How is U(x) integrable if it doesn't have an antiderivative?

[pc2-brazil]

Homework Statement

Show that the function U(x) defined by
U(x)=\begin{cases}<br /> 0 &amp; \text{ if } x&lt;0 \\ <br /> 1 &amp; \text{ if } x\geq0 <br /> \end{cases}
has not an antiderivative in (-∞, +∞).
(Suggestion: Assume U has an antiderivative F in (-∞, +∞) and obtain a contradiction, showing that this contradiction follows from the mean value theorem, according to which there is a number K such that F(x) = x + K if x > 0, and F(x) = K if x < 0).

2. The attempt at a solution

I guess that, if there is an antiderivative for this function, it would have to be some constant C₁ for x < 0, and x + C₂ for x ≥ 0.
If there is a function F(x) whose derivative is U(x) in (-∞, +∞), then F(x) must be differentiable at all points. So, at any particular value of x, the one-sided derivative coming from the left must be equal to the one-sided derivative coming from the right. But, at x = 0, the one-sided derivatives of F(x) are, from the definition of U(x), 0 (coming from the left) and 1 (coming from the right). The derivative of F(x) doesn't exist at x = 0, so U(x) can't be the derivative of F(x).
I'm not sure whether this is correct, but I didn't use the suggestion. I don't know exactly what this suggestion is asking me to do.

Thank you in advance.

 

[Hariraumurthy]

U(x) is discontinuous at x=0 and therefore by the Fundemental Theorem of Integral Calculus, U(x) is not integrable.

 

[micromass]

U(x) is discontinuous at x=0 and therefore by the Fundemental Theorem of Integral Calculus, U(x) is not integrable.

Firstly, the fundamental theorem of calculus has nothing to do with this.

Secondly, your claim is wrong. The function 2x\sin(1/x)-\cos(1/x) (and which attains 0 in 0) is discontinuous in 0. But it is integrable, and it has an antiderivative: x^2\sin(1/x).

To pc2-brazil. Your proof seems to be correct.

 

[pc2-brazil]

Firstly, the fundamental theorem of calculus has nothing to do with this.

Secondly, your claim is wrong. The function 2x\sin(1/x)-\cos(1/x) (and which attains 0 in 0) is discontinuous in 0. But it is integrable, and it has an antiderivative: x^2\sin(1/x).

To pc2-brazil. Your proof seems to be correct.

Thank you for confirming it.
I hadn't noticed that a new answer had appeared. I was just about to send another question here arguing that U(x) is, indeed, integrable.