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I am wondering if "a function $f(x)$ is integrable on a domain $D$" this proposition is equivalent to "$f(x)$ has antiderivative on domain $D$". If it is not the case, give me a counter example. Thank you.

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There is an even simpler example. Take, e.g.,

$$ f(x) = \begin{cases} 1 & x \geq 0 \\ -1 & x < 0 \end{cases} $$

This function is Riemann-integrable on $[-1, 1]$, but it cannot be the derivative of any function because derivatives cannot have jump discontinuities. (See, e.g., this question.)

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Under any sensible definition of "antiderivative", your function $f$ has the antiderivative $F(x)=|x|$. –  TonyK Nov 17 '12 at 19:32
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@TonyK: This is a fair point. By an "antiderivative" of $f$, I mean a differentiable function $F$ such that $F' = f$. The question asks if $f$ "has an antiderivative on domain $D$". By this, I assume he means on the entire domain $D$. With this interpretation, since $F(x) = |x|$ fails to be differentiable at $x = 0$, it is not an antiderivative for $f$ at $x = 0$ and thus not on all of $[-1, 1]$. –  user48944 Nov 17 '12 at 21:29
    
THank you so much for a simple and clear example. –  Jiddu Krishnamurti Nov 18 '12 at 1:46
    
@TonyK Well, that is not entirely true. $F'(0)$ doesn't exist, while he's setting $f(0)=1$. –  Pedro Tamaroff Nov 18 '12 at 2:13
    
@Peter: This is a sensible definition: $F$ is the antiderivative of $f$ if $F$ is continuous everywhere, differentiable almost everywhere, and $F' = f$ wherever $F'$ exists. –  TonyK Nov 18 '12 at 10:47
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Not really. The function

$$f(x)=\begin{cases} 0\text{ if } x\in\Bbb R\setminus \Bbb Q\\\frac 1 q \text{ if }x= \frac p q \in\Bbb Q, (p,q)=1\end{cases}$$

is defined on $\Bbb R$, has period one, and it is Riemann integrable over $[0,1]$ with $$\int_0^1 f=0$$ but the function has no antiderivative at all (it is not differentiable either). It is known as Thomae's function.

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$x=p/q\in\mathbb{Q}$ –  Seyhmus Güngören Mar 9 at 15:52
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There are integrable functions that are not antiderivatives: Any function that is continuous except at a single point, where it has a jump discontinuity, is an example. (Derivatives have the intermediate value property.)

More interestingly, we can ask whether the existence of an antiderivative ensures integrability. The answer depends on what integral you are considering. There are counterexamples if you mean Riemann or Lebesgue integrals, but the result is true for the Henstock–Kurzweil integral. A nice modern reference where this integral is discussed in detail and this fact is verified is A Modern Theory of Integration, by Robert G. Bartle.

To see an example that this fails for Riemann or Lebesgue integrals, see this answer. See also this related MO question.

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If you can handle a little more advanced material, let me provide you with a reference. Let $ I $ be a closed interval. As shown above by Peter, there exists a bounded function on $ I $ that is Riemann-integrable on $ I $ but does not have an antiderivative on $ I $. On the other hand, if you refer to

http://books.google.com/books?id=fXfEG-F2zJUC&pg=PA34&lpg=PA34&dq=derivatives+are+continuous+on+a+dense+set&source=bl&ots=-_8l2Zj5T1&sig=XnFiJniYei69Fbi0IhhzTdQ9Evs&hl=en&sa=X&ei=JMOnULTeDpKI9ASvyoHwCg&ved=0CDgQ6AEwAg#v=onepage&q=derivatives%20are%20continuous%20on%20a%20dense%20set&f=false,

you will see that there also exists a function on $ I $ that has an antiderivative on $ I $ but is not Riemann-integrable on $ I $.

Let me explain further. As mentioned in the reference, if $ A $ is a dense $ G_{\delta} $-subset of $ I $ (by definition, a $ G_{\delta} $-subset is the intersection of countably many open subsets), then there exists a function $ f $ on $ I $ that is (i) the derivative of a function on $ I $, (ii) continuous at all points in $ A $, and (iii) discontinuous at all points in $ I \setminus A $.

Now, Lebesgue's theorem on the necessary and sufficient condition for Riemann-integrability states that a bounded function on $ I $ is Riemann-integrable on $ I $ if and only if it is continuous almost everywhere on $ I $, i.e., the set of discontinuities of the function has measure $ 0 $. Produce a dense $ G_{\delta} $-subset $ A $ of $ I $ that has measure $ 0 $ (take the set of Liouville numbers contained in $ I $ for example). Then there is a function $ f $ on $ I $ that has an antiderivative and is discontinuous on $ I \setminus A $; by Lebesgue's theorem, $ f $ cannot be Riemann-integrable on $ I $.

Conclusion Riemann-integrability does not imply the existence of an antiderivative, and the existence of an antiderivative does not imply Riemann-integrability.

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(M. Spivak. Calculus 3 ed. pp 271)
More general function: (S. Abbott. Understanding Analysis 1 ed. pp 194 question 7.3.5)

$ f(x) = \begin{cases} 0, & x \neq 1 \\ 1, & x = 1 \end{cases} $

This function is Riemann-integrable everywhere, but it cannot be the derivative of any function by the agency of Darboux's Theorem, which signifies 'derivatives cannot have jump discontinuities'.

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As noone mentioned it, let me state a version of the Fundamental Theorem of Calculus (phrasing by Rudin):

Let $f \in \mathscr{R}[a,b]$. For $a \leq x \leq b$, put:
$$ \displaystyle F(x)= \int _a ^x f(t)dt$$ Then $F$ is continuous on $[a,b]$; furthermore, if $f$ is continuous at a point $x_0$ of $[a,b] $, then $F$ is differentiable at $x_0$, and $$F'(x_0)=f(x_0)$$

So, in the case your function $f$ is continuous, Integrability $\Rightarrow$ Having an anti-derivative. By the "other part" of the FTC, Having an anti-derivative $\Rightarrow$ Integrability.

So, if $f$ is continuous, your equivalence holds.

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Continuity already implies having an antiderivative. Integrability is irrelevant here (it happens that continuity also implies integrability, but phrasing it as you did, "[if] $f$ is continuous, Integrability $\Rightarrow$ Having an anti-derivative" may be more confusing than helpful). –  Andres Caicedo Mar 9 at 16:16
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