= Countable additivity
= Countably additive
{synonym}
For pairwise disjoint <measurable sets> $E_n$, countable additivity means
$$
\mu\!\left(\bigcup_{n=1}^{\infty}E_n\right)
=\sum_{n=1}^{\infty}\mu(E_n).
$$
The infinite <series> is the <limit> of its nonnegative partial sums. This controls an entire <countable union>, not merely each finite stage. If $q_1,q_2,\ldots$ enumerate the <rational numbers> in $[0,1]$, their finite collections have <Lebesgue measure> zero, and <countable additivity> gives their whole <union> <Lebesgue measure> zero. Their being a <dense set> does not change this calculation.
The axiom makes no assertion about an arbitrary <uncountable set> of disjoint pieces. In particular, $[0,1]$ is the <union> of its <singletons>, each of <Lebesgue measure> zero, yet its <Lebesgue measure> is one. Extending additivity to all such <unions> would contradict interval length.
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