TITLE: Proving the FKG Inequality : $\mathbb{E}[f(X)g(X)] \geq \mathbb{E}[f(X)]\mathbb{E}[g(X)]$ using the Chebychev Sum Inequality
QUESTION [5 upvotes]: $f$ and $g$ are nondecreasing functions, then for any random variable $X$, show that: 
$$\mathbb{E}[f(X)g(X)] \geq \mathbb{E}[f(X)]\mathbb{E}[g(X)]$$
I know this question has been addressed here before: 
Show $\mathbb{E}[f(X)g(X)] \geq \mathbb{E}[f(X)]\mathbb{E}[g(X)]$ for $f,g$ bounded, nondecreasing
Inequality for Expected Value of Product
However, what I am interested in attempting to do is prove it using the Chebychev Sum Inequality which states: 
Given $a_1 \geq a_2 \geq \cdots \geq a_n$ and $b_1 \geq b_2 \geq \cdots \geq b_n$ then $$ {1\over n} \sum_{k=1}^n a_k \cdot b_k \geq \left({1\over n}\sum_{k=1}^n a_k\right)\left({1\over n}\sum_{k=1}^n b_k\right).$$
Similarly, if $a_1 \leq a_2 \leq \cdots \leq a_n$ and $b_1 \geq b_2 \geq \cdots \geq b_n,$, then the opposite inequality holds. 
Logically, I feel such an argument should be possible since $f$ and $g$ being monotone non-decreasing are known to be measurable and $X$ being a random variable i.e. a measurable function, their respective compositions with it should also be measurable. If we take $F := f \circ X$ and $G:= g \circ X$ then the problem is asking us to show : 
$$\int FG d\mu \geq \int Fd\mu \int Gd\mu $$
It is also known that the product of two simple functions is also a simple functions. So, we can represent each of these integrals as a limits of integrals of simple functions (increasing limits of simple functions if we want to use MCT). What I want to do is prove the desired inequality using Chebychev for the integrals of each of the simple functions and use a limiting argument to extend the result generally. I am familiar with the other proofs (Coupling, Fubini-Tonelli, Covariance operator as guiding intuition etc), admittedly simpler, but this is something that I have been trying to do for a while -- knowing that it should technically be possible -- but haven't been able to.

REPLY [5 votes]: Concerning the approximation by simple functions, my answer under the first link does exactly that.
Here is another "statistical" way to get the statement from the inequality for sums. 
Let $X_1,\dots,X_n$ be independent copies of $X$. By Chebyshev, 
$$
\frac1n \sum_{k=1}^n f(X_k) g(X_k) \ge \frac1{n^2} \sum_{k=1}^n f(X_k)\sum_{k=1}^n g(X_k).
$$
As $n\to \infty$, by the law of large numbers
$$
\mathrm{E}[f(X)g(X)] \ge \mathrm{E}[f(X)]\mathrm{E}[g(X)].
$$