Counting Measure not $\sum$-finite

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.

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