Prop: If $\mu$ is a measure on a ring $\mathcal{R}$, ${\mu }^{\ast }$ is the induced outer measure on $\mathcal{H}\mathcal{(}\mathcal{R}\mathcal{)}$ and $\bar{\mu }$ is the measure induced by ${\mu }^{\ast }$ on the $\sigma$-ring $\bar{\mathcal{S}}$ of all ${\mu }^{\ast }$-measurable sets, then every set in $\mathcal{S}(R)$ is ${\mu }^{\ast }$-measurable.

Th: If $E\in \mathcal{H}\mathcal{(}\mathcal{R}\mathcal{)}$, them $$\begin{align*}
\mu^(E) &= \inf{\overline\mu(F) \mid E \subseteq G \in \overline{\cal S}} \
& = \inf{\overline\mu(F) \mid E\subseteq F \in \mathcal{S(R)}}.
\end{align}$$This is equivalent that the outer measure induced by $\bar{\mu }$ on $\mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$ and the outer measure induced by $\bar{\mu }$ on $\bar{\mathcal{S}}$ both coincide with ${\mu }^{\ast }$.

Def: If $E\in \mathcal{H}\mathcal{(}\mathcal{R}\mathcal{)}$ and $F\in \mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$, we shall say that $F$ is measurable cover of $E$ if $E\subseteq F$ and if, for every set $G\in \mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$ for which $G\subseteq F\setminus E$, we have $\bar{\mu }(G)=0$. Loosely speaking, a measurable cover of a set $E$ in $\mathcal{H}\mathcal{(}\mathcal{R}\mathcal{)}$ is a minimal set in $\mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$ which covers $E$.

Th: If a set $E\in \mathcal{H}\mathcal{(}\mathcal{R}\mathcal{)}$ is a $\sigma$-finite measure, then there exists a set $F\in \mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$ such that ${\mu }^{\ast }(E)=\bar{\mu }(F)$ and such that $F$ is a measurable cover of $E$.

Th: If $E\in \mathcal{H}\mathcal{(}\mathcal{R}\mathcal{)}$ and $F$ is a measurable cover of $E$, then ${\mu }^{\ast }(E)=\bar{\mu }(F)$; if both $F_1$ and $F_2$ are measurable covers of $E$, then $\bar{\mu }(F_1\mathrm{△}{F}_2)=0$.

Cor: If $\mu$ on $\mathcal{R}$ is $\sigma$-finite, then so are the measure $\bar{\mu }$ on $\mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$ and $\bar{\mu }$ on $\bar{\mathcal{S}}$.

Def: Let ${\mu }^{\ast }$ is an outer measure, $\bar{\mu }$ be the induced measure, and ${\bar{\mu }}^{\ast }$ is the outer measure induced by $\bar{\mu }$. In general we have that ${\mu }^{\ast }$ and ${\bar{\mu }}^{\ast }$ are not the same, if, however, the induced outer measure ${\bar{\mu }}^{\ast }$ does coincide with the original outer measure ${\mu }^{\ast }$, then ${\mu }^{\ast }$ is called regular.

Prop: If ${\mu }^{\ast }$ is a regular outer measure on a hereditary $\sigma$-ring $\mathcal{H}$ and if $\{E_n:n<\omega \}$ an increasing sequence of sets in $\mathcal{H}$ with $\underset{n\to \mathrm{\infty }}{lim}{E}_n=E$, then ${\mu }^{\ast }(E)=\underset{n\to \mathrm{\infty }}{lim}{\mu }^{\ast }(E_n)$. This result cannot be generalised to non regular outer measure.

Extension, Completion and Approximation

Th: If $\mu$ is a $\sigma$-finite measure on a ring $\mathcal{R}$, then there is a unique measure $\bar{\mu }$ on the $\sigma$-ring $\mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$ such that, for $E$ in $\mathcal{R}$, $\bar{\mu }(E)=\mu (E)$; the measure in $\bar{\mu }$ is $\sigma$-finite.

The measure $\bar{\mu }$ is called the extension of $\mu$.

Th: If $\mu$ is a measure on a $\sigma$-ring $\mathcal{S}$, then the class $\bar{\mathcal{S}}$ of all sets of the form $E\mathrm{△}N$, where $E\in \mathcal{S}$ and $N$ is a subset of a measure zero set in $\mathcal{S}$, is a $\sigma$-ring, and the set function $\bar{\mu }$ defined by $\bar{\mu }(E\mathrm{△}N)=\mu (E)$ is a complete measure on $\bar{\mathcal{S}}$.

The measure $\bar{\mu }$ is called the completion of $\mu$.

Prop: Suppose that $\mu$ is a measure on a $\sigma$-ring $\mathcal{S}$ and that $\bar{\mu }$ on $\bar{\mathcal{S}}$ is its completion. If $A,B\in \mathcal{S}$ and if $A\subseteq E\subseteq B$, and $\mu (B\setminus A)=0$, then $E\in \bar{\mathcal{S}}$.

Th: If $\mu$ is a $\sigma$-finite measure on a ring $\mathcal{R}$, and if ${\mu }^{\ast }$ is the outer measure induced by $\mu$, then the completion of the extension of $\mu$ to $\mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$ is identical with ${\mu }^{\ast }$ on the family of ${\mu }^{\ast }$-measurable sets.

Prop: If $\mu$ is a $\sigma$-finite measure on a ring $\mathcal{R}$, then, for every set $E$ of finite measure in $\mathcal{S}\mathcal{(}\mathcal{R}\mathcal{)}$ and for every $\epsilon >0$, there exists a set $E_0\in \mathcal{R}$ such that $\mu (E\mathrm{△}{E}_0)\le \epsilon$.