A question regarding a statement in Tao's note on independence: https://terrytao.wordpress.com/2015/10/12/275a-notes-2-product-measures-and-independence/comment-page-1/#comments
More generally, if ${X}$ and ${Y}$ are random variables taking values in ranges ${R, R'}$ respectively, then
$\displaystyle {\bf E} F(X) G(Y) = ({\bf E} F(X)) ({\bf E} G(Y))$
for any scalar functions ${F,G}$ on ${R,R'}$ respectively, provided that ${F(X)}$ and ${G(Y)}$ are either both unsigned, or both absolutely integrable. This is the property of ${X}$ and ${Y}$ which is equivalent to independence (as can be seen by specialising to those ${F,G}$ that take values in ${\{0,1\}})$.
I wonder why is this condition equivalent to the fact that $X$ and $Y$ are independent, and what the author means by "specialising to those ${F,G}$ that take values in ${\{0,1\}}$".
Fix measurable subsets $A,B$ of $\mathbb{R}$ and let $F(x)=1(x\in A)$, which is $1$ if $x\in A$ and $0$ otherwise. Let $G(y)=1(y\in B)$. Then $$F(x)G(y)=1(x\in A,y\in B).$$ Assuming the equality in the question, we have $$\mathbb{P}(X\in A)\mathbb{P}(Y\in B)=\mathbb{E}(F(X))\mathbb{E}(G(Y))=\mathbb{E}(F(X)G(Y))=\mathbb{P}(X\in A,Y\in B).$$ This is exactly independence. Also, note that $F,G$ only have $0$ and $1$ as outputs (which is what the author means by "specializing to functions $\ldots$). You don't need the full strength of the inequality. You only need it for all (measurable) $F,G$ which have only $0$ and $1$ as outputs.
Next, suppose they're independent. Let $F(x)=\sum_{i=1}^m a_i1(x\in A_i)$ be a measurable simple function and let $G(y)=\sum_{i=1}^n b_i1(y\in B_i)$ be another measurable simple function. Then \begin{align*} \mathbb{E}(F(X)G(Y)) & = \sum_{i=1}^m \sum_{j=1}^n a_ib_j\mathbb{E}(1(X\in A_i,Y\in B_j)) = \sum_{i=1}^m \sum_{j=1}^n a_ib_j\mathbb{E}(1(X\in A_i,Y\in B_j)) \\ & = \sum_{i=1}^m \sum_{j=1}^n a_ib_j\mathbb{P}(X\in A_i,Y\in B_j) = \sum_{i=1}^m\sum_{j=1}^n a_ib_j\mathbb{P}(X\in A_i)\mathbb{P}(Y\in B_j) \\ & = \sum_{i=1}^m\sum_{j=1}^n a_ib_j\mathbb{E}(1(X\in A_i))\mathbb{E}(1(Y\in B_j)) \\ & = \Bigl(\sum_{i=1}^m a_i\mathbb{E}(1(X\in A_i))\Bigr)\Bigl(\sum_{j=1}^n b_j\mathbb{E}(1(Y\in B_j))\Bigr) \\ & = \mathbb{E}(F(X))\mathbb{E}(G(Y)).\end{align*}
From there you can take limits and use, for example, monotone convergence theorem or density of simple functions among integrable functions, to extend from simple functions to any of those in the question.