Separate null-set completion from arbitrary measurability

Try

A Borel measure may leave some subsets of null Borel sets outside its domain. Completion adds those subsets without changing the size of any previously measurable set. Lebesgue measure is complete, while the Borel sigma-algebra equipped with ordinary length is not the completed domain. Saying that a property holds almost everywhere therefore always refers to a specified measure space and its null sets.

Predict

Predict whether completion makes every subset measurable. It does not. Completion only fills in subsets of sets already known to have measure zero. A Vitali construction shows that translation-invariant countably additive length cannot consistently be defined on every subset of the line. The obstruction is structural, not a failure to find a sufficiently fine interval cover.

Experiment

Keep three claims separate.

Claim	Status for Lebesgue measure
Every Borel set is measurable	Yes
Every subset of a null set is measurable	Yes, by completeness
Every subset of the line is measurable	No, under the usual choice framework

A set can have a well-defined outer measure while failing measurability. Outer measure was defined on all subsets; the splitting equality is the additional test.

Understand

For a measure space (X, F,μ), its completion consists of sets E for which E differs from some A∈F only inside a measurable null set N. Assign the completed measure μbar (E)=μ(A). This is well-defined: two possible representatives differ only on a null set, so their measures agree by decomposing their common part and null remainders, without subtracting infinities. Complements and countable unions preserve this representation, using a countable union of null envelopes. Disjoint additivity follows by replacing representatives with disjoint measurable versions after removing null overlaps. Thus the completion is a complete measure extending the original one.

Apply

For nonmeasurability, declare x equivalent to y in[0,1] when x−y is rational. Choose one representative from each equivalence class, using the usual axiom of choice, and call the resulting set V. For rational q in[−1,1], the sets V+q are pairwise disjoint: equality v+q=w+r forces representatives v, w to be equivalent, hence equal, then q=r. Their union contains[0,1], since every point differs rationally from its representative, and it lies in[−1,2]. If V were Lebesgue measurable with measure zero, the union would be null, contradicting its containment of[0,1]. If V had positive measure, countable additivity and translation invariance would make the union infinite in measure, contradicting bounded containment. Thus V is not measurable.

Continue

The argument does not produce an explicit decimal membership algorithm for V, and it depends on selecting representatives. It also does not say every non-Borel set is nonmeasurable: completion supplies many measurable non-Borel subsets of null sets. In probability, completing the space makes modifications on null events measurable, but product constructions and conditional expectations still require careful domain statements. Use almost-everywhere equality to ignore null discrepancies in integrals, not to claim the underlying functions agree at every point or on every arbitrary subset.

Continue to Approximate measurable values from below.