Relation between measure of sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:31:27Z http://mathoverflow.net/feeds/question/112157 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112157/relation-between-measure-of-sets Relation between measure of sets Lorenzo 2012-11-12T09:20:02Z 2012-11-12T14:05:52Z <p>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,</p> <p>$$\mu (A \cap B ) = \sum_{\sigma}\mu(A \cap C_{\sigma}\cap B) \leq \sum_{\sigma} \mu(A \cap C_{\sigma} ) \frac{\mu( C_{\sigma} \cap B)}{\mu(C_{\sigma})}, ?$$ </p> <p>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!</p> <p>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</p> <p>$$Prob(T^t(x) \in D | x \in A )\leq\sum _{\sigma}Prob (T^{t-1}(x) \in C{\sigma}| x\in A ) Prob (T^t(x) \in D | T^{t-1}(x) \in C\sigma).$$ 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.</p> http://mathoverflow.net/questions/112157/relation-between-measure-of-sets/112161#112161 Answer by Sergei Ivanov for Relation between measure of sets Sergei Ivanov 2012-11-12T10:26:25Z 2012-11-12T14:05:52Z <p>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 $1/10$.</p> <p><strong>Added later.</strong> 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 $\mu(A)\le 1$.</p>