Existence of MGF in an interval containing $0$ implies that moment of all orders exist This is not a duplicate of this (although it refers to the same problem), and maybe a very basic question so kindly bear with me.
Suppose the Moment Generating Function (MGF) $M(s)$ is finite for some interval $(-s_o,s_0)$ for $s_0>0$. My book states:

Since $e^{|sx|}\le e^{sx}+e^{-sx}$ and the latter function is integrable $\mu$ for $|s|<s_0$, so is $\sum_{k=0}^\infty|sx|^k/k!=e^{|sx|}$. By the corollary to Theorem 16.7, $\mu$ has finite moments of all orders and $M(s)=\sum_{k=0}^\infty\frac{s^k}{k!}E(X^k)=\sum_{k=0}^\infty\frac{s^k}{k!}\int_{-\infty}^\infty x^k\mu(dx)$, $|s|<|s_0|$.

The Corollary in question is as follows:

If $\sum \int |f_n|d\mu<\infty$ then $\sum_n f_n$ converges almost everwhere and is integrable, and $\int \sum_n f_n d\mu=\sum_n\int f_n d\mu$.

I don't understand (1) The inequality $e^{|sx|}\le e^{sx}+e^{-sx}$, and (2) What is $f_n$ of the corollary in our context, i.e. how does the corollary apply. The book is Probability and Measure by Billingsley. The proofs I have looked up on the net are all slightly different.
 A: Notice that depending on whether $sx \geq 0$ or $sx < 0$, then $e^{|sx|}$ equals either $e^{sx}$ or $e^{-sx}$. Since both of these quantities are positive, we can then bound $e^{|sx|}$ by their sum, i.e. $e^{|sx|} \leq e^{sx} + e^{-sx}$.
To apply the corollary, take $f_k(x) = \frac{s^k}{k!} x^k$. It's important to note that the previous sentence in the textbook reads

Suppose that $M(s) = \int_{-\infty}^\infty e^{sx} \mu (dx)$ is defined throughout an interval $(-s_0, s_0)$ for $s_0 > 0$

In particular, that makes $e^{sx} + e^{-sx}$ $\mu$-integrable for $-s_0 < s < s_0$, which implies that $e^{|sx|} = \sum_{k=0}^\infty \frac{|sx|^k}{k!}$ is also $\mu$-integrable.
Thus,
$$\sum_{k=0}^\infty \int_{-\infty}^\infty \left| \frac{(sx)^k}{k!} \right|=  \int_{-\infty}^\infty \sum_{k=0}^\infty  \frac{|sx|^k}{k!} = \int_{-\infty}^{\infty}e^{|sx|}dx < \infty$$
where the exchange of integration and summation can be done since the integrand is non-negative.
Using the corollary in question,
$$
\begin{align*}
M(s) &= \int_{-\infty}^\infty e^{sx} \mu (dx) \\
&= \int_{-\infty}^\infty \sum_{k=0}^\infty f_k \mu (dx) \\
&= \sum_{k=0}^\infty \int_{-\infty}^\infty f_k \mu (dx) \\
&= \sum_{k=0}^\infty \frac{s^k}{k!}\int_{-\infty}^\infty x^k \mu (dx) \\
&= \sum_{k=0}^\infty \frac{s^k}{k!}\mathbb{E}(X^k)
\end{align*}$$
for $|s| < |s_0|$, as desired.
