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How is it possible that this function: $$ f(x)=\left\{\begin{array}{ll} 0, & -1\le x < 0 \\ 1, & 0\le x \le 1\end{array}\right. $$

is R-integrable in $[-1,1]$ , but does not have an antiderivative there?

From the mean value theorem for derivatives, it can not be that the derivative has a jump at 0 , so it can not have a antiderivative?

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Why would $1$ not be the integral of this function? That's what it is at least for Riemann and Lebesgue integration. – 1015 Feb 2 '13 at 21:52
This function is Riemann integrable and its integral is 1. – Mercy King Feb 2 '13 at 21:56
I think you are confusing integral with antiderivative. – Karolis Juodelė Feb 2 '13 at 21:59
yes i meant the german word stammfunktion which is antiderivative in english.... sry – bakabakabaka Feb 2 '13 at 22:02
What about: $$ F(x)=\left\{\begin{array}{ll} 0, & -1\le x < 0 \\ x, & 0\le x \le 1\end{array}\right. $$ ? – Darius Apr 4 '14 at 12:54
up vote 3 down vote accepted

One way to see that $f$ doesn't have an antiderivative, is to use Darboux's theorem which states that every derivative has the intermediate value property, even if it (the derivative) is not continuous.

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A function does not have to be differentiable in order to be Riemann integrable.

Recall one more time Lebesgue's criterion for Riemann integrability from my previous answer: a function $f: [a, b] \longrightarrow \Bbb R$ is Riemann integrable if and only if it is bounded and continuous almost everywhere on $[a,b]$.

In the present case, $|f(x)| \leq 1$ on $[-1, 1]$. Furthermore, it is continuous everywhere except at the point $x = 0$, and the set $\{0\}$ has Lebesgue measure zero. Hence $f$ is continuous almost everywhere on $[-1,1]$.

It follows that $f$ is Riemann integrable on $[-1,1]$, and of course its integral is $1$ (the area under the graph of $f$).

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Perhaps say it this way: If $$ F(x) := \int_{-1}^x f(t) dt $$ then we may say $F$ is an indefinite integral of $f$. Certainly $F$ exists. But $F$ fails to be differentiable at the point $x=0$. One of the proofs of the Fundamental Theorem of Caluculs will, indeed, show that $F'(x) = f(x)$ at every point where $f$ is continuous. But $x=0$ escapes from that criterion.

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Which proof of the fundamental Theorem of Calculus od you mean? – bakabakabaka Feb 2 '13 at 22:23

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