Is this function Riemann Integrable on $[0,1]$:

$$f(x) = \frac{\sin\left(\cos\left(\frac{\pi}{2x}\right)\right)}{\sin\left(\cos\left(\frac{\pi}{2x}\right)\right)} \cdot \frac{\sin\left(\cos\left(\frac{\pi}{2\left(1-x\right)}\right)\right)}{\sin\left(\cos\left(\frac{\pi}{2\left(1-x\right)}\right)\right)} $$

As it approaches $0$ the discontinuities appear to approach infinity, and the same as it approaches $1$. That being said it still looks like the Riemann sum would simply equal 1, it's just a step function. But I'm more curious as to whether it's possible to create a function that satisfies, on $[0,1]$:

  1. All of the values of $f$ that exist are constant (say equal to $1$).
  2. The function has at least one one point of continuity.
  3. The function is not Riemann Integrable.

Also, what would be the the value of a definite integral that was only continuous at one point on the interval in question. Like

$$f(x) := -x^2 + 1 \, \, \, x\in\mathbb{Q}$$

$$f(x) := 1 \, \, \, x\in (\mathbb{R} , {\not} \mathbb{Q})$$


The set of discontinuities of the given function has measure zero, hence $f$ is Riemann integrable by the Riemann-Lebesgue Theorem. To construct a similar function which is not Riemann integrable, you may define $f$ as $1$ out of a fat Cantor set, then apply condensation of singularities to make every point of such Cantor set to be a point of discontinuity for $f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.