Expectation of a product of increasing functions of a positive random variable In the proof of Theorem 3.3 in "An introduction to Stein's method" by Barbour and Chen we can read

[...] since both $x^2$ and $x \wedge 1$ are increasing functions of $x \ge 0$, it follows that, for any random variable $\xi$,
  $\mathbb{E}[\xi^2]\mathbb{E}[|\xi| \wedge 1] \le \mathbb{E}[\xi^2|\xi| \wedge 1]$ [...]

I take this to refer to a claim:
Let $f,g : \mathbb{R}^+ \to \mathbb{R}$ be non-decreasing functions and let $X$ be a non-negative random variable. Then $\mathbb{E}[f(X)]\mathbb{E}[g(X)] \le \mathbb{E}[f(X)g(X)]$.
My question is: why is this claim true? I've tried to prove it but I haven't made any progress. I'm very grateful for any answers or hints on how to show it.
 A: One way to see this is as a limiting argument of the rearrangement inequality (see https://en.wikipedia.org/wiki/Rearrangement_inequality ). If you approximate $f,g$ by step functions on intervals of width $\epsilon > 0$, then if you consider $n$ consecutive intervals and let $f_1,\ldots,f_n$ and $g_i,\ldots,g_n$ be the values of your step function approximation on the $n$ intervals, then the left side of your inequality (applied to the step functions) is $(1/n^2) \sum_{ij}f_ig_j$ and the right side of your inequality is $(1/n) \sum_i f_i g_i$. You can view the left side as $n$ sums of $n$ products, where each set of $n$ products has each $f_i$ and each $g_i$ appearing once.  (One of these sums is $\sum_i f_i g_i$ and the others are obtained by circularly shifting the indices of the $g_i$ while keeping the indices of the $f_i$ fixed.) The rearrangement inequality thus says the left side is less than or equal to the right side. Taking the limit as $n \to \infty$ gives you the inequality for the entire step function approximations. Then taking $\epsilon \to 0$ gives the final desired result for $f,g$.
Note that the only assumption needed is that $f,g$ are (weakly) increasing.
UPDATE: The answer given basically assumes $X$ takes on values uniformly in some interval. But even if not, and even if $X$ has point masses, the values that $X$ takes on can be "split" into bins where the probability mass of $X$ within each bin is the same. This may require breaking point masses into fractional parts. As long as the bins for $X$ are ordered (for $i < j$, all the values of $X$ in bin $i$ are less than or equal to the values of $X$ in bin $j$), the argument for $f$ and $g$ will still go through.
