Vitali Constructions on Other Bounded Intervals

Adapting the Construction to an Interval

The rational-translation construction is not tied to \([0,1]\). Its essential feature is that two points in the interval can differ by only a bounded amount, while their difference belongs to \(\mathbb{Q}\) whenever they lie in the same rational-translation class. The interval’s length therefore determines both which rational shifts are needed and how large the union of translates can be.

All measure statements below are in the Lebesgue measure space \((\mathbb{R},\mathcal{L},m)\), which is complete and sigma-finite but not finite. Lebesgue measure is translation-invariant: if \(E\in\mathcal{L}\) and \(q\in\mathbb{R}\), then \(E+q=\{x+q:x\in E\}\) is in \(\mathcal{L}\) and \(m(E+q)=m(E)\). Countable additivity applies to disjoint Lebesgue measurable sets. A bounded interval with endpoints \(a<b\), with any choice of including or excluding endpoints, has measure \(b-a\).

Using \([-L,L]\), rather than only rational shifts from the unit-interval construction, ensures that every difference between two points of \(I\) is covered. This choice also gives a simple containing interval for all the translates. The endpoints of \(I\) need not be included: its closure is contained in \([a,b]\), and all the estimates below use that closed interval.

Coverage and Disjointness

Proof. Take \(x\in I\). Its equivalence class in \(I\) has exactly one representative \(v\in V\). Since \(x\sim v\), the number \(q=x-v\) is rational. Also \(x,v\in[a,b]\), so \(-L\le x-v\le L\). Thus \(q\in D_L\) and \(x=v+q\in V+q\), proving the first inclusion.

For disjointness, suppose \(v_1+q_1=v_2+q_2\), where \(v_1,v_2\in V\) and \(q_1,q_2\in D_L\). Then \(v_1-v_2=q_2-q_1\in\mathbb{Q}\). The points \(v_1,v_2\) are therefore in the same equivalence class in \(I\), so the representative property gives \(v_1=v_2\). The original equality then gives \(q_1=q_2\). Hence distinct translates are disjoint.

Finally, \(V\subseteq[a,b]\) and every \(q\in D_L\) satisfies \(-L\le q\le L\). Therefore \(V+q\subseteq[a-L,b+L]\) for each \(q\in D_L\), which proves the second inclusion. \(\square\)

The shifts form a countably infinite set: \(D_L\) is countable because it is a subset of \(\mathbb{Q}\), and it is infinite because \(L>0\) and there are infinitely many rationals in \([-L,L]\). The containing interval in the proposition has length $$ (b+L)-(a-L)=(b-a)+2L=3L. $$ Thus the translated family covers an interval of measure \(L\), while all its members lie in an interval of measure \(3L\).

The Translate Bounds Force Nonmeasurability

Proof. Suppose, for contradiction, that \(V\in\mathcal{L}\). Every translate \(V+q\) is then Lebesgue measurable and satisfies \(m(V+q)=m(V)\). Write \(c=m(V)\). Because \(V\subseteq[a,b]\), monotonicity of Lebesgue measure gives \(0\le c\le L\), so \(c\) is finite.

Enumerate the distinct elements of the countably infinite set \(D_L\) as \(q_1,q_2,\ldots\). The proposition gives pairwise disjoint translates, so countable additivity and the two inclusions yield $$ L=m(I)\le m\left(\bigcup_{j=1}^{\infty}(V+q_j)\right) =\sum_{j=1}^{\infty}m(V+q_j) =\sum_{j=1}^{\infty}c \le m([a-L,b+L])=3L. $$ Here the equality with the series uses countable additivity for disjoint measurable sets; the bounds use monotonicity. If \(c=0\), the series is \(0\), contradicting \(L>0\). If \(c>0\), its partial sums are \(Nc\), which are unbounded as \(N\to\infty\); this contradicts the upper bound \(3L<\infty\). Both possibilities are impossible. Hence \(V\notin\mathcal{L}\). \(\square\)

The displayed inequalities are conditional on the assumption that \(V\) is measurable. They are not claims that a measure \(m(V)\) exists for the constructed representative set. The contradiction is precisely that measurability would assign it a value \(c\), and neither possible value type—zero or positive—can satisfy the translate bounds.

A Common Error: Keeping the Unit-Interval Shift Range

The allowed shifts depend on the interval’s length. For example, in \([2,5]\), using only shifts in \(\mathbb{Q}\cap[-1,1]\) does not guarantee coverage: the point \(x=5\) and a possible representative \(v=2\) have difference \(x-v=3\), which is excluded from that range. The correction is to use \(\mathbb{Q}\cap[-3,3]\), because the interval has length \(3\). More generally, if \(I\) has endpoints \(a<b\), then \(|x-v|\le b-a\) for \(x,v\in I\), so the range \(\mathbb{Q}\cap[-(b-a),b-a]\) always contains the required rational differences.

A related mistake is to say that all the translates remain inside \(I\). They need not: a shift can move a representative beyond either endpoint. The valid containment is the expanded interval \([a-L,b+L]\), whose measure is \(3L\). It is this finite upper bound, not containment inside the original interval, that supplies the contradiction.

If the interval has length zero, the argument does not apply: \(D_0=\{0\}\), and a representative set for a singleton interval is measurable. The positive-length hypothesis is necessary, both for the infinite family of rational shifts and for the positive lower bound \(m(I)=L\).

The method’s key point is that the numerical constants adjust with the interval: its measure supplies the lower bound \(L\), and the expanded containing interval supplies the upper bound \(3L\). The conflict between these bounds and the common measure of infinitely many disjoint translates is unchanged.