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I know that $F(x)=\int^x_a f(x)dx$ is derivable in $[a,b]$ if $f(x)$ is continuous. Is there a function $f(x)$ which is not continuous, but $F(x)=\int^x_a f(x)dx$ is derivable ?

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up vote 4 down vote accepted

Let $f(x) = 0$ if $x$ is irrational, and $f(x) = \frac{1}{q}$ if $x = \frac{p}{q}$, where $x$ is a rational and $p$,$q$ are co-prime integers.

This is the classical example of a function which is Riemann integrable, but not continuous everywhere (it is continuous at the irrationals).

In fact, there is a general result called the Riemann-Lebesgue theorem which states that:

If $f:[a,b]\to \mathbb{R}$ is a bounded function, then $f$ is Riemann integrable on $[a,b]$ if and only if the set of discontinuities of $f$ is of measure $0$.

So you can pick any continuous function, change its value at a finite number of points, and you have your example.

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Is it Riemann integrable? – 89085731 Feb 22 '12 at 8:15
Can you explain it for me?3ks – 89085731 Feb 22 '12 at 8:20
@Gingerjin: I had one function in mind, and wrote another. See edit. – Aryabhata Feb 22 '12 at 8:22
@Aryabhata: Yes, it certainly wasn't iff. – André Nicolas Feb 22 '12 at 8:29
@Gingerjin: $F(x) = \int_{0}^{x} f(x) dx$ is constantly $0$ and so is derivable. – Aryabhata Feb 22 '12 at 8:29

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