-called "improper integrals" aren't integrals, but limits of integrals. Riemann integration doesn't makes sense if the integrand is unbounded. 2nd: "But is there a way to know if such [so-called] integrals will converge or not?" Yes: tutorial.math.lamar.edu/Classes/CalcII/…. $\endgroup$
    – Martín-Blas Pérez Pinilla
    May 29, 2020 at 15:37

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