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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

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

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

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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 $f$.

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