Saturated measure

In mathematics, a measure is said to be saturated if every locally measurable set is also measurable. A set ${\displaystyle E}$, not necessarily measurable, is said to be a locally measurable set if for every measurable set ${\displaystyle A}$ of finite measure, ${\displaystyle E\cap A}$ is measurable. ${\displaystyle \sigma }$-finite measures and measures arising as the restriction of outer measures are saturated.

If an outer measure ${\displaystyle \mu ^{*}}$ has the property: for any set ${\displaystyle E}$ with finite ${\displaystyle \mu ^{*}(E)}$, there is a finite ${\displaystyle \mu ^{*}}$-measurable set ${\displaystyle A}$ containing ${\displaystyle E}$, then ${\displaystyle \mu ^{*}}$ induces a saturated measure. Moreover, we can induce a saturated measure space ${\displaystyle (X,\Sigma ',\mu ')}$ from any measure space ${\displaystyle (X,\Sigma ,\mu )}$, with ${\displaystyle \Sigma '}$ as the collection of all locally measurable sets in ${\displaystyle (X,\Sigma ,\mu )}$ and ${\displaystyle \mu '}$ as the extension of ${\displaystyle \mu }$ by taking ${\displaystyle \mu '=+\infty }$ on ${\displaystyle \Sigma '-\Sigma }$. However, such extension is not the smallest extension of ${\displaystyle \mu }$ on ${\displaystyle \Sigma '}$ but ${\displaystyle \mu ''(E):=\sup\{\mu (A):A\in \Sigma ,A\subseteq E,\mu (A)<+\infty \}}$.

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