Three Tests for a Vitali Shift Range
A Vitali argument uses translates of a set of representatives, but not every choice of shifts gives the same proof. To get a contradiction, the shifts must do enough to cover an interval of positive measure; infinitely many translates must also fit inside a bounded interval, so their measures cannot all be positive. This tutorial tests those requirements in explicit variations.
Work on \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete and sigma-finite, but not finite. Define \(x\sim y\) for \(x,y\in[0,1]\) if and only if \(x-y\in\mathbb{Q}\). By the axiom of choice, select a set \(V\subseteq[0,1]\) containing exactly one point from each equivalence class. The goal is to test whether \(V\) can be Lebesgue measurable.
For a set \(D\subseteq\mathbb{Q}\), write \(V+D=\bigcup_{q\in D}(V+q)\), where \(V+q=\{v+q:v\in V\}\). Three questions organize the argument: are different translates disjoint, does their union cover an interval, and can infinitely many of the translates be contained in one bounded interval?
What Disjointness and Coverage Require
Proof. Suppose \(z\in(V+q)\cap(V+r)\). Then \(z=v+q=w+r\) for some \(v,w\in V\), so \(v-w=r-q\in\mathbb{Q}\). Thus \(v\sim w\). Since \(V\) contains exactly one representative of each class, \(v=w\), and then \(q=r\). This proves disjointness when the shifts are distinct. For coverage, the representative \(v\) of the class of \(x\) satisfies \(x-v\in\mathbb{Q}\). Since \(x,v\in[0,1]\), \(-1\le x-v\le1\). Taking \(q=x-v\) gives the claimed representation. \(\square\)
The coverage statement is deliberately tied to the full difference range \([-1,1]\): it guarantees that the shift from \(v\) to \(x\) is among the allowed shifts. For bounding translates, if \(D\subseteq[-s,t]\), where \(s,t\ge0\), then \(V\subseteq[0,1]\) gives \(V+D\subseteq[-s,1+t]\). The length of this containing interval is \(1+s+t\). These facts let us calculate the bounds before assuming \(V\) is measurable.
A Sufficient Criterion for the Contradiction
Proof. Assume, for contradiction, that \(V\in\mathcal{L}\). Translation invariance of Lebesgue measure makes each \(V+q\) measurable and gives \(m(V+q)=m(V)\). Put \(a=m(V)\); since \(V\subseteq[0,1]\), monotonicity gives \(0\le a\le1\), so \(a\) is finite. The translates indexed by distinct shifts are pairwise disjoint by the lemma. For every positive integer \(N\), choose \(N\) distinct shifts \(q_1,\ldots,q_N\in D\). Their union is contained in \(J\), so finite additivity on these disjoint measurable sets gives $$ N a=\sum_{j=1}^{N}m(V+q_j) =m\left(\bigcup_{j=1}^{N}(V+q_j)\right) \le m(J)<\infty. $$ The same bound holds for every \(N\). If \(a>0\), choose an integer \(N>m(J)/a\); then \(Na>m(J)\), contradicting the inequality. Hence \(a=0\). Because \(D\) is countable, \(V+D\) is a countable union of measure-zero translates. Countable subadditivity gives \(m(V+D)=0\). But \(I\subseteq V+D\), so monotonicity gives \(0<m(I)\le m(V+D)=0\), another contradiction. Therefore \(V\notin\mathcal{L}\). \(\square\)
The boundedness hypothesis supplies one fixed finite upper bound for every finite collection of translates. Countability is used separately: once \(m(V)=0\), it ensures the full union \(V+D\) is null. In applications, the set of shifts that proves coverage need not itself be bounded, provided it is countable and an infinite bounded collection of allowed shifts is available to force \(m(V)=0\).
Worked Example: The Symmetric Range from −1 to 1
Take \(D=\mathbb{Q}\cap[-1,1]\). Show that this range yields the contradiction and calculate its containing-interval bound.
The lemma shows that \([0,1]\subseteq V+D\): every \(x\in[0,1]\) differs from its representative by a rational in \([-1,1]\). Also \(V\subseteq[0,1]\), so translating by \(q\in[-1,1]\) gives \(V+q\subseteq[-1,2]\). Thus \(J=[-1,2]\) works, and its length is \(2-(-1)=3\). The range \(D\) is countable and contains infinitely many distinct rationals.
If \(V\) were measurable and \(a=m(V)\), then any \(N\) distinct shifts in \(D\) would give \(Na\le3\), since the corresponding disjoint translates lie in \([-1,2]\). If \(a>0\), an integer \(N>3/a\) contradicts this bound, so \(a=0\). Countable subadditivity then gives \(m(V+D)=0\), although \([0,1]\subseteq V+D\) and \(m([0,1])=1\). Check: the interval bound is \(3\), while the covered interval has measure \(1\); a null union cannot contain that interval.
