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## Relation between measure of sets

Is it true that, given a space $X$ and a probability measure $\mu$ on it, given some sets $A, B \subset X$ and a finite number of disjoint sets $C_{\sigma}$ such that $\bigcup_{\sigma} C_{\sigma} =X$, the following inequality holds,

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

Probabily it is wrong in general ( but I am not sure ), but is it true that if each $C_{sigma}$ is contained into $A$, then the equality holds? I think yes!

Context of the question: approximation of a Markov partition with a partition which is non-Markov (Dynamical Systems). In particular $A$ is the set where the initial condition is contained with a probability given by the measure of this set. I want to estimate the probability that at time $t$ the system will be in a set $D$. The set $B$ of the previous expression corresponds then to the sets of points which satisfy $T^t(B) = D$, where $T$ is the map of the dynamical system. If the inequality holds, then I can write

$$Prob(T^t(x) \in D | x \in A )\leq\sum _{\sigma}Prob (T^{t-1}(x) \in C{\sigma}| x\in A ) Prob (T^t(x) \in D | T^{t-1}(x) \in C\sigma).$$ Estimating the right side of the product of the second term and by iteration, I am then able to treat the dynamical system as a stochastic process, although the partition is not Markov.

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The LHS is simply $\mu(A\cap B)$ by finite additivity. – anstei Nov 12 at 9:29
I think this is a fine question. For $A=B$ it is equivalent to the Cauchy-Schwarz inequality, so it is not trivial (if true). – GH Nov 12 at 10:07
Lorenzo, I didn't case the negative vote, but without any context this question looks like homework. Perhaps you could explain the context in which the question arose? You might also like to look at the FAQ for guidelines on asking questions. – HW Nov 12 at 10:08
Lorenzo: The inequality is true for $A=B$. By Cauchy-Schwarz, $\mu(A)^2=(\sum_\sigma\mu(A\cap C_\sigma))^2\leq(\sum_\sigma\mu(C_\sigma))(\sum_\sigma \mu(A\cap C_\sigma)^2/\mu(C_\sigma))$. Rearranging, you get your inequality for $A=B$. – GH Nov 12 at 10:15
Lorenzo: If both $A$ and $B$ is a union of $C_\sigma$'s, then $\mu(A\cap C_\sigma)\mu(B\cap C_\sigma)/\mu(C_\sigma)$ equals $\mu(C_\sigma)$ when $C_\sigma\subset A\cap B$ and zero otherwise, hence the right hand side in your inequality is $\mu(A\cap B)/\mu(A)$. That is, your inequality is trivial in this case, and if you omit the $\mu(A)$ from the denominator then it becomes an identity. – GH Nov 12 at 10:56

The answer is no. Consider $B\subset A$ with $\mu(B)$ very small, $C_1=A\setminus B$ and $C_2=X\setminus C_1$. Then the r.h.s. equals $\mu(B)^2/\mu(A)\mu(C_2)$ which can be less than $\mu(B)$ since $\mu(A)$ and $\mu(C_2)$ can be grater than $1/2$ and $\mu(B)$ less than $1/10$.

Added later. Under the assumption that each $A$ an $B$ is a union of some $C_\sigma$, each $C_\sigma$ is contained in either $A\cap B$ or $A\setminus B$ or $B\setminus A$ or $X\setminus(A\cup B)$. Only those contained in $A\cap B$ contribute to the r.h.s., and hence the r.h.s. equals $$\sum_{\sigma:C_\sigma\subset A\cap B} \frac{\mu(C_\sigma)}{\mu(A)} = \frac{\mu(A\cap B)}{\mu(A)}\ge \mu(A\cap B)$$ because $\mu(A)\le 1$.

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I specified the conditions where I expect this inequality to be true (in particular to be an equality). – Lorenzo Nov 12 at 10:39
@Lorenzo: See my last comment. – GH Nov 12 at 10:56
Sergei, your last inequality should be reversed: $\geq\mu(A\cap B)$. – GH Nov 12 at 11:09
$$\sum_{\sigma}\mu(A \cap C_{\sigma}\cap B) \leq \sum_{\sigma} \mu(A \cap C_{\sigma} ) \frac{\mu( C_{\sigma} \cap B)}{\mu(C_{\sigma})}$$ Yes, if $C_{\sigma} \in A$ then the equality holds. – Lorenzo Nov 12 at 11:13
@GH: thanks, fixed it. – Sergei Ivanov Nov 12 at 14:06