Timeline for Convergence of sets
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 24, 2011 at 8:27 | vote | accept | SBF | ||
| Jan 24, 2011 at 8:26 | comment | added | SBF | 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. | |
| Jan 23, 2011 at 21:52 | comment | added | Did | 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.) | |
| Jan 23, 2011 at 21:26 | comment | added | SBF | 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 $\phi(x,A) = 1$. | |
| Jan 23, 2011 at 21:06 | history | answered | Did | CC BY-SA 2.5 |