Covering a set from outside

Everything on this ladder so far has added up lengths as though the operation needed no defence. It does need one, and supplying it takes a definition that is three lines long and does something startling on its first application.

The definition is Lebesgue’s, from 1901. Given a set $E$ on the line, consider every way of covering it by countably many open intervals, add up the lengths of the intervals in each covering, and take the smallest total that any covering achieves — the infimum. That number is the outer measure of $E$.

Why the definition has the shape it does #

Three choices in that sentence are doing work, and each was wrong in some earlier attempt.

Countably many intervals, not finitely many. With finitely many, the rationals in $[0,1]$ would need a covering that is a finite union of intervals containing a dense set, which must be the whole interval — giving them measure one. Allowing countably many is what makes the theory work, and it is the single change from Jordan’s earlier definition.

Covering from outside, not filling from inside. An inner definition — the largest total length of intervals fitting inside — gives the rationals nothing, since no interval fits inside them. Taking outer and inner together and demanding they agree is Jordan’s approach, and it declares far too many sets unmeasurable.

The infimum, not any particular covering. No covering need achieve the value; the measure is the limit of what coverings can do, and for an interval no covering does better than the interval itself.

Those three choices are the whole design, and the rest of the theory is consequences.

The rationals, covered for nothing #

The first application is the one that convinces or does not.

List the rationals in $[0,1]$ — possible, since they are countable — and call them $q_1, q_2, q_3, \ldots$. Fix any $\varepsilon > 0$. Around $q_n$ put an open interval of length $\varepsilon/2^n$.

Every rational is covered, by construction. The total length is $\sum \varepsilon/2^n = \varepsilon$. So the outer measure of the rationals is at most $\varepsilon$, for every positive $\varepsilon$, and therefore it is zero.

A dense set — one that meets every interval of the line — has length nothing. That is the sentence that either lands or does not, and it is worth sitting with. The rationals are everywhere; between any two reals there are infinitely many; and their total length is zero.

The trick is entirely in the geometric series. Any sequence of lengths that sums to something finite would do, and halving is chosen because it is the easiest to write.

What the picture shows and what it hides #

The hero draws the first fourteen intervals separately and then their union as one band, and the two numbers under it are the point.

The lengths add to $0.18$. The union is also about $0.18$ long, because at this budget the intervals happen not to overlap much. That is a fact about the parameters, not a theorem — with a larger budget or a longer list they would overlap, and the union would be shorter than the sum.

The inequality that always holds is that the union is no longer than the sum, which is countable subadditivity, and it is the one property of outer measure that comes free from the definition. Anything more — that disjoint sets have measures that add — has to be proved, and cannot be proved for all sets at once, which is the subject of the last rung.

The figure also draws the rows in a margin wider than the unit interval, because the interval covering $0$ sticks out to the left of $0$ and the one covering $1$ sticks out to the right. A covering is allowed to do that, and clipping the picture would have hidden it.

Measure zero, and what it is a criterion for #

A set has measure zero exactly when it can be covered by intervals of arbitrarily small total length. That is worth stating as a standalone test because it is used constantly.

Any countable set has measure zero, by the argument above with the points listed instead of the rationals. So the algebraic numbers have measure zero, and the set of endpoints of the removed middles in any Cantor construction has measure zero.

Not every measure-zero set is countable. The middle-thirds set has as many points as the whole line and measure zero — its stage-$k$ approximation is a covering by $2^k$ intervals of total length $(2/3)^k$, which goes to nothing.

So measure zero is a strictly weaker condition than countable, and the gap between them holds the most interesting sets in the subject. Countability is about how many points; measure is about how much room they take.

The fat Cantor set, weighed properly #

The previous rung claimed a length for the fat Cantor set and had no definition to claim it with. With one in hand the claim can be made properly, and the argument is short.

The complement of the set inside $[0,1]$ is a countable union of disjoint open intervals — the removed middles — whose lengths add to about $0.4224$. Those intervals are a covering of the complement, and no covering does better, so the complement has measure exactly that.

