Consider two complex, regular, Borel measures $\mu$ and $\nu$ on some smooth manifold $M$. If $\int _M f \ \Bbb d \mu = \int _M f \ \Bbb d \nu$ for all Schwartz test functions $f$, does it follow that $\mu = \nu$?
A variation of the above: if $M$ is just a topological space and $\int _M f \ \Bbb d \mu = \int _M f \ \Bbb d \nu$ for all bounded continuous functions $f$ (possibly that vanish at infinity), does it follow that $\mu = \nu$?
I believe the answer to be positive: first, one proves the result for positive measures, using that
$$\mu (E) = \sup \left\{ \int \limits _M f \ \Bbb d \mu ; \quad 0 \le f \le 1, \ \text{supp } f \subseteq E \right\} = \\ = \sup \left\{ \int \limits _M f \ \Bbb d \nu ; \quad 0 \le f \le 1, \ \text{supp } f \subseteq E \right\} = \nu (E)$$
for every open subset $E$; next, one shows that $\mu (E) = \nu (E)$ for every compact subset $E$, by outer regularity; next, $\mu (E) = \nu (E)$ for every measurable subset $E$ by inner regularity; finally, using the Jordan decompositions of the measures, one gets the result for complex measures.
Is the above sketch correct? Where is the countable additivity hidden in it (I know that the equality may fail if one of the measures is only finitely additive)? Is finiteness hidden anywhere in it? Is regularity necessary or just convenient?