Let f(t) and g(t) be two real-valued integrable functions defined on the real axis $[0, \infty)$. Define the integer Mellin moments of these two functions as \begin{equation} F(n) = \int_0^\infty dt \,f(t) \, t^{n-1} \end{equation} and \begin{equation} G(n) = \int_0^\infty dt \,g(t) \, t^{n-1}, \end{equation} for $n$ a non-negative integer, respectively. Suppose that these integrals exist for all $n$, i.e. converge absolutely for any $n$. Further suppose that we have no knowledge of the properties of $f(t)$, $g(t)$ other than they are real-valued, or of the Mellin transforms for non-integer $n$.
Given only that $F(n)= G(n)$ for all n, can we conclude that $f(t)=g(t)$? Or, rather, what can we conclude about $f(t)$ and $g(t)$? What are the restrictions on $f(t)$ and $g(t)$, if any, to have $f(t)=g(t)$?
This is related to an old question here: Explicitly reconstructing a function from its moments but it does not seem to me that this was ever satisfactorily answered.
Edit: One plausible answer is the following:
Since $F(n) = G(n)$ for all $n$, we know that \begin{equation} \int_0^\infty dt \,f(t) \, t^{n-1} - \int_0^\infty dt \,g(t) \, t^{n-1} = \int_0^\infty dt \,(f(t)-g(t)) \, t^{n-1} = 0. \end{equation} This holds for all $n$, so we must have $(f(t)-g(t))=0$.