Since the set and its complement partition the interval and both are measurable, their measures add to one, so the set has measure $0.5776$. Every step of that used countable additivity, which is the property the definition does not hand over free and which holds because both sets are measurable.

Compare the same computation for the middle-thirds set: the removed middles add to exactly one, the complement has measure one, and the set has measure zero. The two computations are identical and their answers differ because one series converges to one and the other does not.

Why an interval’s measure is its length #

The definition has to be checked against the one case where the answer is already known, and the check is not trivial.

That the outer measure of $[a,b]$ is at most $b-a$ is immediate: cover it with one interval slightly larger. That it is at least $b-a$ needs work, because it means no clever covering can do better.

The argument uses compactness. Suppose intervals $I_1, I_2, \ldots$ cover $[a,b]$ with total length less than $b-a$. Since $[a,b]$ is closed and bounded, finitely many of them already cover it — that is the Heine–Borel theorem — and a finite cover of an interval by intervals has total length at least the interval’s, which is an induction on the number of intervals.

So the first sanity check on the definition needs a real theorem, and the theorem is the one about finite subcovers. That dependency is worth noticing: the whole of measure theory rests on the fact that a closed bounded interval is compact.

Countable additivity is the hard part #

Outer measure is defined for every set, and it is subadditive: the measure of a union is at most the sum of the measures. What is wanted is equality for disjoint sets, and that is false in general.

Lebesgue’s solution is to restrict attention to sets that behave. A set $A$ is measurable when it splits every other set additively: for every $E$, the measure of $E$ equals the measure of $E \cap A$ plus the measure of $E$ outside $A$. That is Carathéodory’s criterion, and it looks like a technicality until it is used, at which point it does everything.

The measurable sets form a collection closed under complements and countable unions, and outer measure is countably additive on it. Every interval is measurable, so every set that can be built from intervals by countably many unions, intersections and complements is too — which is every set anybody writes down.

The sets that are not measurable have to be constructed, and the construction is the subject of the last rung.

Regularity: every measurable set is nearly an open one #

One theorem makes the definition usable in practice, and it says that measurable sets are not as strange as the definition permits.

For any measurable $E$ and any $\varepsilon > 0$ there is an open set containing $E$ whose measure exceeds $E$’s by less than $\varepsilon$, and a closed set inside $E$ whose measure falls short by less than $\varepsilon$. So every measurable set is squeezed between an open and a closed one that agree to within anything.

That is called regularity, and it is what lets proofs replace an arbitrary measurable set by an open one — a countable union of intervals — at the cost of an $\varepsilon$. Almost every argument in the subject does exactly that at some point.

It also explains the shape of the definition once more. Outer measure was defined by approximating from outside with open sets, so the outer half of regularity is nearly a restatement; the inner half is the content, and it holds because the complement can be approximated from outside too.

A set can be as complicated as one likes and still be within $\varepsilon$ of a union of intervals, which is the reassurance the definition needs after it has admitted the rationals and the Cantor sets.

Almost everywhere, and why it is the right notion #

The phrase that measure theory contributed to the rest of mathematics is almost everywhere: a statement holds almost everywhere when the set where it fails has measure zero.

That phrase is useful because measure-zero sets are negligible in the strongest available sense — a countable union of them is still one, so countably many almost-everywhere statements hold simultaneously almost everywhere. No comparable statement is true of nowhere dense sets: a countable union of nowhere dense sets can be everything, as the rationals show.

Closure under countable unions is why measure zero became the standard notion of negligible rather than any of its competitors. It is the property that lets an argument take a limit.

A limit that forgets to be continuous is the standard trouble with pointwise limits, and almost-everywhere convergence is the notion under which the trouble becomes manageable — which is most of what the Lebesgue integral is for.

What the definition costs to compute with #

A last practical point, since the definition is an infimum over an uncountable family of coverings and that sounds unusable.

