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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?

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  • $\begingroup$ Also, something should be attached to the "arbitrary" subsets. Do you mean measurable? $\endgroup$
    – Daniel Spector
    Commented Nov 14, 2012 at 12:39
  • $\begingroup$ you may be looking for something like en.wikipedia.org/wiki/FKG_inequality which 1) would want you to restrict to increasing sets A,B and 2) would show that the inequality can go either way $\endgroup$
    – mike
    Commented Nov 14, 2012 at 13:41
  • $\begingroup$ Yes. In cany case no inequality holds in general: if for a certain A, B, C a certain inequality holds, then take A, complementary of B, and C: in this case the opposite inequality holds. $\endgroup$
    – QuantumLogarithm
    Commented Nov 14, 2012 at 14:57
  • $\begingroup$ I guess my comment is aimed at the fact that integration of the characteristic function of a set which is not measurable is not defined, so I want to encourage a preciseness with which we use our math language. If indeed we are talking about well-defined quantities, the question can be addressed, and as you say, the answer is no in general :). $\endgroup$
    – Daniel Spector
    Commented Nov 15, 2012 at 8:07

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  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.

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