The answer to your second question is yes. Let $G=(\mathbb{Q}[\sqrt{2}], +).$ This is a countable abelian group, so $\mathbb{R}/G$ is a hyperfinite Borel equivalence relation. In particular, it embeds into $\mathbb{R}/\mathbb{Q}.$ (See section 7 here for a summary of these results: Kechris - The theory of countable Borel equivalence relations, preliminary version May 8, 2019). Thus we can use a transversal for the latter to construct a transversal for the former. Let $X$ be a transversal for $\mathbb{R}/G.$
Let $V=\{(x + q\sqrt{2}) \text{ mod 1}: x \in X, q \in \mathbb{Q}\} \subset [0, 1).$ This is a Vitali set closed under translation by $\mathbb{Q}\sqrt{2}.$ We will show its outer measure is at least $\frac{1}{4}$ (in fact, the argument can be extended to show its outer measure is 1). Suppose towards contradiction that $U$ is an open cover of $V$ with measure less than $\frac{1}{4}.$ For each $n \ge 2,$ we can canonically find an open cover of $V \cap [0, \frac{\sqrt{2}}{n}]$ of measure less than $\frac{\sqrt{2}}{2n}$ by considering the least $m$ such that $\lambda(U \cap [\frac{m\sqrt{2}}{n},\frac{(m+1)\sqrt{2}}{n}])<\frac{\sqrt{2}}{2n}.$ With this we can recursively construct open covers $U_n$ of $V$ of measure less than $2^{-n-2}.$ Letting $\{q_n\}$ enumerate the rationals, we see $\{q_n+U_n\}$ covers $\mathbb{R},$ so $\lambda^*(\mathbb{R}) < 1,$ contradiction.
Also, here's an observation relevant to your first question. If $\mathbb{R}=\bigcup_{n<\omega} X_n$ is a countable union of countable sets, then there is a null subset of $[0, 1]$ which meets every mod $\mathbb{Q}$ class, namely $W=\bigcup_{n<\omega} \{x \in [0, 2^{-n}]: \exists q \in \mathbb{Q} (x+q \in X_n)\}.$ This is null since it countable outside of any $[0, \epsilon].$ Furthermore, if there is a Vitali set, that could be used to separate out a Vitali subset of $W,$ so a model satisfying these two properties would be a counterexample to the first question.
There are at least two different $\sigma$-algebras that Lebesgue measure can be defined on:
- The (concrete) $\sigma$-algebra ${\mathcal L}$ of Lebesgue-measurable subsets of ${\bf R}^d$.
- The (abstract) $\sigma$-algebra ${\mathcal L}/{\sim}$ of Lebesgue-measurable subsets of ${\bf R}^d$, up to almost everywhere equivalence.
(There is also the Borel $\sigma$-algebra ${\mathcal B}$, but I will not discuss this third $\sigma$-algebra here, as its construction involves the first uncountable ordinal, and one has to first decide whether that ordinal is physically "permissible" in one's concept of an approximation. But if one is only interested in describing sets up to almost everywhere equivalence, one can content oneself with the $F_\delta$ and $G_\sigma$ levels of the Borel hierarchy, which can be viewed as "sets approximable by sets approximable by" physically measurable sets, if one wishes; one can then decide whether this is enough to qualify such sets as "physical".)
The $\sigma$-algebra ${\mathcal L}$ is very large - it contains all the subsets of the Cantor set, and so must have cardinality $2^{\mathfrak c}$. In particular, one cannot hope to distinguish all of these sets from each other using at most countably many measurements, so I would argue that this $\sigma$-algebra does not have a meaningful interpretation in terms of idealised physical observables (limits of certain sequences of approximate physical observations).
However, the $\sigma$-algebra ${\mathcal L}/{\sim}$ is separable, and thus not subject to this obstruction. And indeed one has the following analogy: ${\mathcal L}/{\sim}$ is to the Boolean algebra ${\mathcal E}$ of rational elementary sets (finite Boolean combinations of boxes with rational coordinates) as the reals ${\bf R}$ are to the rationals ${\bf Q}$. Indeed, just as ${\bf R}$ can be viewed as the metric completion of ${\bf Q}$ (so that a real number can be viewed as a sequence of approximations by rationals), an element of ${\mathcal L}/{\sim}$ can be viewed (locally, at least) as the metric completion of ${\mathcal E}$ (with metric $d(E,F)$ between two rational elementary sets $E,F$ defined as the elementary measure (or Jordan measure, if one wishes) of the symmetric difference of $E$ and $F$). The Lebesgue measure of a set in ${\mathcal L}/{\sim}$ is then the limit of the elementary measures of the approximating elementary sets. If one grants rational elementary sets and their elementary measures as having a physical interpretation, then one can view an element of ${\mathcal L}/{\sim}$ and its Lebesgue measure as having an idealised physical interpretation as being approximable by rational elementary sets and their elementary measures, in much the same way that one can view a real number as having idealised physical significance.
Many of the applications of Lebesgue measure actually implicitly use ${\mathcal L}/\sim$ rather than ${\mathcal L}$; for instance, to make $L^2({\bf R}^d)$ a Hilbert space one needs to identify functions that agree almost everywhere, and so one is implicitly really using the $\sigma$-algebra ${\mathcal L}/{\sim}$ rather than ${\mathcal L}$. So I would argue that Lebesgue measure as it is actually used in practice has an idealised physical interpretation, although the full Lebesgue measure on ${\mathcal L}$ rather than ${\mathcal L}/{\sim}$ does not. Not coincidentally, it is in the full $\sigma$-algebra ${\mathcal L}$ that the truth value of various set theoretic axioms of little physical significance (e.g. the continuum hypothesis, or the axiom of choice) become relevant.
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I have a (possibly idiosyncratic) view that the natural form of measure theory is for finite measure spaces and bounded functions. Other cases are obviously very important, but we have to work harder to get them. You can see this is many of the proofs, where the finite case is easier, and we have to work a bit more to generalize it. For example, the usual proof of the Radon-Nikodym theorem works that way.
In the finite measure space case, with bounded functions, everything can be made symmetric. The symmetry is broken in the general case, because allowing infinity breaks it. In integration, this asymmetry shows up in the way we have to have separate theorems for the non-negative measurable functions and the integrable functions. For bounded functions on finite measure spaces you don't need to impose any extra conditions.