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.



Relation between measure of sets
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 $\mu(A)\le 1$.


