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Does there exist a function $f$ such that it is continuous a.e. on a measurable subset $E$ $\subseteq$ $R$ (the set of real numbers), Lebesgue integrable, but not Riemann integrable? I was thinking maybe the Dirichlet function, because it is Lebesgue integrable but not Riemann integrable, but it is nowhere continuous! Any ideas?

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    The Riemann integrable functions on E are precisely the functions which are continuous a.e. on E.2011-03-27
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    I'm not sure what $E$ is, but it seems you haven't seen Lebesgue's criterion for Riemann integrability: http://en.wikipedia.org/wiki/Riemann_integral#Integrability2011-03-27
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    @bobobinks How can we prove that if $f$ is continuous a.e. and Lebesgue integrable, then it is Riemann integrable?2011-03-27
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    If there is such a function, it should probably get added to [this list](http://math.stackexchange.com/questions/740/useful-examples-of-pathological-functions), if it isn't already there.2011-03-28

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Firstly, the Riemann integral doesn't even make sense unless the domain of definition of $f$ is a bounded interval. Secondly, no unbounded function is Riemann integrable. These, however, are the only obstructions: Lebesgue's criterion states that a function from $[a,b]$ to $\mathbb{R}$ is Riemann integrable if and only if it is bounded and almost everywhere continuous

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    The domain of definition is not specified to be a bounded interval. How does one handle this kind of situation?2011-03-28
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    @Sachin: you could extend the definition of the Riemann integral using some kind of [improper integral](http://en.wikipedia.org/wiki/Improper_integral).2011-03-28