Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I look for a example of family of atomic measures such that their sum is not atomic. A measure $\mu$ on a $\sigma$-algebra $S$ of subsets of $X$ is called atomic if every measurable set of positive measure contain an atom. Here by an atom we mean a set set $E\in S$ with $\mu(E)>0$ such that for every $F\subset E$ with $F\in S$ we have $\mu(F)=0$ or $\mu(E\setminus F)=0$).

I found the following example (page 651, Example 1.3) but I did not understand everything.

Assume that $X$ is a compact Hausdorff space in which all singletons are not in $G_\delta$ (Examples of such topologies are here). Let $S$ be a $\sigma$ -algebra generated by all compact $G_\delta$ subsets of $X$. For each $x \in X$ we define a measure $\mu_x$ on $S$ by putting $\mu_x(E)=1$ if $x \in E$ and $\mu_x(E)=0$ otherwise. Let $\mu$ be a measure on $S$ defined by $\mu(\emptyset)=0$ and $\mu(E)=\infty$ otherwise.

How to show that $\mu(E)=\sum \{\mu_x(E): x \in X\}$ for all $E\in S$?

It is clear that it suffices to know that every nonempty $E\in S$ is infinite.

Thanks for reading and I would appreciate it if you could solve my problem.

share|improve this question
1  
This equality is true if $E$ is a compact $G_{\delta}$, since such a set cannot be finite by the hypothesis over the topological space. Now the set of $E$ such that this equality is true is a $\sigma$-algebra. –  Davide Giraudo Dec 23 '11 at 13:28
    
Thanks. But maybe there is one problem in proving that sets for which the measures are equal form $\sigma$-algebra. Namely, let $E\neq \emptyset$, $E \neq X$ and measures on the LHS and on RHS be equal on $E$. Why then these measures are equal on $X\setminus E$ or equivalently why $X\setminus E$ is infinite? –  L.T Dec 23 '11 at 15:50
    
Indeed, you're right, because for example $X\setminus\{x_0\}$ satisfies the last equality since $X$ is infinite, but not its complement. So maybe we have to show that $\mathcal A:=\{E\subset X, \mu(E)=\sum_{x\in X}\mu_x(E)\mbox{ and }\mu(E^c)=\sum_{x\in X}\mu_x(E^c)\}$ is a $\sigma$-algebra which contains $S$ –  Davide Giraudo Dec 23 '11 at 15:53
add comment

1 Answer

up vote 3 down vote accepted

Let $\tau$ be the topology of $X$. Let $\mathscr{G}$ be the set of non-empty compact $G_\delta$ subsets of $X$; $\mathscr{G}\,$ is a base for a topology $\tau'$ on $X$. Let $\mathscr{V}=\{V\in\tau':X\setminus V\in\tau'\}$, the family of $\tau'$-clopen subsets of $X$; clearly $\mathscr{V}\;$ is closed under complementation. Moreover, every non-empty member of $\mathscr{V}\;$ is infinite.

Suppose that $G\in\mathscr{G}$. If $G=X$, then certainly $G\in\mathscr{V}$. Otherwise, $X\setminus G$ is a non-empty $\tau$-open subset of $X$. Let $x\in X\setminus G$ be arbitrary; there is a sequence $\langle U_n:n\in\omega\rangle$ of $\tau$-open nbhds of $x$ such that $U_0\subseteq X\setminus G$ and $U_n\supseteq\operatorname{cl}U_{n+1}\supseteq U_{n+1}$ for each $n\in\omega$. Let $$H=\bigcap_{n\in\omega} U_n=\bigcap_{n\in\omega}\operatorname{cl}U_n\;;$$ then $H$ is clearly a closed (and hence compact) $G_\delta$ in $\langle X,\tau\rangle$ with $x\in H\subseteq X\setminus G$. It follows that $X\setminus G\in\tau'$ and hence that $G\in\mathscr{V}$. Thus, $\mathscr{G}\subseteq \mathscr{V}$.

Now suppose that $\{V_n:n\in\omega\}\subseteq\mathscr{V}$, and let $V=\bigcup_n V_n$. Certainly $V\in\tau'$, so to show that $V\in\mathscr{V}$, it suffices to show that $X\setminus V\in\tau'$. Let $x\in X\setminus V\,$ be arbitrary. For each $n\in\omega$ there is a $G_n\in\mathscr{G}$ such that $x\in G_n\subseteq X\setminus V_n$. Let $G=\bigcap_n G_n$; then $x\in G\subseteq X\setminus V$, and $G\in\mathscr{G}$, so $X\setminus V\in\tau'$, $V\in\mathscr{V}$, and $\mathscr{V}\;$ is closed under countable unions.

Thus, $\mathscr{V}\;$ is a $\sigma$-algebra containing $\mathscr{G}$ whose non-empty members are infinite, and it follows that the non-empty members of the $\sigma$-algebra generated by $\mathscr{G}$ are infinite as well.

share|improve this answer
    
Many thanks. It is a very nice proof. –  L.T Dec 23 '11 at 21:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.