One sided derivative of the moment generating function Let $Z$ be a random variable with one-sided heavy tailed distribution, that it, the moment generating function of $Z$ is infinite for every $t<0$, but is finite on interval $[0,z)$ for some $z>0$. In this case, does the moment generating function have one-sided derivative at $0$? If yes, does this derivative is equal to (possibly negative infinite) expectation of $Z$? If yes, reference to a book where this is rigorously proved would be appreciated.
 A: 
In this case, does the moment generating function have one-sided derivative at 0?

The moment-generating function is convex on its domain of definition (just differentiate twice in the expectation). Its derivative is well-defined on $(0,z)$ and nondecreasing, and thus has a right-limit at $0$. This limit can be finite or infinite.

If yes, does this derivative is equal to (possibly negative infinite) expectation of Z?

Yes. Writing $\psi(t) := \mathbb{E} (e^{tZ})$, for all $t \in (0,z)$,
$$\psi'(t) = \mathbb{E} (Ze^{tZ}) = \mathbb{E} (Z e^{tZ} \mathbf{1}_{Z \leq 0}) + \mathbb{E} (Z e^{tZ} \mathbf{1}_{Z > 0}).$$
Then $\lim_{t \to 0^+} \mathbb{E} (Z e^{tZ} \mathbf{1}_{Z \leq 0}) = \mathbb{E} (Z \mathbf{1}_{Z \leq 0})$ and  $\lim_{t \to 0^+} \mathbb{E} (Z e^{tZ} \mathbf{1}_{Z > 0}) = \mathbb{E} (Z \mathbf{1}_{Z > 0})$, both by the monotone convergence theorem. Finally, their sum is well-defined (potentially $-\infty$) since $Z \mathbf{1}_{Z > 0}$ is integrable, so that
$$\lim_{t \to 0^+} \psi'(t) = \mathbb{E} (Z).$$
