Let $X$ be the d-dimensional hypercube $X=[0,1]^d$ and let $f$ and $g$ be such that $f(x) = 1$ if $x \in A$ and $0$ otherwise, $g(x)=1$ if $x \in B$ and $0$ otherwise, where $A$ and $B$ are generic subsets of $X$. Can I say something general about the relation between the two following expressions? 


$$\int\_X f(x) g(x) dx$$ and $$\int\_X f(x) dx \int\_X g(x) dx$$


(for instance that the first one is always smaller or equal to the second one). If not, is there a way to prove that,


$$\frac{\int\_X f(x) g(x) h(x)dx}{\int\_X f(x)dx} \leq \frac{\int\_X f(x) g(x) dx}{\int\_X f(x)} \frac{\int\_X g(x) h(x) dx}{\int\_X g(x) dx},$$ where $h(x)=1$ if $h(x) \in C \subset X$ and $0$ otherwise?



Integral inequality
1. Your first integral is $|A|$ (volume), and similarly the other integrals.
Cauchy-Schwarz inequality implies $|A\cap B|^2\leq |A||B|$ and nothing else can be said.
2. Your inequality with 3 sets is not true. Take $A,B,C$ independent. That is the volume of
each product set is equal to the product of the volumes. Such sets can be easily constructed,
and can have any given volumes. Then your inequality says
$$|A\cap B\cap C||B|\leq |A||A\cap B||A\cap C|,$$
that is
$$|A||B|^2|C|\leq |A|^3|B||C|,$$
that is $|B|\leq|A|^2$, which is in general not true.


