Sets of Unambiguous Measure Here is a fictionalised history of measure on $[0,1]$. We started with intervals and assigned them lengths. We noticed the properties of these lengths regarding disjoint unions and complements, and, wanting these to hold for arbitrary sets, came to the Borel measure $\mu$. We then asked if there was a natural way to extend this measure. The Lebesgue measure was natural because its sets $S$, in some sense, had to have the measure $\lambda(S)$. They are all disjoint unions of Borel sets $B$ and null sets $N$, which in any possible extension of the Borel measure, had to be assigned measure $\mu(B)$. My question is, can we go any further in this fashion?
Formally, let $(\Omega,\mathcal{F},\mu)$ be a measure space and $(\Omega,\mathcal{F}_0,\mu_0)$ its completion. Call a measure space $(\Omega,\mathcal{F}',\mu')$ such that $\mathcal{F}\subseteq\mathcal{F}'$ and $\mu'|_\mathcal{F}=\mu$ an extension of $(\Omega,\mathcal{F},\mu)$. Suppose that for the set $S\subseteq\Omega$ there exists an extension $(\Omega,\mathcal{F}',\mu')$ of $(\Omega,\mathcal{F},\mu)$ such that $S\in\mathcal{F}'$ and for any two such extensions $(\Omega,\mathcal{F}',\mu')$ and $(\Omega,\mathcal{F}'',\mu'')$, $\mu'(S)=\mu''(S)$. Then is it necessarily the case that $S\in\mathcal{F}_0$? If not, is it at least true for $\mathbb{R}$ as the Borel or Lebesgue measure space?
 A: For a counterexample, let $\Omega$ be an uncountable set,
$\mathcal{F}$ be the $\sigma$-algebra on $\Omega$ consisting of all
the countable subsets of $\Omega$ and their complements, and $\mu$
the counting measure on $\mathcal{F}.$ Clearly, $\mu$ is complete,
i.e., $\mathcal{F}_0 = \mathcal{F}.$ Let $S$ be a subset of $\Omega$
that does not belong to $\mathcal{F}.$ (Incidentally, it is not
trivial to prove that such a set exists!  It does, assuming the
Axiom of Choice: see Kaplansky, Set Theory and Metric Spaces
(1972), Theorem 13.) Proper
extensions of $\mu$ exist, for example the counting measure on the
$\sigma$-algebra of all subsets of $\Omega.$ Because $S$ has
$\mu$-measurable subsets of arbitrarily large measure, the measure
of $S$ in every extension of $\mu$ must be $+\infty.$ Therefore, $S$
has "unambiguous measure" in the terms of the question; but we do
not have $S \in \mathcal{F}_0.$
On the other hand, in the general situation, if we suppose that $S$
has unambiguous measure and $\mu^*(S) < +\infty,$ then it is
necessarily the case that $S \in \mathcal{F}_0.$
Proof. Let $S$ be a subset of $\Omega$ such that
$\mu^*(S) < +\infty.$ By Proposition 1.5.5 of D. L. Cohn,
Measure Theory (2nd ed. 2013), we have $S \in \mathcal{F}_0$
if and only if $\mu_*(S) = \mu^*(S).$ Let $X$ be a subset of
$\Omega$ satisfying $S \subseteq X \in \mathcal{F}$ and
$\mu(X) < +\infty.$ Let $\hat{\mathcal{F}}$ be the collection of all
$\mu$-measurable subsets of $X,$ and $\hat{\mu}$ the restriction of
$\mu$ to $\hat{\mathcal{F}}.$ Then
$(X, \hat{\mathcal{F}}, \hat{\mu})$ is a finite measure space,
$\hat{\mu}^*(S) = \mu^*(S),$ and $\hat{\mu}_*(S) = \mu_*(S).$ Let
$\mathcal{C}$ be the $\sigma$-algebra on $X$ generated by
$\hat{\mathcal{F}} \cup \{S\}.$ By
Cohn, Exercise 1.5.12 (c), for every number $\alpha$ between
$\mu_*(S)$ and $\mu^*(S),$ there exists a measure $\nu$ on
$(X, \mathcal{C})$ that extends $\hat{\mu}$ and satisfies
$\nu(S) = \alpha.$
I have posted a detailed proof of the latter result:
A measure is completely determined by its values on borel
sets?. (Although the
question mentions only Borel sets, the answer applies generally.)
The proof has received no comments or votes (up or down!), so it is
hard to be entirely confident of its validity, but I see nothing
wrong with it.  The answer mentions in passing that a theorem in
Royden, Real Analysis (3rd ed. 1988) seems to show that the
hypothesis of finiteness is unnecessary; if that's true, it would
simplify the present argument considerably.
Let $\tilde{\mathcal{F}}$ be the collection of all $\nu$-measurable
subsets of $\Omega$ disjoint from $X,$ and $\tilde{\mu}$ the
restriction of $\mu$ to $\tilde{\mathcal{F}}.$ Then
$(\Omega \setminus X, \hat{\mathcal{F}}, \hat{\mu})$ is a measure
space.  Define:
$$
\mathcal{A} = \{C \cup D : C \in \mathcal{C}, \
D \in \tilde{\mathcal{F}}\}.
$$
Then $\mathcal{A}$ is a $\sigma$-algebra on $\Omega.$ For all
$A \in \mathcal{A},$ define:
$$
\rho(A) = \nu(A \cap X) + \tilde{\mu}(A \setminus X).
$$
Then $(\Omega, \mathcal{A}, \rho)$ is a measure space such that
$\rho$ extends $\mu$ and $\rho(S) = \alpha.$ Therefore, if $S$ has
unambiguous measure, we must have $\mu_*(S) = \mu^*(S),$ i.e.,
$S \in \mathcal{F}_0. \ \square$
The constraint $\mu^*(S) < +\infty$ cannot be dropped, even when
$\Omega = \mathbb{R},$ $\mathcal{F}$ is the Borel algebra
or the algebra of Lebesgue measurable subsets of
$\mathbb{R},$ and $\mu$ is Lebesgue measure, $\lambda,$ on
$\mathcal{F}.$ For a counterexample in this case, let $X = [0, 1],$
and let $V$ be a subset of $X$ that is not Lebesgue measurable, such
as a Vitali set (or a van Vleck set, as used in the other answer).
Repeat the preceding construction of $(\Omega, \mathcal{A}, \rho),$
but with $V$ in place of $S$; now let $S = \mathbb{R} \setminus V.$
Because $S$ has subsets of arbitrarily large Lebesgue measure, it
must have measure $+\infty$ in any extension of $\lambda$ for which
it is measurable.  Also, $S \in \mathcal{A},$ so $S$ is measurable
in the proper extension $(\mathbb{R}, \mathcal{A}, \rho)$ of
$(\mathbb{R}, \mathcal{F}, \lambda).$ That is, in the terms of the
question, $S$ has "unambiguous measure"; but it is not Lebesgue
measurable, i.e., $S \notin \mathcal{F}_0.$
