Measurable cardinal existence condition 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.$&ltk$ -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 $ &lt\omega_1$ -additive ? How can we show that this weaker condition still implies the existence of a measurable cardinal?
Thanks a lot!
 A: You can find the answer to this in Jech's book. I sketch a proof below:
Let $\kappa$ be the least cardinal with a $\sigma$-additive measure. Let $X_\xi$ $\xi&lt\lambda$ (and $\lambda&lt\kappa$) be a sequence of disjoint subsets of $\kappa$. We want to show that $\sum_{\xi\in\lambda}\mu(X_\xi)=\mu(\bigcup_{\xi\in\lambda}X_\xi)$. By the definition of the sum, it has to be the case that there are countable many sets with non-zero measure (otherwise, uncountable many would have measure some $\frac{1}{n}$ hence the measure would be infinite). Hence we just need to show that the union of the zero sets is a zero set (it has measure $0$). Without loss of generality let's assume that all $X_\xi$ have measure zero.
Heading towards a contradiction let's assume that the union has non-zero measure. We define the following measure on $\lambda$: For $Y\subseteq\lambda$ $$\nu(Y)=\frac{\mu(\bigcup_{\xi\in Y}X_\xi)}{\mu(\bigcup_{\xi\in\lambda}X_\xi)}$$
Observe that this is a $\sigma$-additive measure on $\lambda$. First of all the fact that all $X_\xi$ have measure $0$ implies that singleton have measure $0$. That $\lambda$ has measure $1$ is trivial. Finally for the $\sigma$-additivity we simply use the $\sigma$-additivity of $\mu$. We have reached a contradiction since $\kappa$ is the least cardinal with a $\sigma$-additive measure. Hence $\kappa$ is $&lt\!\kappa$-additive.
