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Let $E$ be a compact subset of $\mathbb{R}^n$. Let the density function $\phi(x,y)$ be Lipschitz continuous and such that $$ \int\limits_E \phi(x,y)dy=1 $$ for all $x\in E$. Let us consider the non-increasing sequence of non-empty compact sets $A_n$ such that for all $x\in A_{n+1}$ we have $$ \int\limits_{A_n} \phi(x,y)dy=1. $$ Since $A_n$ are compacts, there exists a non-empty limit set $A = \bigcap\limits_n A_n$

Do we have for all $x\in A$ that $$ \int\limits_A \phi(x,y)dy = 1? $$

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    $\begingroup$ What do you mean by "Since $A_n$ are compacts, there is a limit $A=\lim_n A_n$"? The set of compact sets is by no means compact itself. $\endgroup$
    – fedja
    Commented Jan 23, 2011 at 13:34
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    $\begingroup$ Could you make the question more precise? From the hypothesis it seems that taking $A_n=E$ for all $n$ would satisfy the condition. Also what do you mean by a limit set? In the Hausdorff metric? the set of points that belong to all $A_n$ for $n$ greater than some $n_0$? Also what is $A_0$? What I think this might mean is this: think of $\phi$ as the transition kernel for a MC. Let $A_0=E$; Let $A_1$ be the support of the random variable $X_1$ given that $X_0$ has some distribution with support $E$. Now let $A_2$ be the support of $X_2$ etc. Is this right? $\endgroup$
    – Anthony Quas
    Commented Jan 23, 2011 at 17:24
  • $\begingroup$ I've just mentioned that I forgot to write that the sequence $A_n$ is non-increasing. Since the limit is understood as a set which belongs to all $A_n$. $\endgroup$
    – SBF
    Commented Jan 23, 2011 at 21:21

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Anthony's interpretation of the question is that $A_n$ is the support of $X_n$, where $(X_n)$ is any Markov chain of transition kernel $\phi$ such that the distribution of $X_0$ has support $E$. If this interpretation is correct, the result holds. To see this, note that, under the assumption that $A_1\subset E=A_0$, the sequence $(A_n)$ is nonincreasing. Let $A$ denote the intersection of all the sets $A_n$ and let $x\in A$. Since $x\in A_n$ for every $n$, $\phi(x,A_{n+1})=1$ [writing $\phi(x,B)$ for the integral of $\phi(x,\cdot)$ over $B$]. Since $A_{n+1}$ decreases to $A$, by bounded convergence, $\phi(x,A_{n+1})$ decreases to $\phi(x,A)$, and this proves that $\phi(x,A)=1$. (Thus compactness would be irrelevant. Note that the indices $n$ and $n+1$ are mixed up in the question.)

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  • $\begingroup$ I edited the topic. Please note 1. $A_n$ is an arbitrary non-increasing sequence of non-empty compact sets. I need a compactness to be sure that the limit set is non-empty. 2. for all $x\in A_{n+1}$ we have $\phi(x,A_n) = 1$. That is all we have. Is it sufficient to hold that for all $x\in A$ we have $\phi(x,A) = 1$. $\endgroup$
    – SBF
    Commented Jan 23, 2011 at 21:26
  • $\begingroup$ Still seems like basic measure theory to me... Fix $x\in A$ and consider the nonnegative functions $f_n$ and $f$ defined on $E$ by $f_n(y)=\phi(x,y)$ if $y\in A_n$, $f_n(y)=0$ otherwise, and by $f(y)=\phi(x,y)$ if $y\in A$, $f(y)=0$ otherwise. You know that $f_n\to f$ pointwise and you want to prove that the integrals of $f_n$ converge to the integral of $f$. Does that ring a bell? (By the way, if $A$ is empty, every statement beginning by "For every $x\in A$, one has..." is true. Ergo exit compactness.) $\endgroup$
    – Did
    Commented Jan 23, 2011 at 21:52
  • $\begingroup$ Thank you for these nice comment about empty set, but I need compactness to say also that the set $A$ is non-empty, it is important. Could we make the proof smaller? Say, for all $x\in A$ and all $n$ we have $K(x,A_n) = 1$ (here I use $K(x,A_n)$ rather then $\phi(x,A_n)$ to stress that it is a measure). I think that by continuity of measure we have $$ K(x,A) = \lim\limits_{n\to\infty}K(x,A_n) = 1 $$ because the sequence $A_n$ is non-increasing and $A$ is an intersection of all these sets. $\endgroup$
    – SBF
    Commented Jan 24, 2011 at 8:26

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