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Let $f$ be a bounded function on a close interval, $[0,1]$ e.g.. Can it be everywhere discontinuous and integrable?

Thanks!

P.S. It isn't a homework for me and I asked this question just out of curiosity.

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Is this homework? –  Ryan Budney Apr 22 '13 at 7:45
    
No. I'm just curious. –  Henry Wen Apr 22 '13 at 8:39
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closed as off topic by Ryan Budney, Did, Pietro Majer, Dan Petersen, R W Apr 22 '13 at 12:13

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3 Answers

up vote 3 down vote accepted

Dpends on which type of integration you are asking for.

If you mean Lebesgue integral, then yes, a function can be nowhere continuous but still have a Lebesgue integral, as Henr.L points out.

If you want Riemann integral, then the answer is no. A theorem sometimes called Lebesgue's theorem states that a bounded function on the reals has a Riemann-integral if and only if the set of points where it is discontinuous has zero measure. The proof of this theorem is quite long but not technically difficult (in my opinion).

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I haven't learned real analysis yet. But still thanks for your answer. –  Henry Wen Apr 22 '13 at 9:36
    
The Dirichlet function, $\chi_{[0,1]\cap\mathbb{Q}}$, is Lebesgue integrable and continuous nowhere. However, it is equal a.e. to a continuous function (the zero function). For the sake of a stronger example, there are Lebesgue integrable functions that are continuous nowhere, even if we allow to modify them on a null set. For instance, the characteristic function of a measurable set $C\subset (0,1)$ such that $0 < |C \cup A| < |A|$ for any nonempty open set $A\subset (0,1)$ (there exists such a set $C$). –  Pietro Majer Apr 22 '13 at 16:19
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Take the characteristic function of $\mathbb Q$: it is not Riemann integrable, everywhere discontinuous, almost everywhere 0 and thus Lebesgue integrable with integral 0.

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Cantor set's charateristic function? Homework?

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But the characteristic function of a Cantor set is continuous on the complement of the Cantor set, so it isn't anywhere discontinuous. P.S. It isn't a homework for me and I asked this question just out of curiosity. –  Henry Wen Apr 22 '13 at 8:29
    
@Hentry Wen: Henr.L is right, that function is everywhere discontinuous, because it takes the value 0 and 1 in every interval. –  Zsbán Ambrus Apr 22 '13 at 8:40
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But I still think it is it is discontinuous on the Cantor set but continuous everywhere else. –  Henry Wen Apr 22 '13 at 8:50
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Ok, then, take the characteristic functions of $[0,1]\cap \mathbb Q$. –  Loïc Teyssier Apr 22 '13 at 9:05
    
Thanks! Dirichlect function is Lebesgue integrable. –  Henry Wen Apr 22 '13 at 9:33
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