In measure theory, one usually (almost always?) defines a measure on a space X only for some subset B2X of the set of all subsets of X. The sets of B are called measurable sets; the sets of 2X\B are called unmeasurable. The measure function is μ:BR requires that B be a σ algebra to attain its name. But we usually get some unmeasurable sets.

Why? What prevents us from extending, say, Lebesgue measure to be defined on all sets?

Theorem. Let I=[0,1] (the argument works equally well for any interval, and immediately extends to show the same result for R). [Assuming the axiom of choice,] there is no measure m:2IR which is invariant to translation (i.e. m(x+A)=m(A) for every set A⊆I and x∈I).

Proof. Define an equivalence relation ~ on the real numbers: x~y iff x-y∈Q is a rational number. It is easy to see that ~ is an equivalence relation. For s∈I and T⊆I, we will use the notation s+T={s+t mod 1:t∈T} to denote the set T, shifted right cyclically by s.

Let X⊂I be a set of representatives of the equivalence classes I/~. That is,

• For every x,y∈X, x-y∉Q (every 2 elements of X are an irrational distance apart);
• For every a∈I (or indeed a∈R), there is a (necessarily unique) x∈X for which a~x, i.e. a-x is rational.
The existence of this X follows from the axiom of choice, and requires it or something similar.

Consider the sets a+X, for all a∈Q. On the one hand, these sets are disjoint (or we have x,y∈X and a≠b∈Q for which a+x=b+y; this means x-y=b-a∈Q, in contradiction to the 1st property of X). On the other, we have that I is the disjoint union of the sets a+X, a∈QI (for if some y∈R is in no a+X, then the second property of X fails). So I is the countable union of the disjoint sets {a+X:a∈QI}.

We claim that X must be an unmeasurable set. For if X were measurable, then either m(X)=0 or m(X)>0. In the first case, by countable additivity and invariance of m we would have 1 = m(I) = ∑a∈QIm(a+X) = ∑a∈QIm(X) = 0. In the second case, by these same properties we would have 1 = m(I) = . Neither is possible.

QED.

Suppose we give up on invariance to translation. We won't have Lebesgue measure on R, we'll just have a measure. That is, we decide to ignore all properties of R except for cardinality. IS there a countably additive measure μ:2RR?

We don't know. AxelBoldt points out that we can find some, but they're not satisfying replacements for Lebesgue measure. For instance, if we pick x∈R, we can define a measure μx

μx(A) = 1, if x∈A;
μx(A) = 0, if x∉A
which is countably additive. Indeed, as μx makes no distinction between elements of R, except whether or not they are x, it is "ℵ-additive": Additivity holds for any collection of disjoint sets. It's just not very interesting... (Of course, similar results hold when replacing "1" with "17", or indeed with "∞" in the measure above).

It's probably not what we want, though. Measures are supposed to quantify "how much" there is of something, and μx quantifies just "how much x" is in a set. We can go a little further, by selecting a sequence x1, x2, ...∈R and positive coefficients c1, c2, ..., and looking at the measure

μ(A) = ∑i≥1 ci⋅μxi(A)
which counts how many of xi (suitable weighted) are in a set A. μ still doesn't distinguish between more than countably many classes of elements of R, though, so it's still not very interesting. In particular, it doesn't satisfy anything of what we'd expect from generalising "length", and it cannot be made to do so.

The situation is quite dissatisfying. It looks like the ugliness of σ-algebras from measure theory is here to stay.

If you really want to feel better about it (despite the true hopelessness of the situation), you can go the other way and look at smaller σ-algebras. Think of "true" physical systems. Can you hope to tell the volume of anything? Probably not. But you can definitely tell the volume of any Borel set. And Borel sets are the smallest interesting σ-algebra in Rn (they're the smallest that contains all intervals...), so we could argue that we're not missing anything really important. The concept of a measurable function follows from measurability. Going further, a physical function must be measurable. Indeed, in stochastic calculus we use intentionally-limited measures to be able to talk about things like "functions which don't know the future"; Itô's lemma is true only for such functions. The limitation of measurability can be used to our advantage. But it's a poor excuse for what every mathematics undergraduate would love to be able to do: to measure every set.

Thanks to AxelBoldt for correcting some horrible errors in the original version.

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