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Let $X$ be a compact space and $\mu$ the Lebesgue measure, with $\mu(X)=1$. Let A and B be two subsets of $X$ with positive measure. What can I say about the relations between $\mu (A \cap B)$ and $\mu (A) \mu(B)$? Is one of the two quantities always larger of equal to the other one?

If not, how can I prove that, given a finite number of disjoint sets $C_\sigma$ such that $\bigcup_{\sigma} C_{\sigma}= X$, then

$$\sum_{\sigma}\mu(A \cap C_{\sigma}\cap B) \leq \sum_{\sigma} \frac{ \mu(A \cap C_{\sigma} ) } {\mu(A)} \frac{\mu( C_{\sigma} \cap B)}{\mu(C_{\sigma})}, ? $$

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As you've noted in your taggings, you can formulate this in terms of probability. $\mu(A \cap B)$ is the probability that two events both occur, and $\mu(A) \mu(B)$ would be the probability that those two events occur given that they are independent – Christopher A. Wong Nov 12 '12 at 0:21

To answer your first question, there can be no inequality relationship between $\mu(A\cap B)$ and $\mu(A)\mu(B)$. Consider a set $A\subset X$ with $\mu(A)>0$. Then, take $B=A$ so that $\mu(A\cap B)=\mu(A)$, while $\mu(A)\mu(B)=(\mu(A))^2<\mu(A)$ since $\mu(X)=1$. Thus in this case we have $\mu(A\cap B)>\mu(A)\mu(B)$. Now take $A,B$ such that $A\cap B=\emptyset$ and $\mu(A),\mu(B)>0$. Then $0=\mu(A\cap B)<\mu(A)\mu(B)$.

To prove your statement, you'll probably find the law of total probability and various forms of conditional probability useful.

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