A set that has no size at all

The definition of measure applies to every subset of the line, and it is countably additive on the sets it calls measurable. The obvious question is whether the qualification is necessary, and the answer is that it is — provably, and by an argument that is a page long.

Giuseppe Vitali published the construction in 1905, four years after Lebesgue’s definition, and it settled the matter immediately: there is no way to assign a length to every subset of the line that is translation invariant and countably additive.

The classes #

Declare two real numbers equivalent when their difference is rational. That is an equivalence relation — it is reflexive, symmetric, and transitive because the rationals are closed under subtraction — so it cuts the line into classes.

Each class is a copy of the rationals shifted: the class of $x$ is $\{x + q : q \in \mathbb{Q}\}$. So every class is countable and dense, meeting every interval however short.

How many classes are there? Uncountably many. If there were countably many, then the line would be a countable union of countable sets and so countable, which it is not. So there are as many classes as there are real numbers.

Each class is dense and countable; there are uncountably many of them; and every real number is in exactly one. That is the whole setup, and none of it is controversial.

The choice #

Now form a set $V$ by taking exactly one point from each class, choosing the representative inside $[0,1]$ — which is possible since each class is dense and so meets that interval.

There is no rule for making that choice. The classes are not ordered in any way a selection could exploit; there is no smallest element, no canonical representative, nothing to prefer. The selection is asserted to exist by the axiom of choice, which says precisely that a choice from each of a family of non-empty sets can be made.

The figure draws four classes and picks one point from each, which is a selection from four sets and needs no axiom. The construction needs uncountably many simultaneous choices, and that is where the axiom enters and where the whole argument’s status comes from.

The translates tile #

The key property is that the rational translates of $V$ are pairwise disjoint and together cover a great deal.

Disjoint: if $V + q$ and $V + r$ share a point for rationals $q \neq r$, then $v + q = w + r$ for some $v, w \in V$, so $v - w = r - q$ is a non-zero rational, so $v$ and $w$ are in the same class — and $V$ has only one point per class, so $v = w$, so $q = r$. Contradiction.

Covering: take any $x \in [0,1]$. Its class has a representative $v \in V$, and $x - v$ is rational and lies in $[-1, 1]$. So $x \in V + q$ for that rational $q$, and the translates by rationals in $[-1,1]$ cover $[0,1]$.

Contained: every $V + q$ with $q \in [-1,1]$ sits inside $[-1, 2]$, since $V \subseteq [0,1]$.

So writing $Q$ for the rationals in $[-1,1]$, which is a countable set,

$$[0,1] \subseteq \bigcup_{q \in Q} (V + q) \subseteq [-1, 2].$$

The arithmetic that cannot work #

Suppose $V$ had a measure $\ell$. Translation does not change measure, so every $V + q$ has measure $\ell$. The translates are disjoint and countable in number, so by countable additivity the union has measure equal to the sum of countably many copies of $\ell$.

If $\ell = 0$, that sum is $0$. But the union contains $[0,1]$, whose measure is $1$, and measure is monotone. So the union has measure at least $1$, and $0 < 1$ is a contradiction.

If $\ell > 0$, that sum is infinite, since countably many copies of a positive number add to infinity. But the union sits inside $[-1,2]$, whose measure is $3$. Infinity is not at most $3$, and that is a contradiction too.

There is nothing left. A measure is a non-negative number and both cases fail, so $V$ has no measure — and the assumption that produced the contradiction was that every set has one.

The outer measure of the set, which does exist #

There is a wrinkle worth clearing up, because has no measure is easy to misread.

Outer measure is defined for every subset of the line, including $V$, and it is a perfectly good number. What fails is not the definition but additivity: $V$ is not measurable in Carathéodory’s sense, so its outer measure and the outer measures of its complement do not add correctly, and the number cannot be treated as a length.

The value itself is not determined by the construction. Different selections give sets of different outer measure, and one can be arranged to have outer measure as small as one likes above zero or as large as one — the choice inside each class is free enough to move it. So there is no the outer measure of a Vitali set, which is another sign that the object is a family of objects rather than one.

