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 $\mu^*(A\setminus S) = 0$?

  • $\begingroup$ 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

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 $\mu^*(\{1\}) = 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.