Worked Example: An Asymmetric Range from −1 to 2
Keep the same representative set \(V\), but take \(D=\mathbb{Q}\cap[-1,2]\). Check that the extra positive shifts preserve coverage and compute the new bound.
For each \(x\in[0,1]\), the shift \(q=x-v\) from its representative still belongs to \(\mathbb{Q}\cap[-1,1]\), which is a subset of this \(D\). Hence \([0,1]\subseteq V+D\), even though \(D\) is asymmetric. For every \(q\in[-1,2]\), \(V+q\subseteq[-1,3]\): the smallest possible left endpoint is \(0+(-1)=-1\), and the largest possible right endpoint is \(1+2=3\). The containing interval has length \(3-(-1)=4\).
Under the assumption that \(V\) is measurable, \(N\) distinct shifts give disjoint translates inside \([-1,3]\), and therefore \(Na\le4\), where \(a=m(V)\). If \(a>0\), choosing an integer \(N>4/a\) is impossible, so \(a=0\). Since this \(D\) is countable, \(m(V+D)=0\); since \([0,1]\subseteq V+D\), monotonicity would also imply \(1\le m(V+D)\). The contradiction shows that the wider asymmetric range works as well. Check: the endpoints used in the bound are \(0-1=-1\) and \(1+2=3\), so its length is exactly \(4\).
Worked Example: A Narrow Range Does Not Guarantee Coverage
Take \(D=\mathbb{Q}\cap[-\tfrac12,\tfrac12]\). Explain why the coverage step cannot be inferred for every valid choice of representatives.
Choose the representative \(0\) for the equivalence class \(\mathbb{Q}\cap[0,1]\), and choose one representative from every other class as usual. This is a valid selection. If \(1\in V+D\), then \(1=v+q\) for some \(v\in V\) and \(q\in D\). Both \(1\) and \(q\) are rational, so \(v=1-q\) is rational and belongs to \([0,1]\). Thus \(v\) is in the class represented by \(0\), and the selection forces \(v=0\). The equation then requires \(q=1\), which does not belong to \([-\tfrac12,\tfrac12]\). Therefore \(1\notin V+D\) for this valid selection.
This example does not assert that every narrow-range union misses \(1\) for every selection. It shows that the selection rule alone does not ensure coverage by this range. The criterion cannot be applied without proving coverage for the actual shifts and representatives. Check: the only possible selected representative in the rational class is \(0\), and the resulting required shift is \(1\), strictly greater than \(\tfrac12\).
When the Bounds Are Sufficient—and a Common Error
The numerical upper bound need not be close to the length of the covered interval. In the first example, the containing interval has length \(3\), and in the second it has length \(4\); both are enough because each one is a single finite bound that applies to arbitrarily many disjoint translates. The proof does not require the upper bound to be smaller than the measure of the covered interval. It first forces \(m(V)=0\), and only then uses countable subadditivity to make the entire union null.
A common error is to treat a bounded range as if it automatically covered the target interval. For the narrow range above, the incorrect step would be: “Every \(x\in[0,1]\) is equivalent to a representative \(v\), so \(x=v+q\) with \(q\in D\).” Equivalence gives only \(q=x-v\in\mathbb{Q}\); the bound \(|q|\le\tfrac12\) does not follow. For the point \(x=1\) and the specified representative \(v=0\), the required shift is \(q=1\notin D\). The correction is to verify both \(q\in\mathbb{Q}\) and the actual range condition \(q\in D\) before claiming coverage.
A second error is to conclude \(m(V)=0\) from disjointness alone. Disjointness says that the measures add for a finite family; the bounded containing interval supplies the uniform estimate \(Na\le m(J)\). Without a fixed finite \(m(J)\), this argument does not force \(a=0\). Conversely, boundedness alone is not enough either: the shift range must contain infinitely many distinct shifts to make \(N\) arbitrarily large, and the countable union used after \(a=0\) must remain countable.
For each target point, write it as its selected representative plus a rational shift, and verify that the shift belongs to the chosen range.
Use uniqueness of representatives: an intersection between two translates would make their representatives equivalent, hence equal, forcing the shifts to be equal.
Find one bounded interval containing the translates, then apply its finite measure to every finite collection of distinct shifts.
After the finite-family bound forces \(m(V)=0\), use countability of the full shift range to conclude that its union is null.
Check Your Understanding
For each question, state which coverage or measure-bound condition is needed, rather than relying only on the fact that the shifts are rational.
- For \(D=\mathbb{Q}\cap[-2,2]\), find an interval containing \(V+D\), and calculate its length.
- For \(D=\mathbb{Q}\cap[-1,1]\), identify the rational shift that represents a point \(x\in[0,1]\) from its selected representative \(v\).
- Why does \(Na\le m(J)\) for every positive integer \(N\), with \(m(J)<\infty\), force \(a=0\)?
- In the narrow-range example, why does \(1=v+q\) force \(v=0\) and \(q=1\)?
- Could an uncountable shift family be used in the same final null-union step? Explain which hypothesis of the proposition would then be missing.