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)?
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.
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 $\sum_{i \in I} a_i &lt; +\infty$ 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 $\mu(\cup_{\alpha \in A} E_{\alpha}) &lt; +\infty$ then only many countable of $E_{\alpha}$ could have positive measure. This can in general produce contradictions.