Tail probability bound on the expected value of measurable function of a random variable I was reading Allan gut's graduate probability book and came across this problem,
Let $X$ be a non-negative random variable and $g$ be a non-negative, strictly increasing, differentiable function. Then,
$ Eg(X) < \infty \iff \sum_{n=1}^{\infty}g'(n)\textbf{P}(\textbf{X}>n) <\infty$  .
I am stuck and any help will be appreciated.
 A: This result is false. As indicated by my comments, we can define a function $g$ that is strictly increasing, differentiable, and satisfies $g(n)=n$ and $g'(n)=0$ for all $n \in \{0, 1, 2, ...\}$.   Then $\sum_{n=1}^{\infty} g'(n)P[X>n]=0$ regardless of $X$, but if $X$ is non-negative and integer-valued then $g(X)=X$.
I found a PDF of the book here (see page 76):
https://www.usb.ac.ir/FileStaff/5678_2018-9-18-12-55-51.pdf
Theorem 12.3 is wrong but can be fixed as follows:
Let $X$ be a nonnegative random variable.  Let $g:[0, \infty)\rightarrow\mathbb{R}$ be a function that is nonnegative, continuous, differentiable with a continuous derivative for all $x>0$, and satisfies $g'(x)\geq 0$ for all $x>0$. Then
a) $E[g(X)] = g(0) + \int_0^{\infty} g'(t) P[X>t]dt$
b) For $n \in \{1, 2, 3, ...\}$ define
\begin{align}
L_n &= \inf_{t \in [n, n+1]}g'(t)\\
H_n &= \sup_{t \in [n, n+1]}g'(t)
\end{align}
In addition to the conditions of part (a), suppose that $L_n/H_n\rightarrow 1$.
Then
$$E[g(X)]<\infty \iff \sum_{n=1}^{\infty} g'(n)P[X>n]<\infty$$

The conditions for part (b) are satisfied by functions $g$ of the form $g(x) = x^c$ for any $c> 0$.

To prove part (b) I use the fact that if $\{a_n\}_{n=1}^{\infty}$ and $\{v_n\}$ are nonnegative sequences and $v_n\rightarrow 1$ then $$ \sum_{n=1}^{\infty} a_n<\infty \iff \sum_{n=1}^{\infty} a_nv_n < \infty$$

Proof of part (b): First suppose $\sum_{n=1}^{\infty} g'(n)P[X>n]<\infty$.  Let $m$ be such that $L_n>0$ for all $n \geq m$. Then from part (a)
\begin{align}
E[g(X)] &= g(0) + \int_0^{\infty} g'(t)P[X>t]dt\\
&= g(0) + \int_0^m g'(t)P[X>t]dt + \int_m^{\infty} g'(t)P[X>t]dt\\
&\leq g(0) + \int_0^m g'(t)P[X>t]dt + \sum_{n=m}^{\infty} H_nP[X>n]\\
&\leq g(0) + \int_0^m g'(t)P[X>t]dt + \sum_{n=m}^{\infty} \frac{H_n}{L_n}g'(n)P[X>n]\\
\end{align}
where we have used the fact $g'(n)/L_n\geq 1$. The final expression is finite since we can utilize the previous fact with $v_n=\frac{H_n}{L_n}$ and $a_n=g'(n)P[X>n]$. Thus $E[g(X)]<\infty$.  The other direction is similar.
