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 ?
Find a primitive function that is derivable
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calculus
1 Answers
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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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0Is it Riemann integrable? – 2012-02-22
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0Can you explain it for me?3ks – 2012-02-22
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0@Gingerjin: I had one function in mind, and wrote another. See edit. – 2012-02-22
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0@Aryabhata: Yes, it certainly wasn't iff. – 2012-02-22
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1@Gingerjin: $F(x) = \int_{0}^{x} f(x) dx$ is constantly $0$ and so is derivable. – 2012-02-22
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0@Aryabhata why $F^{'}(x)$ is not equal to f(x) when x is rational – 2012-02-22
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0@AndréNicolas why F′(x) is not equal to f(x) when x is rational – 2012-02-22
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0@Aryabhata This result is usually called either "Lebesgue integrability condition" or "Lebesgue's criterion for Riemann integrability". Riemann was not directly responsible for this result. I say this because it could lead to confusion with the famous [Riemann-Lebesgue Lemma](http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma). Also, the Lebesgue criterion actually states the stronger result: $f$ is Riemann integrable on $[a,b]$ if and only if $f$ bounded almost everywhere and continuous almost everywhere. – 2012-02-22
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0@RagibZaman why F′(x) is not equal to f(x) when x is rational – 2012-02-22
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0@Gingerjin: You understand that $F(x)=0$ for all $x$. (Informally, the set of rationals has measure $0$, the values of $f(x)$ at rationals make no difference to the "area.") So $F'(x)=0$ for all $x$. From the definition of $f(x)$, this means that $F'(x)=f(x)$ when $x$ is irrational. But $f(x)\ne 0$ when $x$ is rational, so $F'(x)\ne f(x)$ if $x$ is rational. For example, $f(5/12)=1/12$ but $F'(5/12)=0$. – 2012-02-22
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0@RagibZaman: Can you give me a reference to a text which considers $f$ being bounded almost everywhere? I haven't seen that. Also, the way I learnt it, it had been called the Riemann Lebesgue theorem! Perhaps that is why the other result is called a Lemma and not a theorem! But you are right, it could be confusing (and hence I chose to state the whole theorem instead of just mentioning it). – 2012-02-22
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0@Aryabhata [Here](http://en.wikipedia.org/wiki/Riemann_integral#Integrability) is a link to the result. Also go to the "Notes" section, as notes 2,3,4 and 5 refer you to various statements and proofs of the theorem. – 2012-02-23
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0@RagibZaman: That wiki does not talk about bounded a.e. In fact, now I remember! If $f$ is Riemann Integrable, then $f$ is bounded. See here: http://books.google.com/books?id=5hM0DcdEGf4C&pg=PA5#v=onepage&q&f=false. – 2012-02-23