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Michael Spivak, in his "Calculus" writes

Although it is possible to say precisely which functions are integrable,the criterion for integrability is too difficult to be stated here

I request someone to please state that condition.Thank you very much!

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    I guess, you mean the Riemann integral, so the criterion is given [here](http://en.wikipedia.org/wiki/Riemann_integral#Integrability): the function $f$ is Riemann integrable on $[a,b]$ if and the subset of $[a,b]$ where $f$ is discountinuous has [Lebesgue measure](http://en.wikipedia.org/wiki/Lebesgue_measure) zero.2012-05-29

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This is commonly called the Riemann-Lebesgue Theorem, or the Lebesgue Criterion for Riemann Integration (the wiki article).

The statement is that a function on $[a,b]$ is Riemann integrable iff

  • It is bounded
  • It is continuous almost everywhere, or equivalently that the set of discontinuities is of zero lebesgue measure
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    Thank you.I shall revisit this when I understand something about zero measure sets.2012-05-30
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Just a small clarification (as an answer as I cannot comment):

The way in which mixedmath wrote the answer might lead to confusions, a different way would be:

A bounded function $f$ on $[a,b]$ is Riemann integrable iff it is continuous almost everywhere.

Note that this assumes the function to be bounded and it is not an implication of the theorem (For instance think of $f=\frac{1}{\sqrt{(x)}}$, whose integral in $[0,1]$ is 2 but it is not bounded).

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    @Wojowu thanks you are right :), in fact Riemann integrability is defined only for bounded functions in most of the texts. Confusion might arise when referring to improper Riemann integrable as "integrable" also.2017-08-04