In other words, if F(a) = F(b), can I assume that a = b? I can see how this would always be the case if the function were strictly positive, but may not be the case for, say, a symmetric function with both positive and negative values.
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The answer is no unless the functions are continuous. An example of functions that violate this are $f(x)=0$, $g(x)=\begin{cases}1&x=0\0&x\neq0\end{cases}$. – Rushabh Mehta Jan 12 '20 at 19:44
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Consider $\sin(x)$ ...
The point is that $\sin$ satisfies the identity $\sin(\pi+x)=-\sin(x)$. This means that every integral of $\sin$ over an interval of length an integer multiple of $2\pi$ is zero; for example, $$\int_0^{2\pi}\sin(x)dx=\int_{17}^{17+6\pi}\sin(x)dx=\int_{\pi-e}^{3\pi-e}\sin(x)dx=0.$$ So the antiderivative of a function is generally not injective.
And with that in mind you should be able to think of many other examples.
Noah Schweber
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