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Let $\Omega$ be a non-empty set, and let $\mathcal{A}$ be an algebra of subsets of $\Omega$, i.e. let $\mathcal{A}$ be a set consisting of subsets of $\Omega$ such that

  1. $\emptyset \in \mathcal{A}$.
  2. $A^c \in \mathcal{A}$ for every $A \in \mathcal{A}$ (where $A^c = \Omega\setminus A$).
  3. $A \cup B \in \mathcal{A}$ for every $A, B \in \mathcal{A}$.

Let $\mu$ be a finite measure on $\mathcal{A}$, i.e. let $\mu:\mathcal{A}\rightarrow[0,\infty)$ be such that

  1. $\mu(\emptyset) = 0$.
  2. $\mu(\cup_{i = 1}^\infty A_i) = \sum_{i = 1}^\infty\mu(A_i)$ for every sequence $A_1, A_2, \dots \in \mathcal{A}$ of pairwise disjoint sets such that $\cup_{i = 1}^\infty A_i \in \mathcal{A}$.

Denote by $\sigma_\Omega(\mathcal{A})$ the sigma-algebra on $\Omega$ generated by $\mathcal{A}$, and denote by $\mu^*$ the outer measure on $\Omega$ induced by $\mu$.

Let $S \subseteq \Omega$.

A measurable cover of $S$ is defined to be a set $A \in \sigma_\Omega(\mathcal{A})$ satisfying:

  1. $S \subseteq A$.
  2. $\mu^*(S) = \mu^*(A)$.

Let $A$ be a measurable cover of $S$. Is it necessarily the case that $\mu^*(A\setminus S) = 0$?

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  • $\begingroup$ Some notes: I'm fairly confident you would usually call $\mu$ a (finite) premeasure. Also if you're going to give all the definitions of basic objects, you might as well include the definition of the induced outer-measure, that's the one I'm least likely to remember personally at least. $\endgroup$ – Physical Mathematics Jan 30 at 20:52
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Let $\Omega = \{0,1\}$ and let $\mathcal A = \{\emptyset, \Omega\}$, so $\sigma_\Omega(\mathcal A) = \mathcal A$. Then let $\mu : \mathcal{A} \to [0,\infty)$ given by $\mu(\emptyset) =0$ and $\mu(\Omega) = 1$. Then $\mu^*(\{0\}) = \mu^*(\{1\}) = 1$. But consider a measurable cover of $\{0\}$, it clear must be $\Omega$, and $\Omega \backslash \{0\} = \{1\}$ and $\mu^*(\{1\}) = 1$.

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