Does the difference between a subset and a measurable cover thereof have outer measure $0$?

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

• 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. – 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 $$\mu^*(\{1\}) = 1$$.