Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Which of the following sentences is wrong?

$A.$ The function $\displaystyle{x^{\frac{1}{2}}}$ is integrable at $[-1,1]$.

$B.$ The function $f$ is integrable, where $f(x)=1 \ \ $, $\ 0 \leq x \leq 1 \ , \ \ f(0)=12$.

$C.$ Each continuous function is integrable.

$D.$ Each bounded function is integrable.


I think that the sentence $B$ is wrong, because this function is not continuous at $0$.

Do my thoughts make sense?

share|cite|improve this question
Discontinuous does not imply that the function isn't integrable. – user61527 Jun 26 '14 at 0:43
The sentences $C$ and $D$ stand,or not? – user159870 Jun 26 '14 at 0:46
One of them does. – user61527 Jun 26 '14 at 0:46
What is $\left(-\frac12\right)^{1/2}$? – Emily Jun 26 '14 at 0:47
@T.Bongers: I think both are wrong. Just take $f=1$ constant, not integrable over $\mathbb{R}$. – Gina Jun 26 '14 at 0:47

I think this question is missing some context (it sound like exactly 1 is wrong, but here we have more than one, so there are presumably some extra conditions here). 3 of these can be wrong or right depend on the situation at hand.

A. As long as we do not define $x^{\frac{1}{2}}$ in a weird way it is integrable. Basically, when $x<0$ the function can take $i\sqrt{-x}$ or $-i\sqrt{-x}$ and if the set of point where the function take each choice is measurable, then the function is integrable.

B. If you meant to write $0<x$ then yes is integrable (you wrote $0\leq x$ then it is not even a function).

C. Not necessarily. The function is certainly measurable, but might be not integrable because it increase too fast, or the space is infinite.

D. Not necessarily. The function might not be measurable. Or perhaps we integrate over a space with infinite measure.

share|cite|improve this answer

I will try to answer somewhat without giving it away.

For A, try and actually do the integration. Do you run into any trouble? If so, can you explain what the problem is? (Depending on context, i.e. if you include complex valued functions as integrable in your definition, then you may have no problem.)

For B, the comments left by T.Bongers give good indication what the answer is. In performing the integration, the discontinuity does not effect the answer. This holds more generally for any function with a countable number of discontinuities.

For C it is important to know what the definition of integration is. When this is used in the context of Riemann integration, often the only sets allowed in the integration are compact (i.e. intervals [a,b]). If this is the case, C is true (this should have been a fundamental concept from Calculus). If unbounded domains are also allowed, your counterexample makes sense.

For D I would appeal to a well-known theorem in Real Analysis which states that a bounded function $f$ on a compact interval $[a,b]$ is Riemann integrable if and only if its set of discontinuities has measure zero (for instance, they are countable). Therefore, creating a bounded function with a set of discontinuities which do not have measure zero is a counterexample to this statement.

It seems like you are reviewing integrability conditions (based on your other recent question), but you need to have a precise definition of integrability in hand and some familiarity with the standard theorems (i.e. those for C and D that I have used aboved). There are different notions of integrability, but from the other questions you have mentioned I have assumed you meant Riemann integrable. I would return to the definition you are working with, and see if it implicitly assumes that the function is being integrated over an interval $[a,b]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.