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
$$
\int\limits\_A \phi(x,y)dy = 1?
$$



Convergence of sets
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.)