The failure is that a countable disjoint union of translates has outer measure strictly less than the sum of theirs. Subadditivity holds, as it always does; equality is what breaks, and it breaks by an infinite amount.

What the argument actually assumed #

It is worth listing the assumptions, because the contradiction shows they are jointly impossible and gives no opinion on which to drop.

Every subset has a measure. Used to give $V$ one.

Translation invariance. Used to give every $V + q$ the same measure.

Countable additivity. Used to add the translates.

Normalisation: an interval’s measure is its length. Used to get $1$ and $3$.

The axiom of choice. Used to build $V$.

Any four of the five are consistent. Lebesgue’s theory keeps the last four and drops the first, which is why measurable appears in every statement of the subject. That is the standard choice, and the reason it is standard is that the sets it excludes are exactly the ones nobody can write down.

The alternative nobody takes, and the one somebody did #

Dropping countable additivity gives back a measure on all sets — finitely additive ones exist, by a theorem of Banach, and in one and two dimensions there is even a translation-invariant one. But a finitely additive measure cannot take limits, and taking limits is what the whole apparatus is for. Nobody works in that setting for analysis.

Dropping the axiom of choice is more interesting. Solovay proved in 1970 that there is a consistent world in which every set of reals is Lebesgue measurable, at the cost of choice — a world with a weaker principle of dependent choice, enough to do most of analysis. So every set is measurable is not absurd; it is a genuine alternative with real costs elsewhere.

The construction therefore does not prove that non-measurable sets exist. It proves that they exist if choice does, which is a conditional, and the condition is one most mathematicians accept for reasons that have nothing to do with measure.

How much choice the argument needs #

The axiom of choice comes in strengths, and it is worth knowing which one is required, because the answer is not the full one.

What the construction needs is a choice function on a family of sets indexed by the reals — uncountably many simultaneous choices. That is more than countable choice, which allows a choice from each of a sequence of sets, and more than dependent choice, which allows a sequence of choices each depending on the last.

Countable choice and dependent choice are the versions ordinary analysis uses constantly and quietly: proving that a sequentially compact set is compact, that a countable union of countable sets is countable, that a function continuous in the sequential sense is continuous in the epsilon sense. All of those are safe, and none of them builds a Vitali set.

So the line between the choice analysis needs and the choice that produces pathology falls between countable and uncountable, and Solovay’s world is the precise statement of that: dependent choice holds, full choice fails, every set is measurable, and analysis goes through.

That is a satisfying resolution and it is not the one working mathematicians adopt, because full choice is used elsewhere — for a basis of every vector space, for the existence of maximal ideals, for Tychonoff’s theorem. Giving it up to save measurability trades one inconvenience for several.

Why the set cannot be described #

The strongest thing known about this is worth stating, because it explains why no picture of $V$ exists.

Every set that can be described by a formula in the ordinary language of analysis — every Borel set, every analytic set, and a good deal beyond — is measurable. The hierarchy of definable sets goes a long way up, and it is measurable all the way.

So $V$ is not definable in any of those senses. It is not that nobody has found a description; there provably is none of the kinds that have been catalogued. The set exists as an object the axiom of choice asserts, and every attempt to say which points are in it fails.

A choice nobody can write down is the general phenomenon and this is its sharpest instance: the axiom produces an object, and the object has no name.

The Banach–Tarski construction, and how it differs #

The famous relative of this argument is worth distinguishing, because the two are often run together.

Banach and Tarski showed in 1924 that a solid ball can be cut into five pieces and reassembled, by rotations and translations alone, into two balls of the same size. That also uses choice, and the pieces are also non-measurable.

The difference is which group is acting. Vitali’s argument uses translations of the line, which form a commutative group, and gets a contradiction with countable additivity. Banach and Tarski use rotations of three-space, which form a group containing a free subgroup on two generators, and get a much stronger conclusion — a paradoxical decomposition with finitely many pieces.

In one and two dimensions no such decomposition exists, precisely because the relevant groups are too commutative to contain a free subgroup. So the line’s pathology is Vitali’s and no worse, and three dimensions is where the group grows enough for the stronger statement.

