Find $r>0$ such that $\mathsf{E}\mathrm{e}^{r\xi} = 1$ Let $\xi:\Omega\to\mathbb{R}$ be a random variable such that $\mathsf{E}\xi < 0$. I would like to find $r>0$ such that $g(r):=\mathsf{E}\mathrm{e}^{r\xi} = 1$.
Maybe you can help me to derive some necessary of sufficient conditions for such $r$ to exist? Say, if $\xi = -1$ then there is no such positive $r$. I tried to investigate the behavior of  $g(r)$ since $g(0) = 1$, but I was not successful in it.
Edited: maybe one of you can help with an example when $\mathsf{P}(\xi>0)>0$ but there is no positive  root of $g(r) = 1$?
 A: Let us forget integrability questions for a moment and assume that $\mathrm{e}^{r|\xi|}$ is integrable for every positive $r$. Then the function $g$ is well defined and continuous everywhere.
First step: as you noted, $g(0)=1$ and $g'(0)=E(\xi)$, hence $g'(0)<0$ and $g(r)<1$ for $r>0$, $r$ close to $0$.
Second step: assume that $P(\xi>0)\ne0$ (otherwise, as shown by Theo, the result is false), and choose $x>0$ such that $P(\xi\ge x)=p>0$. Then $\mathrm{e}^{r\xi}\ge\mathrm{e}^{rx}\mathbf{1}_{\{\xi\ge x\}}$ for every nonnegative $r$, hence $g(r)\ge\mathrm{e}^{rx}p$ for every nonnegative $r$, and one sees that $g(r)\to+\infty$ when $r\to+\infty$.
Conclusion: since the function $g$ is continuous, the intermediate value theorem implies that there exists a positive $r$ such that $g(r)=1$.
When $\mathrm{e}^{r|\xi|}$ is not integrable for at least some positive $r$, three different things may happen: $g(r)$ may be infinite except at $r=0$, or finite for every $0\le r<r_0$ and infinite for $r\ge r_0$, for a given finite $r_0$, or finite for every $0\le r\le r_0$ and infinite for $r>r_0$. 
The conclusion that there exists a positive $r$ such that $g(r)=1$ would still hold if for example, there exists a positive $r_1$ such that $g(r_1)$ is finite and $g(r_1)>1$. It would fail, for example, if $g(r)$ is finite for every $0\le r\le r_0$ and infinite for $r>r_0$, with $g(r_0)<1$.