In practice nobody searches over coverings. For an open set the measure is the sum of the lengths of its component intervals, which is a series. For a closed set it is one minus the measure of its complement, which is open. For anything built from those by countable operations the measure comes from countable additivity, one step at a time.

So the definition is used once, to establish the properties, and then never again — every actual computation goes through additivity and the value on intervals. That is the usual shape for a definition by infimum: it exists to prove the rules, and the rules do the work.

The area under a curve is the same arrangement one level up. The integral is defined by a supremum over approximations, and nobody computes an integral by searching approximations; the definition establishes linearity and the fundamental theorem, and those are what get used.

A definition nobody computes with directly is not a bad definition. It is a definition doing its job, which is to make theorems provable rather than to make arithmetic easy.

The outer measure of a set nobody can describe #

Since outer measure is defined for every set, it is defined for sets that cannot be written down, and its value there is not always what intuition suggests.

There are sets whose outer measure is one and whose complement in $[0,1]$ also has outer measure one — so both halves of a partition of the interval appear to be the whole thing. That is not a contradiction, because outer measure is only subadditive, and it is a sign that at least one of the two sets is not measurable.

Constructing such a set needs the axiom of choice, and it is exactly what the last rung does. What is worth noticing here is that the definition permits the pathology without producing it: outer measure is a perfectly good function on all sets, and the trouble is only that it fails to add up.

A definition that applies to everything and behaves on most things is the usual outcome, and drawing the line between the two is the technical content of the subject.

Translation invariance, and the property that has to give #

One more property is worth naming now because the last rung is about losing it.

Outer measure is translation invariant: sliding a set along the line does not change its measure, because sliding a covering slides its intervals and leaves their total length alone. That is immediate from the definition and it is the property that makes the notion a length rather than an arbitrary weighting.

So outer measure has three desirable properties. It is translation invariant. It gives an interval its own length. And it is countably subadditive, with countable additivity on the measurable sets.

The last rung shows those three cannot be strengthened to hold for every subset of the line at once: assuming countable additivity on all sets, together with translation invariance and the interval’s length, produces a contradiction. Something has to be given up, and what Lebesgue gives up is the demand that every set be measurable.

That is a deliberate choice and not a failure, and it is worth knowing which of the alternatives were available. Giving up translation invariance produces a theory nobody wants. Giving up countable additivity gives back a theory that cannot take limits, which is what the integral is for. Giving up the axiom of choice makes every set measurable and costs a great deal elsewhere. Restricting the sets is the cheapest of the four.

What the pictures cannot show #

The infimum is over all coverings and only one is drawn. The hero shows a particular covering with a particular budget. The measure is the smallest total any covering achieves, and no picture holds all of them.

The union’s length is computed and not visible. At the scale drawn, fourteen intervals of rapidly shrinking length are mostly a few pixels wide, and whether two of them overlap cannot be seen. The generator merges them and reports the total.

And the list of rationals is finite in the drawing and infinite in the argument. Forty rationals leave gaps that a reader can see; the full list leaves none, and the total length is still under the budget. There is no stage of the picture at which the density becomes visible.

Where the ladder goes next #

The definition is now in place, and the next rung asks what it was built for. Riemann’s integral fails on some perfectly ordinary functions, and the exact criterion for it to work is that the discontinuities have measure zero — a statement that could not even be phrased before this rung.

Sideways: a dense countable set is the object whose measure came out at nothing, and the fat Cantor set is the object this rung was needed to weigh.

What is worth carrying away #

A definition earns its keep by what it makes true, not by how obvious it looks.

Covering from outside with countably many intervals is not the only conceivable definition of length, and it is not the first one that was tried. What recommends it is that intervals get their own length, countable sets get nothing, the notion is closed under countable operations, and the sets it fails on have to be built deliberately.

The dense set of measure zero is the fact to keep. Being everywhere and taking up no room are compatible, and any argument that treats dense as though it implied substantial has already gone wrong.