A finite shadow of the same argument #

The construction has a finite cousin that is not paradoxical at all, and putting the two side by side shows exactly where infinity enters.

Take the integers modulo $12$ and the subgroup $H = \{0, 4, 8\}$ of multiples of four. It has three elements and index four, so choosing one representative from each of its four cosets gives a set $S$ of four elements, and the three translates $S$, $S + 4$, $S + 8$ are disjoint and cover all twelve.

Give each element weight $1/12$. Then $S$ has weight $1/3$, there are three translates, and $3 \times 1/3 = 1$. Everything adds up, no choice is needed, and no paradox appears.

The finite case never fails because the number of translates is finite, so the total is a finite sum and can land on any positive number at all. Vitali’s argument fails because the number of translates is countably infinite, and a countably infinite sum of equal terms has only two possible values: nought if the term is nought, infinity otherwise.

That gap — between the finitely many values a finite sum can take and the two a countable sum of equal terms can — is where the entire contradiction lives. Nothing about the classes or the choice would matter if the sum could land at $1$.

Countable additivity is what makes measure theory able to take limits and is also what makes this argument work; the two are the same property used for and against.

What the picture does and does not draw #

The hero is unusually limited even by this collection’s standards, and it is worth saying exactly what it manages.

It draws four classes as scatterings, each at fourteen points, generated from seeds that are differences of square roots of distinct non-squares — so their differences are irrational and the classes are genuinely different. That irrationality is not something the generator can check, and it does not claim to: what it checks is that no seed sits a small rational away from another, which would mean a seed had been written twice.

It draws six rational translates of a selection from those four, and checks that no two translates carry a point to the same place. That disjointness is exact, because each point is carried as a seed plus a rational and equality is decided in integers rather than by comparing decimals.

What it cannot draw is the selection, which needs one point from each of uncountably many classes. Four is a picture of four choices; the construction needs a choice for every real number, and the axiom asserting that has no diagram.

What the pictures cannot show #

Uncountably many classes, four drawn. The classes are as numerous as the reals and each is dense, so any honest picture of even two of them would fill the row.

The selection. Every point drawn in the lower half is a specific number with a formula. The set $V$ has no formula, and the rows are an illustration of a procedure rather than of the object.

Nor can it certify that the classes are different. Two seeds are in different classes exactly when their difference is irrational, and no finite computation decides that — every real is within a millionth of some rational with a denominator under a thousand, so a numerical test for irrationality is a test that always fails. The seeds are irrational by construction and the generator says so rather than pretending to verify it.

And the arithmetic that produces the contradiction. Countably many copies of one length adding to nought or to infinity is a fact about series, checked in the generator over candidate lengths and not visible in any arrangement of dots.

Where the ladder goes next #

This closes the ladder. It began with a set that has as many points as the line and no length, and it ends with a set that has no length in a stronger sense — not zero but undefined, and provably so.

What remains unwritten belongs elsewhere. The Cantor function, which is continuous, rises from nought to one and has zero derivative outside a set of measure zero, is a rung this ladder names and does not write. So is the Lebesgue integral in its own right, which is a different anchor rather than a further rung on this one. Sideways: the axiom that built the set is where its strangeness comes from, and the criterion for integrability is what the measurable sets were needed for.

What is worth carrying away #

When a list of desirable properties turns out to be inconsistent, the interesting question is which one is cheapest to give up.

Length should apply to every set, should not change when a set is slid along, should add up over countably many disjoint pieces, and should give an interval its own length. Those four cannot all hold. Every version of the subject is a choice among them, and the standard version gives up the first because the sets it loses are the ones nobody can name.

The general habit is to treat an impossibility theorem as a menu rather than as a defeat. Vitali’s argument is usually presented as showing that non-measurable sets exist. What it actually shows is that four reasonable demands are jointly unsatisfiable, and everything after that is a decision.

The same shape recurs across the collection. Three demands on a rule for dividing seats cannot all be met, and every real apportionment method is a choice among them. Four conditions on a way of splitting a cost pin down exactly one rule, which is the same theorem with the opposite sign. In each case the interesting content is the list, and the theorem’s job is to prove the list cannot be extended.