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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.

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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
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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.

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Many thanks. It is a very nice proof. –  L.T Dec 23 '11 at 21:52
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