If a measure $\mu$ arises from the restriction of an outer measure $\mu^{*}$, is it true that $\mu$ is saturated? A measure $\mu$ on a measurable space $(X, \mathcal{M})$ is called saturated if given a subset $E$ of $X$ the condition $E \cap A \in \mathcal{M}$ for every $A \in \mathcal{M}$ with $\mu(A) < \infty$ implies that $E \in \mathcal{M}$ (that is if any locally measurable subset of $X$ is measurable). The Wikipedia article on saturated measures states that "measures arising as the restriction of outer measures are saturated".
I know how to prove this result when the outer measure itself arises from a pre-measure (a nice proof can be found in the answer to this question), but not every outer measure is generated by a pre-measure so I'd like to know if the previous assertion is actually true and, if it is, how can I prove the result in the general case, that is how can I show that any measure $\mu$ resulting from the restriction of an outer measure $\mu^{*}$ is saturated?
I wasn't really sure about the validity of the result so I tried to construct a counterexample by considering the set $X = \{0,1\}$ and the outer measure $\mu^{*} : \mathcal{P}(X) \rightarrow [0, \infty]$ given by $\mu^{*}(\emptyset)=0$, $\mu^{*}(\{0\}) = 2$, $\mu^{*}(\{ 1 \})=2$ and $\mu^{*}(X)=3$ which is not generated by a pre-measure and has $\emptyset$ and $X$ as the only two $\mu^{*}$-measurable subsets of $X$. The problem is that the measure induced by this outer measure $\mu^{*}$ is saturated (since $\mu^{*}(X) is finite), so if a counterexample does exist it would be nice to know it.
If every measure obtained by the restriction of an outer measure is actually saturated, I'd like to know a proof of this result and my attempt was to consider the measure space $(X, \mathcal{M}^{*}, \overline{\mu})$ where $\mathcal{M}^{*}$ is the $\sigma$-algebra on $X$ consisting of the $\mu^{*}$-measurable subsets of $X$ and $\overline{\mu}$ is the restriction of the outer measure $\mu^{*}$ to $\mathcal{M}^{*}$ and then considering the outer measure, say $\mu^{+}$, induced by the measure $\overline{\mu}$. Since the outer measure $\mu^{+}$ is induced by the measure $\overline{\mu}$, we know that the measure obtained by restricting $\mu^{+}$ to the collection of $\mu^{+}$-measurable subsets of $X$, say $\hat{\mu}$, is a saturated measure and I believe I would be able to finish the proof if $\mathcal{M}^{*}$ is equal to the $\sigma$-algebra of $\mu^{+}$-measurable subsets of $X$ and $\mu^{+}= \mu^{*}$ (but I think this is only true if the original outer measure $\mu^{*}$ is induced by a pre-measure which leads us back to the original problem of proving the desired result when we drop this assumption).
Any hints or ideas would be greatly appreciated and thank you all in advance for your answers.
 A: OK I think I found a counterexample which shows the result is false in the general case (when the outer measure is not induced by a pre-measure). If $\mu^{*} : \mathcal{P}(\mathbb{R}) \rightarrow [0, \infty]$ is the outer measure defined by $\mu^{*}(\emptyset) = 0$, $\mu^{*}(A) = 1$ if $A$ is countable and nonempty, and $\mu^{*}(A) = \infty$ if $A$ is uncountable.
It is easy to see that $\mu^{*}$ is an outer measure on $\mathcal{P}(\mathbb{R})$.
Now, if $E$ is a proper nonempty subset of $\mathbb{R}$, $x \in E$ and $y \in \mathbb{R} \setminus E$, then $\{x,y\}$ is clearly a nonempty countable subset of $\mathbb{R}$, $\{x,y\} \cap E = \{x\}$, and $\{x,y\} \cap E^{c} = \{y\}$, so that $\{x,y\} \cap E$, and $\{x,y\} \cap E^{c}$ are nonempty countable subsets of $\mathbb{R}$ as well. From the definition of $\mu^{*}$ we obtain that $\mu^{*}(\{x,y\}) = 1 < 1+1 = \mu^{*}(\{x,y\} \cap E) + \mu^{*}(\{x,y\} \cap E^{c})$ and consequently $E$ is not a $\mu^{*}$-measurable subset of $\mathbb{R}$.
The above argument shows that $\emptyset$ and $\mathbb{R}$ are the two only $\mu^{*}$-measurable subsets of $\mathbb{R}$, that is $\mathcal{M}^{*} = \{ \emptyset , \mathbb{R} \}$ (where $\mathcal{M}^{*}$ denotes the $\sigma$-algebra of $\mu^{*}$-measurable subsets of $\mathbb{R}$). But then any proper nonempty subset $F$ of $\mathbb{R}$ is locally measurable in the measure space $(X, \mathcal{M}^{*}, \mu^{*}|_{\mathcal{M}^{*}})$ since $\emptyset$ is the only element of $\mathcal{M}^{*}$ with finite measure $\mu^{*}|_{\mathcal{M}^{*}}$ and $F \cap \emptyset = \emptyset \in \mathcal{M}^{*}$. Since said nonempty proper subset $F$ of $\mathbb{R}$ is not $\mu^{*}$-measurable we conclude that $\mu^{*}|_{\mathcal{M}^{*}}$ is not a saturated measure.
