probability - Expectation of nonnegative random variable when passed through nonnegative increasing differentiable function. Part II: Electric Boogaloo

This is a follow up to my previous question:

Expectation of nonnegative random variable when passed through nonnegative increasing differentiable function

I am now wanting to establish a follow up to the above problem. Specifically, if $X$ is a nonnegative random variable and $g:\mathbb{R}\rightarrow\mathbb{R}$ is a nonnegative, strictly increasing, differentiable function, then

$$\mathbb{E}g(X)<\infty \iff \sum_{n=1}^{\infty}g^{\prime}(n)\mathbb{P}(X>n)<\infty$$

I believe I can show the inequality when $g(x)=x^{p}$ for $p\in\mathbb{N}$, but the case of a general $g$ is more mysterious to me.

My attempt for the converse proceeds in the following way: If you assume that the series converges then (by the linked question)

\begin{equation}
\mathbb{E}g(X) = g(0)+\int_{0}^{\infty}g^{\prime}(X)\mathbb{P}(X>x)dx \\
= g(0)+\sum_{n=0}^{\infty}\int_{n}^{n+1}g^{\prime}(x)\mathbb{P}(X>x)dx \\
\leq g(0)+\sum_{n=0}^{\infty}(g^{\prime}(n+1)+g^{\prime}(n))\mathbb{P}(X>n) \\
= g(0)+\left(\sum_{n=0}^{\infty}g^{\prime}(n+1)\mathbb{P}(X>n)\right)+\left(\sum_{n=0}^{\infty}g^{\prime}(n)\mathbb{P}(X>n)\right).
\end{equation}

However I am unsure how to proceed from here. I don't see how the middle series would converge without more assumptions on $g$.

Any help with the equivalence in general would be appreciated.

Answer

This answer elaborates on my comments (to show the claim is false): Define $g:[0,\infty)\rightarrow \mathbb{R}$ by
$$g(x) = 1 + x + \frac{1}{2\pi} \cos(2\pi x + \pi/2)$$
Then
$$g’(x) = 1 - \sin(2\pi x + \pi/2) \geq 0 \quad \forall x \geq 0$$
and for $n\geq 0$ we get
$$ g'(n) = 0 \quad \mbox{ if and only if $n$ is an integer}$$
It follows that $g$ is nonnegative and strictly increasing over $x \geq 0$.

Furthermore $g(x)\geq 1 + x -1/(2\pi)\geq x$ and so
$$ g(x) \geq x \quad \forall x \geq 0$$
Let $X$ be any nonnegative random variable that satisfies $E[X]=\infty$. We get:
$$ g(X)\geq X \implies E[g(X)] \geq E[X] = \infty$$
but
$$ \sum_{n=1}^{\infty} g’(n) P[X>n] = 0$$
You can easily extend $g$ to have domain over all real numbers while preserving the non negativity and strictly increasing properties.

*Note: This shows that one direction of the "if and only if" claim is false. The pre-kidney answer shows the other direction is also false.