Let $$(X, \Sigma, \mu)$$ be a measure space ($$X$$ is a set, $$\Sigma$$ is a $$\sigma$$-algebra and $$\mu$$ is a (positive) measure, i.e. a non-negative, countably-additive function $$\mu : \Sigma \to \mathbb{R}_{\geq 0} \cup \{ + \infty \}$$). I have the impression every measure $$\mu$$ satisfies the following condition which I call superadditivity:

$$\mu \left( \bigcup_{i \in I} E_i \right) \geq \sum_{i \in I} \mu ( E_i )$$

where the $$E_i$$ are pairwise disjoint measurable sets for all $$i \in I$$ and $$I$$ is a set (not necessarily countable, in which case the right-hand side is defined as the supremum of sums over countable subsets of $$I$$) such that $$\cup_{i \in I} E_i$$ is measurable . I have in mind the case where $$X$$ is an uncountable set and $$\mu$$ is a non-atomic measure, i.e. $$\mu ( \{ x \} ) = 0$$ for all $$x \in X$$. Then, if for instance $$I=X$$ and $$E_x = \{x \}$$, the left-hand side above is $$\mu(X)$$ while the right-hand side is 0, satisfying the inequality. The definition is then an extension of countable additivity to non-countable unions, for which you cannot ensure equality.

Is every measure superadditive? Is every probability measure superadditive? In each case, can you provide a proof/counterexample?

• you probably want to assume that the union is measurable Commented Jan 19, 2023 at 20:37
• Yes @user8268, editted Commented Jan 19, 2023 at 20:39
• @peek-a-boo Correct me if I'm wrong but I think countable subadditivity holds for not-necessarily disjoint sets. If the sets are disjoint, all I know of is countable additivity, which is in no contradiction with my "uncountable superadditivity". Commented Jan 19, 2023 at 20:43
• Something else to think about... you should be able to prove that if $a_i > 0$ for each $i$ ranging in an uncountable set, then $\sum_i a_i = +\infty$. So whenever the uncountable union of measurable sets is measurable and has finite measure, but each set has positive measure, the inequality you propose must fail. Commented Jan 19, 2023 at 20:44
• @MBolin yes I deleted my comment because I noticed you required disjointness in your definition (unlike what I expected, comparing with the definition of countable subadditivity) Commented Jan 19, 2023 at 20:47

Suppose $$\{E_i\}_{i\in I}$$ is a collection of pairwise disjoint measurable sets such that their union $$E:=\bigcup\limits_{i\in I}E_i$$ is measurable. For any countable subset of indices $$J\subset I$$, we have due to countable additivity and monotonicity of measures that \begin{align} \sum_{j\in J}\mu(E_j)&=\mu\left(\bigcup_{j\in J}E_j\right)\leq \mu(E). \end{align} Taking the supremum over all countable $$J\subset I$$ thus gives \begin{align} \sum_{i\in I}\mu(E_i)\leq \mu(E). \end{align}