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?
Best Answer
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}