Measure with uncountable union of sets I have a question about one of the properties included in the measure definition:
$\mu(\cup_{k=1}^\infty E_k)=\sum_{k=1}^\infty \mu(E_k)$, with $\{E_k\}_{k=1}^{\infty}$ being pairwise disjoint sets.
If we replace this condition with:
$\mu(\cup_{\alpha \in A} E_{\alpha})=\sum_{\alpha \in A} \mu(E_{\alpha})$, where $A$ is some uncountable set of indicators.
Then is it possible that the whole power set of $\mathbb{R}$: $P(\mathbb{R})$ could be measurable (we should also change the condition in the $\sigma$-algebra definition)? Is it even possible to define the right side of the new condition (uncountable sum)?
 A: It's not clear to me whether you're talking about Lebesgue measure or an arbitrary measure.
Lebesgue Measure on $\Bbb R.$
Lebesgue measure on $\Bbb R$ is not uncountably additive. Say $S_x=\{x\}$. Then $\bigcup_xS_x=\Bbb R$, but $$\sum_xm(S_x)=0\ne\infty=m(\Bbb R).$$
On the Other Hand,
let $\mu$ be counting measure on $\Bbb R$. Then $\mu$ is defined on the entire power set, and it's easy to see that it's additive for disjoint unions of any cardinality.
A: Let us define $\mu:\mathscr{E} \to [0,+\infty]$ where $\mathscr{E} \subset \mathcal{P}(X)$.
Recalling the following:
Lemma: If $(a_i)_{i \in I} \subset [0,+\infty[$ and $\sum_{i \in I} a_i < +\infty$ then only a many countable $a_i$ are strictly positive.
We observe that there is something that does not work, in general, if we assume $\mu(\cup_{\alpha \in A} E_{\alpha})=\sum_{\alpha \in A} \mu(E_{\alpha})$ for any $A$ with arbitrary cardinality.
Whenever $\mu(\cup_{\alpha \in A} E_{\alpha}) < +\infty$ then only many countable of $E_{\alpha}$ could have positive measure. This can in general produce contradictions.
