I am reading Stopple's *A Primer of Analytic Number Theory*. On page 234, the Mellin Transform $\mathcal{M} f$ of a function $f$ is defined as
$$\mathcal{M} f (s) = \int\_1^{\infty} f(x) x^{-s - 1} dx$$


His goal is to use the injectivity of the Mellin Transform (if $\mathcal{M} f = \mathcal{M} g$ then $f = g$) to prove Von Mangoldt's Formula (also on page 234). However, one of the functions he considers has infinitely many jump discontinuities. Therefore, as far as I can see, it looks like he could have replaced it by any other function agreeing with it up to a zero measure set.


So, my question is about the nature of the injectivity of the Mellin Transform: in which sense are $f$ and $g$ equal if $\mathcal{M} f = \mathcal{M} g$?



Injectivity of the Mellin Transform and discontinuities
The answer is implicit in what you wrote. One can define an equivalence relation $f\sim g$ if $f$ and $g$ are equal except on a set of measure 0, then define Mellin transform on equivalence classes instead of functions. For context, the book was intended for undergraduates who have seen no more than a typical calculus sequence, certainly not measure theory. There are much better books for advanced students. (I like Montgomery & Vaughan "Multiplicative Number Theory").


By the way @Brunalt is correct that typically the Mellin transform is defined as an integral from $0$ to $\infty$, not $1$ to $\infty$. In the book it is only applied to functions which are $0$ for $0\le x\le 1$, so including that in the integral only obfuscates. For the typical reader (who is not going on to a PhD in Mathematics) this alternate definition suffices.


