In Proposition 1.7 (p. 3), it says that
If $\mu$ is [a positive Borel measure on $\mathbb{R}$] such that $$ \int_{\mathbb{R}}\exp(\epsilon|\lambda|)\,\mathrm{d}\mu(\lambda)<\infty $$ for some $\epsilon>0$, then $\mu$ is determinate.
It provides a proof on page 5, which doesn't make so much sense. I'll comment what the author says. Firstly, it defines $$ \phi(\xi)=\int_{\mathbb{R}}\exp(i\xi \lambda)\,\mathrm{d}\mu(\lambda) $$ It doesn't say what $\xi$ belongs to. Next ,it says that this integral converges for every complex numbers $|\Im{\xi}|<\epsilon$. I check it myself: we have $|\exp(i\xi \lambda)|=\exp(-\lambda\Im{\xi})\leq \exp(|\lambda\Im{\xi}|)<\exp(\epsilon|\lambda|)$ whenever $|\Im\xi|<\epsilon$. I think it's fine, alright. Then, magically, it "can be extended to an analytic function in this strip". What? What was $\xi$ defined in anyway? The author defines another function $\psi(\xi)$ as above, just replace $\mu$ by $\nu$, and so it also reaches the same conclusion about analyticity as before. Then, since $\nu$ and $\mu$ were assumed to have the same moments, we get $\phi=\psi$ on this strip. This makes sense here, if we look at the Taylor series. Why does it then imply that $\phi=\psi$ on $\mathbb{R}$ by Identity Theorem?