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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: the function $f$ is Riemann integrable on $[a,b]$ if and the subset of $[a,b]$ where $f$ is discountinuous has Lebesgue measure zero. – SBF May 29 '12 at 07:43

2 Answers2

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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
davidlowryduda
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  • It seems I have to wait till I gain some understanding of point-set topology and a bit of analysis before I understand it.Can it be translated to a "language" a high school student (who is not really exceptional and hence has little background in analysis) can understand? –  May 29 '12 at 16:47
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    @Sabyasachi: zero lebesgue measure somehow says that the set of discontinuities can be covered by a collection of line segments of essentially zero combined length. What this means is that they are somehow 'not too dense.' A good heuristic is that if the set of points of discontinuity are countable (able to be put into an ordered list 1,2,3,...), then their lebesgue measure is 0. Unfortunately, there are uncountable zero measure sets, and many pathological examples. But that's the idea to keep in mind. – davidlowryduda May 29 '12 at 22:32
  • Thank you.I shall revisit this when I understand something about zero measure sets. –  May 30 '12 at 07:12
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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).

Diego F Medina
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  • Using the definition from the Wikipedia article linked, $x^{-1/2}$ is not Riemann integrable over $[0,1]$ - if we choose any $0=x_0<\dots<x_n=1$, then taking $t_0\in[0,1]$ arbitrarily close to zero we get arbitrarily high Riemann sum. In the same way, no unbounded function is Riemann integrable. – Wojowu Aug 04 '17 at 15:57
  • @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. – Diego F Medina Aug 04 '17 at 17:12