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Yes. A well-known example is the function defined as:

f(x) =

* 1, if x is rational

* 0, if x is irrational

Since this function has infinitely many discontinuities in any interval (it is discontinuous in any point), it doesn't fulfill the conditions for a Riemann-integrable function. Please note that this function IS Lebesgue-integrable. Its Lebesgue-integral over the interval [0, 1], or in fact over any finite interval, is zero.

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Q: Can you Give an example of bounded function which is not Riemann integrable?
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