Exercise 3.13e in Cinlar's Probability and Stochastics:

Consider the measurable space $(E,B(E))$, where $E=[0,1]$ and $B(E)$ is the set of all Borel subsets of $E$.

Show that the counting measure $\mu$ on it is not $\sigma$-finite and also not $\sum$-finite, where $\sum$-finite means that there is a sequence of finite measures $\mu_1, \mu_2, ...$ such that $\mu=\sum{\mu_i}$.

I proved the not $\sigma$-finite part by showing that that implies the countability of $[0,1]$, a contradiction, but I'm not sure how to prove non-$\sum$-finiteness.

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    $\begingroup$ A countable union of finite sets is still countable. $\endgroup$ – Umberto P. Nov 6 '18 at 0:50
  • $\begingroup$ @Umberto Yes, I used that result to prove non-$\sigma$-finiteness, but what about non-$\sum$-finiteness? Thanks. $\endgroup$ – Hugh Abrams Nov 6 '18 at 0:57
  • $\begingroup$ Does supp refer to supremum? $\endgroup$ – Hugh Abrams Nov 6 '18 at 1:08
  • $\begingroup$ Look at the support of the $\mu_i$'s, note that they must be countable. $\endgroup$ – user10354138 Nov 6 '18 at 1:10
  • $\begingroup$ I'm not familiar with supports; I'll look them up.Thanks. $\endgroup$ – Hugh Abrams Nov 6 '18 at 1:11

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