Question posed in Spivak Chapter 14 that $f$ cannot be a derivative I'm having trouble working out the reasoning behind a question posed in Spivak's Calculus Chapter 14, where he discusses the Fundamental Theorem of Calculus. The excerpt where this is from is as follows:
... A function $f$ may be integrable without being the derivative of another function. For example, if $f(x) = 0$ for $x \ne 1$ and $f(1) = 1$, then $f$ is integrable, but $f$ cannot be a derivative (why not?)
I tried working it out to verify whether $f$ is differentiable by using the definition of a derivative, but realised that the statement was that $f$ cannot be a derivative, not that $f$ is not differentiable (unless there's something I'm missing out here?). Any insights would be greatly appreciated!
 A: You're right, the statement made is that $f$ cannot be a derivative. For the reason, look back to Chapter 11 (somewhere near theorem $7$ if I remember correctly), where Spivak explains why a derivative cannot have a jump discontinuity.
Actually if you look at the problems of chapter $11$, there's one (due to Darboux I think) which says that derivatives of functions satisfy the intermediate-value property (this is a much stronger assertion, but is not needed for this particular question).
A: Suppose $g'(x)=f(x).$  Since $g'(x)=0$ for $x<1$, we have $g(x)=c_1$ if $x<1$ for some constant $c_1$.  Similarly, $g(x)=c_2$ if $x>1$ for some constant $c_2$.  Now either $c_1=c_2$ and $f(1)=g'(1)=0$ or $c_1\neq c_2$ and $g'(1)$ does not exist.
In general, a derivative has no jump discontinuity.
A: You do not need Darboux's Theorem for this. Assume $f$ is defined on $(1-\delta,1+\delta)$ for some $\delta>0$ and that there is a function $F$ that satisfies $F'=f$. Then, if $x\neq 1,\ F=c$ for some constant $c$ (why?). Then, by continuity, $F(1)=c$, so in fact $F'(1)=0\neq 1=f(1)$ and we have a contradiction.
A: Assume, for proof by contradiction, that there is a function $g$ such that $g'=f$.
Since $g$ is differentiable everywhere by assumption (since $f$ is defined everywhere), it must be continuous.
For any $x \in (1,\infty)$ or $x \in (\infty,1)$, we have $f(x)=g'(x)=0$. Thus, $g$ is constant on these intervals. Let's call the constant value $c_-$ on $(\infty,1)$ and $c_+$ on $(1,\infty)$.
It must be that $c_-=c_+$ since by continuity of $g$
$$\lim\limits_{x\to 1^+} g(x)= \lim\limits_{x\to 1^+} c_+ = g(1)$$
$$\lim\limits_{x\to 1^-} g(x)=\lim\limits_{x\to 1^-} c_-=g(1)$$
$$\implies c_-=c_+=g(1)$$
But then $g$ is constant everywhere and $g'(1)=f(1)=0$, which is a contradiction since $f(1)=1$ by assumption.
Therefore, by proof by contradiction, there is no function $g$ such that $g'=f$. Ie, $f$ is not the derivative of any function.
