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We know that a cardinal $k > \omega$ is measurable if there is a measure function $\mu :2^k\mapsto \{0, 1\}$ that satisfies the following 3 conditions:

1.$<k$ -additive: for every set I of indices with card(I) < k, and for every family of pairwise disjoint sets $z_i$, where $i\in I$, we have $\displaystyle \mu(\bigcup_{i\in I} z_i)=\sum_{i\in I} \mu(z_i)$.

2.$\mu(k)=1$

3.$\mu (s)=0$ for every singleton.

Ok, now what if there exists a cardinal $k>\omega$, and a measure function $\mu :2^k\mapsto \{0, 1\}$ that satisfies conditions 2 and 3 above, but is only $ <\omega_1$ -additive ? How can we show that this weaker condition still implies the existence of a measurable cardinal?

Thanks a lot!

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    Writing from my phone it is hard to write a complete answer. However in Jech "*Set Theory*" he shows how to deduce that the least cardinal with such measure is actually measurable. It is in chapter 10 I think.2012-04-20
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    Is Apostolos' answer satisfactory? If so, you should upvote and accept it. If not, please add a comment or edit your question to reflect what might be missing.2012-05-10

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