In the definition of an outer measure, they state the sub-additivity condition as

$\mu_{*}(\bigcup A_{n}) \leq \sum\mu_{*}(A_{n})$ for any sequence of sets $A_{n} \subset X$

My question is does this inequality turn to an equality when the $A_{n}$ are disjoint sets and if it does then what would the difference be between an Outer measure and a measure?

  • $\begingroup$ No it does not automatically change to equality if the sets are disjoint. In some outer measures it may happen that it does but not in general. $\endgroup$ – Someguy Apr 19 '15 at 14:13
  • $\begingroup$ Ok, so suppose we have a sequence of disjoint sets such that the inequality turns to an equality, then would those disjoint sets be considered as $\mu*$ measurable $\endgroup$ – user1314 Apr 19 '15 at 14:15
  • $\begingroup$ Do you know what Caratheodorys measurability criterion is? $\endgroup$ – Someguy Apr 19 '15 at 14:39
  • $\begingroup$ oh is it that a set A is $\mu_{*}$ measurable if for any set $E \subset X$ we have $\mu_{*}(E) = \mu_{*}(E \bigcap A) + \mu_{*}(E \bigcap A^{c} )$ $\endgroup$ – user1314 Apr 19 '15 at 14:42
  • $\begingroup$ Yes, so equality above does not really give us information about the particular $A_n$'s with which we can show measurability. $\endgroup$ – Someguy Apr 19 '15 at 14:43

In addition to the comments above, for an explicit counterexample to your first question (does this inequality turn to an equality when the $A_n$ are disjoint sets...), consider $\mu$ defined on $(\mathbb{N},2^{\mathbb{N}})$ (notation: $2^{\mathbb{N}} = $ power set of $\mathbb{N}$) by \begin{equation} \mu(A) = \begin{cases} 1 & \text{if}\quad A \neq \emptyset\\ 0 & \text{if} \quad A = \emptyset, \end{cases} \quad\quad \text{where} \quad A \in 2^{\mathbb{N}}. \end{equation}

It is easy to see that $\mu$ is an outer measure, but $\mathbb{N} = \bigcup_{n \in \mathbb{N}} \{n\}$ is a disjoint union and

\begin{equation} 1 = \mu(\mathbb{N}) < \sum_{n = 1}^{\infty}\mu(\{n\}) = + \infty \end{equation}


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