In measure theory, one usually (almost always?) defines a measure on a space X only for some subset **B**⊂2^{X} of the set of all subsets of X. The sets of **B** are called *measurable* sets; the sets of 2^{X}\**B** are called *unmeasurable*. The measure function is μ:**B**→**R** 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:2^{I}→**R** 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∈**Q**∩**I** (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∈**Q**∩**I**}.

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∈Q∩I}m(a+X) = ∑_{a∈Q∩I}m(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:2^{R}→**R**?

~~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 μ_{x}

μ_{x}(A) = 1, if x∈A;

μ_{x}(A) = 0, if x∉A

which is countably additive. Indeed, as μ

_{x} 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 μ_{x} quantifies just "how much x" is in a set. We can go a little further, by selecting a sequence x_{1}, x_{2}, ...∈**R** and positive coefficients c_{1}, c_{2}, ..., and looking at the measure

μ(A) = ∑_{i≥1} c_{i}⋅μ_{xi}(A)

which counts how many of x

_{i} (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 **R**^{n} (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.