Proof of the equivalence of the two definitions of heavy-tailed distributions

Question

Heavy-tailed distributions have two equivalent definitions:

For any $\lambda>0$, the moment generating function of the distribution (denoted by $F$) blows up: $$\int_{\mathbb{R}}e^{\lambda x}\mathrm{d}F(x)=+\infty$$ or

For any $\lambda>0$, the decrease of the tail probability of $F$ is slower than exponentially increase: $$\lim_{x\to+\infty}e^{\lambda x}(1-F(x))=+\infty$$ That the second definition implies the first one is not difficult to prove, but how to give a proof to its converse, i.e. how to derive the second definition from the first one?

Answer 1

Suppose that

$$\lim_{x\to+\infty}e^{\lambda x}(1-F(x))=+\infty$$.

Fix $y>0$ to be a point in $\mathbb{R}$. The idea being that we place $y$ to be on the positive tail of the distribution. We can say that the integral is at least as big as $\int_y^\infty e^{\lambda x}f(x)dx \geq e^{\lambda y}(1-F(y))$. Now we know this quantity tends to infinity as we increase $y$. We chose $y$ arbitrarily so the full integral must be infinite.

$$\int_{\mathbb{R}}e^{\lambda x}\mathrm{d}F(x)=+\infty$$

But this is (2) $\implies$ (1).

$(1) \implies (2)$ to follow...

Answer 2

May be the first definition does not imply the second one, considering this question, the exponential distribution with parameter 1 and $\lambda$ = 1.