A positive random variable, $Z$ has a MGF iff the distribution of $Z$ has a tail which is exponentially bounded.
I am looking at a proof for this. The converse part goes as follows.
Suppose the distribution of $Z$ has an exponentially bounded tail i.e., $$P(Z>x)\le Ke^{-cx}$$ for some $c,K$ and all $x$. Then for $\theta<c$, $$E(e^{\theta Z})<\infty~\text{iff}~E(e^{\theta Z}-1)<\infty.$$ However, $$E(e^{\theta Z}-1)=E\int_0^\infty \theta e^{\theta u}\mathbb{I}_{[u<Z]}du.$$ I don't understand how this last integral came to be. Seems to be coming from Fubini's theorem maybe, but kind of confused how.
There is a typo in $$E(e^{\theta Z-1})=E\int_0^\infty \theta e^{\theta u}\mathbb{I}_{[u<Z]}du.$$ It should read $$E(e^{\theta Z})=E\int_0^\infty \theta e^{\theta u}\mathbb{I}_{[u<Z]}du.$$
For any positive r.v. $Y$ we have $EY=\int_0^{\infty} P(Y>y)dy$. Take $Y=e^{\theta Z}$. You get $E(e^{\theta Z})=\int_0^{\infty} P(e^{\theta Z}>y)dy$. Make the substitution $y=e^{\theta v}$. You get $E(e^{\theta Z})=\int P(Z>v) \theta e^{\theta v}dv$. Write $P(Z>v)$ as $E\mathbb{I}_{Z>v}$ and interchange the integral and the expecation.