III.55 Measures

To understand measure theory, and to see why it is useful and important, it is instructive to start with a problem about lengths. Suppose that we have a sequence of intervals in [0, 1] (the closed interval from 0 to 1), of total length less than 1. Can they cover [0, 1]? In other words, given intervals [a₁, b₁] [a₂, b₂], . . . , with Σ(b_n - a_n) < 1, is it possible that their union equals [0, 1]?

One is tempted to answer “no, as the total length is too small.” But this is just to restate the question. After all, why should “total length less than 1” actually imply that the intervals cannot cover [0, 1]? Another tempting answer is to say “just rearrange the intervals so that they go from the left to the right, and then we never get to the right-hand end of [0, 1].” In other words, if the nth interval has length b_n-a_n = d_n, then just translate the intervals to be the intervals [0, d₁], [d₁, d₁ + d₂], . . . . In this rearrangement it is indeed true that we never cover any point beyond Σd_n, and so do not cover all of [0, 1], but why does this imply that the original intervals do not cover [0, 1]?

It is quite easy to see that this rearrangement argument works for a finite number of intervals, but it does not work in general. Indeed, suppose we ask our original question again, but this time for the rationals: that is, let us replace the interval [0, 1] by the rational interval [0, 1] ∩ . If our intervals have lengths for example, so that the total length is only , then certainly the left-to-right intervals will cover only the interval [0, ] ∩ , but it is possible for the original intervals to cover all of [0, 1] ∩ , since we can just enumerate the rationals as q₁, q₂, . . . (see COUNTABLE AND UNCOUNTABLE SETS [III.11]), and then put an interval of length around q₁, one of length around q₂, and so on.

This observation shows that the answer to our problem must involve properties of the reals that are not shared by the rationals, which wrecks any kind of it is obvious” argument. In fact, the result is true for the reals, but its proof is a good exercise.

Why is this an important fact? It stems from a wish to define “length” for general sets of reals (for simplicity, we will concentrate on [0, 1], just to avoid some technicalities about “infinite length”). What should the “length” of a set be? For intervals the answer is clear, and it is also clear for finite unions of intervals. But what about sets like {, , , . . .}, or itself?

A natural first attempt would be to use finite unions of intervals: one could take the length of a set A to be the least value of the length of a finite union of intervals that covers A. More precisely, one could define the length of A to be the infimum of (b₁ - a₁) + ··· + (b_n - a_n), taken over all finite unions of intervals [a₁, b₁] ∪ ··· ∪ [a_n, b_n] that cover A.

Unfortunately, this definition has some very undesirable properties. For example, the length of the set of all rational numbers in the interval [0, 1] would then be 1, as would the length of all irrational numbers in [0, 1]. We would thus have two disjoint sets (and very natural ones at that) such that the length of their union is not the sum of their lengths. So this form of “length” is not really well-behaved for such sets.

What we want is a notion of length that applies to all the sets we know and are used to, and is additive, meaning that the length of A∪B is the sum of the lengths of A and B whenever A and B are disjoint. Remarkably, this can be achieved, and the key idea is to allow countable covers. That is, we modify the above definition as follows: the length (or measure, to give it its usual name) of a set A is the infimum of (b₁-a₁)+(b₂-a₂)+···, taken over all unions of intervals [a₁, b₁] ∪ [a₂, b₂] ∪ ··· that cover A. Note that, thanks to the puzzle discussed earlier, the measure of the interval [a, b] is b - a, just as we would hope.

It is also not hard to see that the measure of the set of rationals in [0, 1] is zero, and it turns out that the measure of the irrationals in [0, 1] is 1. Indeed, any countable set has measure zero. In many contexts, sets of measure zero are regarded as “negligible” or “of no importance.” It is worth mentioning that there are also sets of measure zero that are uncountable (an example is the CANTOR SET [III.17]).

It turns out that, even with this definition, there are pairs of disjoint sets A and B such that the measure of A ∪ B is not the sum of the measures of A and B. However, it can be shown that for all “reasonable” sets the measure is additive. More precisely, one says that a subset of [0, 1] is measurable if the measures of it and its complement add up to 1, as they should. If A and B are disjoint measurable sets, then the measure of their union is the sum of their measures.

This is a very useful fact, since it can be shown that every set that arises naturally in mathematics, or that has an explicit definition, is measurable: intervals, finite unions of intervals, countable unions of intervals, Cantor sets, things involving rationals or irrationals, and so on. In fact, the union of any countable family of measurable sets is again measurable: one says that the measurable sets form a sigma-algebra. Even better, for measurable sets the measure is countably additive, meaning that the measure of a disjoint union of countably many measurable sets is the sum of the measures of the individual sets.

More generally, in many other settings, one wants to end up with a sigma-algebra, containing all the sets one is interested in, on which we can define a countably additive measure, or “length function.” The above example is called Lebesgue measure on [0, 1]. In general, whenever one wishes to define a countably additive measure, one always needs a result like the puzzle above in order to get started.

An important sigma-algebra is the algebra of all Borel sets. This is the smallest sigma-algebra that contains all open and closed intervals. Roughly speaking, it is the collection of all sets that you can build out of open and closed intervals using countable unions and intersections. (However, this masks the fact that the building up process can be very complicated: there is in fact a transfinite hierarchy of Borel sets.) The sigma-algebra of all Borel sets is smaller than the sigma-algebra of all Lebesgue-measurable sets because an arbitrary set of measure zero does not have to be a Borel set. Borel sets are one of the basic notions of DESCRIPTIVE SET THEORY [IV.22 §9]: in a certain technical sense they are “easily describable.”

Here is another example of a sigma-algebra with a countably additive measure: we could work in [0, 1]² (the unit square in the plane), and base our ideas upon rectangles instead of intervals. So we would define the measure of a set as the least total area of a sequence of rectangles that covers the set. This gives an elegant and powerful approach to integration: the integral of a function f (defined on [0, 1], say, and taking values in [0, 1]) is just defined to be the “area under its graph”: that is, the measure of the set {(x,y) : y ≤ f(x)}. Many complicated-looking functions can now be integrated: for example, the function f that is 1 on the rationals and 0 on the irrationals is easily checked to have an integral, namely 0, whereas in earlier theories such as Riemann integration that function would be too rapidly varying to be integrable.

This approach to integration gives rise to the so-called Lebesgue integral (further discussed in the article on LEBESGUE [VI.72]), which is one of the fundamental concepts in mathematics. It allows one to integrate a wide range of functions that are not Riemann integrable, but the main reason for its importance is not so much this as the fact that the Lebesgue integral has very good limiting properties that the Riemann integral lacks. For example, if f₁, f₂,. . . is a sequence of Lebesgue-integrable functions from [0, 1] to [0, 1] and f_n(x) converges to f(x) for every x, then f is Lebesgue-integrable and the Lebesgue integrals of the functions f_n converge to the Lebesgue integral of f.