What Must Be Checked in a Vitali Argument
Vitali’s construction is short, but its contradiction depends on several distinct facts: a representative is selected from each rational-equivalence class, certain translates are pairwise disjoint, the relevant family of translates is countable, and their union lies in a bounded interval. Confusing any of these steps can make an invalid argument look persuasive. In particular, the construction does not assign a measure to its representative set; rather, it proves that the set cannot be Lebesgue measurable.
Work in \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete and sigma-finite, but not finite. On \([0,1]\), define \(x\sim y\) if and only if \(x-y\in\mathbb{Q}\). By selecting one point from each equivalence class that meets \([0,1]\), obtain a set \(V\subseteq[0,1]\). The selection uses the axiom of choice. All set inclusions and equalities below are pointwise; no measurability of \(V\) is assumed unless explicitly stated.
Disjointness Is a Statement About the Shifts
The key disjointness claim concerns two distinct rational shifts. It does not say that every pair of distinct real translates of \(V\) is disjoint.
Proof. Suppose \(x\in(V+q_1)\cap(V+q_2)\). Then there are \(v_1,v_2\in V\) such that \(x=v_1+q_1=v_2+q_2\). Hence \(v_1-v_2=q_2-q_1\in\mathbb{Q}\). By the definition of \(\sim\), \(v_1\sim v_2\). Since \(V\) contains only one representative from each equivalence class, \(v_1=v_2\). Substituting this equality into \(v_1+q_1=v_2+q_2\) gives \(q_1=q_2\), contradicting the assumption. Thus the intersection is empty. \(\square\)
Worked Example: Why Distinct Real Shifts Can Overlap
Let \(\alpha=\sqrt{2}-1\), which belongs to \([0,1]\), and let \(v_0,v_\alpha\in V\) be the representatives of the equivalence classes of \(0\) and \(\alpha\), respectively. The classes are distinct because \(\alpha\notin\mathbb{Q}\), so \(v_0\ne v_\alpha\).
Since \(v_0-0\in\mathbb{Q}\) and \(v_\alpha-\alpha\in\mathbb{Q}\), the difference \(d=v_\alpha-v_0\) equals \(\alpha\) plus a rational number. Therefore \(d\) is irrational, and in particular \(d\ne0\). But \(v_0+d=v_\alpha\), so \(v_\alpha\in V\) and \(v_\alpha\in V+d\). Thus \(V\cap(V+d)\ne\varnothing\) although the shifts \(0\) and \(d\) are distinct.
Check: The disjointness proposition applies to distinct rational shifts. Here \(d\) is irrational, so its hypotheses do not hold. The overlap is not a counterexample to that proposition; it is a counterexample to the broader, incorrect claim that all distinct translates are disjoint.
Countability and the Range of the Translates
The set \(D\) is countable because \(\mathbb{Q}\) is countable, and it is infinite. Countability is essential: countable additivity applies to a countable sequence of pairwise disjoint measurable sets, not to an arbitrary uncountable family. Also, because \(V\subseteq[0,1]\) and \(-1\le q\le1\), every \(v+q\) lies between \(-1\) and \(2\). Thus \(V+q\subseteq[-1,2]\). There is no reason for these translates to remain in \([0,1]\).
Worked Example: Correcting an Incorrect Translate Bound
A mistaken argument might say that \(V+q\subseteq[0,1]\) for every \(q\in D\), and therefore that the union of the translates has measure at most \(1\). Test this claim at the allowed shifts \(q=-1\) and \(q=1\). From \(V\subseteq[0,1]\), translation gives \(V-1\subseteq[-1,0]\) and \(V+1\subseteq[1,2]\), not subsets of \([0,1]\).
The correct common bound is \(\bigcup_{q\in D}(V+q)\subseteq[-1,2]\). The interval \([-1,2]\) has length \(2-(-1)=3\), so its Lebesgue measure is \(3\). Consequently, if the union is measurable, its measure is at most \(3\), not at most \(1\). This bound is enough for the contradiction; it is not necessary to find a smaller interval.
Check: The shift \(q=-1\) subtracts \(1\) from both endpoints of \([0,1]\), giving \([-1,0]\); the shift \(q=1\) gives \([1,2]\). Both are contained in \([-1,2]\), whose measure is \(3\).
Worked Example: Why Uncountable Additivity Cannot Be Used
For each \(x\in[0,1]\), consider the singleton \(\{x\}\). These sets are pairwise disjoint, their union is \([0,1]\), and each has Lebesgue measure \(0\). The family is uncountable. Its union has measure \(m([0,1])=1\), even though every individual set has measure zero.
The invalid calculation would be “the measure of the union is the sum of the measures, hence \(1=\sum_{x\in[0,1]}0=0\).” Countable additivity does not authorize this uncountable sum. In the Vitali argument, countability is supplied by the rational shift set \(D\); it cannot be replaced by a family of all real shifts.
Check: The interval \([0,1]\) has measure \(1-0=1\), while every singleton has measure \(0\). The apparent conflict disappears because countable additivity makes no assertion of this kind for uncountable families.
