College Math Teaching

January 13, 2011

A Non-Measurable Set

In our previous post, we talked about how to define a measure on a subset of the real line (or on [0,1] ).

In today’s post, we’ll give an example of a type of set for which no reasonable measure can be defined.
Let us recall what we wanted in a measure:

1.m([a,b])=b-a (and the same for open, and half open intervals). After all, this is supposed to be a length function. Of course,
b > a .

2. m(E+r)=m(E): that is, measure should be “translation invariant” that is, if you translate a set E by adding the same constant to every element of E, the translated set should have the same measure.

3. m(\{x\})=m(\emptyset )=0 (the measure of a single point and the empty set ought to be zero)

4. If F\subset E then m(F)\leq m(E) and we’d like to have some additivity properties:

5. Suppose E=\cup _{i=1}^{\infty }E_{i}. Then we’d want m(E)\leq\sum_{i=1}^{\infty }m(E_{i}) provided the left hand sum converges (of course, this convergence is absolute since all terms are positive; hence the order of the terms is of no consequence; this uses our calculus results).

Of course, we are talking about a countable union; sums don’t make sense otherwise.

And, if the sets E_{i} are disjoint, we’d love to have m(E)=\sum_{i=1}^{\infty }m(E_{i})

Now let’s define a set for which we could never have a measure which meets the above properties. Yes, we will be using the Axiom of Choice, which, roughly speaking, states that given an infinite disjoint collection of sets, we can “choose” one element from each set.

Start by selecting an irrational number x\in \lbrack 0,1] . Then let the set E consist of all y\in \lbrack 0,1] in which the following is true: y - x is irrational and if z\in E then y - z is irrational. One way to think of this is to establish one point in E and then consider a maximal set with respect to equality modulo the irrationals. Or one can think of building this set inductively with an uncountable number of steps: start with x , choose y so that y - x is irrational, then add z so that both x - z and x - y are irrational and so on. Of course, all arithmetic is done modulo 1.

Now lets look at the following: the rational numbers are countable so we can order them q_{1},q_{2}...,q_{k}. Then consider the translates of E : E_1 =E+q_{1}, E_2 = E + q_{2}.... .
Then we can establish: E is disjoint from each E_{i} because, say, if y\in E and y = z + q_{i} with z\in E then y - z would be rational with both y, z\in E . That is impossible. For a similar reason, all of the E_{i} are disjoint.

Now if w\in E' (the complement of E in [0,1] ), then there is a z in E such that w - z is rational; hence the collection of E_{i} forms the complement of E in [0,1].

So now we have the following situation: if measure is invariant under translation, E and each E_{i} have the same measure. Notice also that all of these sets are disjoint and that \lbrack 0,1]=E\cup \{\cup _{i=1}^{\infty }E_{i}\}

If each set has measure zero and we have countable additivity of disjoint sets, we’d have that [0,1] has measure zero. But if E has a non-zero measure and the measure is translation invariant, then we’d have [0,1] having infinite measure (remember we can’t form a positive infinite sum because each E_{i} has to have exactly the same measure.)

So, no matter how we define measure, we will have to exclude at least some sets (such as E ).

One other thing to note: E has as it’s compliment a countable number of translates of itself, hence it is impossible for E to meet the Caratheodory characterization requirement that E and its complement have measures that add up to the measure of [0,1]


  1. […] the next section, we construct a non-measurable set. Comments (1) LikeBe the first to like this post.1 Comment […]

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  3. […] there are measurable sets that aren’t formed in this manner, and there are such things as non-measurable sets. See here for the definition of “measurable […]

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