I cannot figure out the following:
Give an example of integrable $f:]-1,1[ \rightarrow \mathbb{R}$ such that $G(x)=\int_0^x f(t)dt$ is not $f$'s primitive?
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What does $] \cdot, \cdot [$ mean? – MathMajor Apr 25 '16 at 19:48
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@MathMajor Open interval. – mavavilj Apr 25 '16 at 19:48
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Wow. No offence to you of course since you did not come up with this, but that is the worst notation i have ever seen haha. – MathMajor Apr 25 '16 at 19:49
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5@MathMajor: It is totally standard European notation. Notice that the American notation of $(a,b)$ is really worse. How do you know if it's an open interval or an ordered pair? – Ted Shifrin Apr 25 '16 at 19:52
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1@mavavilj: What condition on $f$ do you need at a point $a\in ]-1,1[$ to be sure that $G'(a)=f(a)$? Can you think of a function $f$ where $G'(x)$ will not equal $f(x)$ for some $x$? – Ted Shifrin Apr 25 '16 at 19:53
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@TedShifrin $f$ must be continuous at point $a$? In order for its integral function to be differentiable at $a$. So if I take $f(x)=1$ for $x \ge 0$ and $f(x)=0$ for $x<0$, then $G(x)=\int_0^x f(t)dt=$ $x$ for $x \ge 0$ and $0$ for $x<0$ and $G$ is not differentiable at $0$ and therefore it's not $f$'s primitive? – mavavilj Apr 25 '16 at 20:00
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1@mavavilj: It is sufficient (but not necessary) for $f$ to be continuous at $a$, yes. But your example will work. Can you possibly invent an example when $G$ is everywhere differentiable and still not a primitive for $f$? – Ted Shifrin Apr 25 '16 at 20:08
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@A.Sh: I'm pretty sure we're in the context of Riemann-integrable functions. – Ted Shifrin Apr 25 '16 at 20:08