# Can a nowhere continuous function be integrable ? [closed]

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.

-
Is this homework? –  Ryan Budney Apr 22 '13 at 7:45
No. I'm just curious. –  Henry Wen Apr 22 '13 at 8:39

## closed as off topic by Ryan Budney, Did, Pietro Majer, Dan Petersen, R WApr 22 '13 at 12:13

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

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).

-
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

Take the characteristic function of $\mathbb Q$: it is not Riemann integrable, everywhere discontinuous, almost everywhere 0 and thus Lebesgue integrable with integral 0.

-
Ok, then, take the characteristic functions of $[0,1]\cap \mathbb Q$. –  Loïc Teyssier Apr 22 '13 at 9:05