The Contradiction Does Not Presuppose \(m(V)\)
The reasoning about the measure of \(V\) is conditional. If \(V\) were Lebesgue measurable, then its translates would be measurable by translation invariance of Lebesgue measure, and \(m(V)\) would be defined. Since \(V\subseteq[0,1]\), this hypothetical measure would satisfy \(0\le m(V)\le1\). Countable additivity can then be applied to the countable, pairwise disjoint family of rational translates.
Proof. Assume, for contradiction, that \(V\in\mathcal{L}\). Set \(W=\bigcup_{q\in D}(V+q)\). For any \(x\in[0,1]\), let \(v\in V\) be the representative of the equivalence class of \(x\). Then \(x-v\in\mathbb{Q}\), and \(x-v\in[-1,1]\) because \(x,v\in[0,1]\). Thus \(x-v\in D\), and \(x=v+(x-v)\in W\). This proves \([0,1]\subseteq W\). Also, \(V+q\subseteq[-1,2]\) for every \(q\in D\), so \(W\subseteq[-1,2]\). The set \(W\) is measurable as a countable union of measurable translates, and therefore \(1=m([0,1])\le m(W)\le m([-1,2])=3\).
Choose any \(N\) distinct elements \(q_1,\dots,q_N\) of \(D\). Their translates are pairwise disjoint by the proposition, are measurable by translation invariance, and all lie in \([-1,2]\). Countable additivity, applied to this finite disjoint family, and translation invariance give $$ N\,m(V)=\sum_{j=1}^{N}m(V+q_j) =m\!\left(\bigcup_{j=1}^{N}(V+q_j)\right)\le 3. $$ This holds for every positive integer \(N\). Since \(m(V)\) is a nonnegative finite number under the assumption, the inequalities \(m(V)\le3/N\) for all \(N\) imply \(m(V)=0\): if \(m(V)>0\), an integer \(N>3/m(V)\) would contradict \(N\,m(V)\le3\). But then translation invariance gives \(m(V+q)=0\) for every \(q\in D\), and countable subadditivity gives \(m(W)=0\). This contradicts \(1\le m(W)\). Hence \(V\notin\mathcal{L}\). \(\square\)
Worked Example: Testing Claims About the Measure of \(V\)
Suppose someone asserts that \(m(V)=1\). This assertion already assumes that \(V\) is measurable. Under that assumption, choose four distinct rational shifts in \(D\), for example \(0,\tfrac13,\tfrac23,1\). The four translates are disjoint, and translation invariance gives measure \(1\) to each. Their union would therefore have measure \(4\), by finite additivity on disjoint measurable sets. Yet that union is contained in \([-1,2]\), whose measure is \(3\). The claim \(m(V)=1\) is incompatible with the construction under the measurability assumption.
Likewise, assuming \(m(V)=0\) would make every rational translate null; countable subadditivity would then give \(m(W)=0\), contradicting \([0,1]\subseteq W\). The argument proves that \(V\) is not measurable. It does not assign \(V\) a Lebesgue measure, and it does not license an unsupported claim that its measure is positive, zero, or one.
Check: The four specified shifts are distinct members of \(D\). If \(V\) were measurable with measure \(1\), the finite union would have measure \(4\), while containment in \([-1,2]\) forces its measure to be at most \(3\).
A Reliable Checklist
When checking a Vitali proof, separate the set-theoretic statements from the measure-theoretic ones. First establish which shifts are being used and prove disjointness for precisely that family. Next check that the family is countable before invoking countable additivity or countable subadditivity. Then calculate a common containing interval from the actual range of the shifts. Finally, treat \(m(V)\) only under an explicit measurability assumption. The contradiction rules out that assumption; it does not begin with a known value for \(m(V)\).
A common invalid shortcut is to combine “the translates are disjoint,” “there are many translates,” and “the representative set has positive measure” without proving each statement and its hypotheses. In the standard construction, disjointness is proved for distinct rational shifts, the family is countably infinite, and the representative set’s measure is considered only conditionally. The bounded-union estimate then rules out both possible cases for a measurable \(V\): zero measure contradicts coverage, while positive measure contradicts the uniform finite bound on arbitrarily large finite unions.
Check Your Understanding
For each question, identify the exact hypothesis that supports the proposed measure step.
- In the disjointness proof, why does \(v_1-v_2\in\mathbb{Q}\) imply \(v_1=v_2\)?
- Give a reason that the proof for distinct rational shifts does not establish disjointness for all distinct real shifts.
- Why is it valid to apply finite additivity to \(N\) rational translates, but invalid to sum measures over all real shifts?
- For \(V\subseteq[0,1]\) and \(q\in\mathbb{Q}\cap[-1,1]\), give a common interval containing \(V+q\) for every such \(q\), and state its measure.
- Under the temporary assumption that \(V\) is measurable, explain why the inequalities \(N\,m(V)\le3\) for every positive integer \(N\) force \(m(V)=0\).