Undergraduates with some exposure to measure theory may know that the Lebesgue measure is the unique, up to multiplication by a constant, translation-invariant Borel measure. If not, then no worries, as today’s post will cover the proof of this result. My motivation for selecting this topic for a blog post was I was trying to remember the proof that Lebesgue measure is invariant under transformation by an orthogonal matrix. The proof of this invariance result, along with a more general result about the behavior of the Lebesgue measure under linear transformations, will be proven in a later post.
Let be a set. A family is a Dynkin system if
- pairwise disjoint .
It is immediate from the definition that , and is closed under finite disjoint union. It is evident that every -algebra is a Dynkin system, but the following example shows the converse is false:
Let for some fixed . Set . That is a Dynkin system contains and is closed under countable disjoint union is obvious. For closure under relative complement, is the difference of two even numbers and therefore even. Evidently, is not a -algebra, since , but .
Proposition 1. Let . Then there is a smallest (also minimal, coarset) Dynkin system containing . is called the Dynkin system generated by . Moreover, .
Proof. Define to be the intersection of all Dynkin systems containing . Then exists and is non-empty since is a Dynkin system and . That is minimal is immediate from the definition. Since every -algebra is a Dynkin system and , we have that .
Lemma 2. A Dynkin system is a -algebra if and only if it is closed under finite intersection: .
Proof. If is a -algebra, then a fortiori it is closed under finite intersection. Suppose is closed under finite intersection. Let be an arbitrary sequence in . Define
By our -stable hypothesis, for each , and is pairwise disjoint. Hence, . So is closed under countable union, and therefore a -algebra.
Proposition 3. If is stable under finite intersections, then .
Proof. Tautologically, , so by definition of , it suffices to show that is a $\sigma$-algebra. By the preceding lemma, it suffices to show that is stable under finite intersection. Fix and define the family
I claim that is a Dynkin system. . Observe that if , then
by the Dynkin axioms applied to . Closure under countable disjoint union is obvious. Observe that since is -stable, . Since is a Dynkin system, we have that , when . Hence,
which is equivalent to saying that is closed under finite intersections.
The proof of the following proposition is obvious from our preceding work.
Proposition 4.Let be a -algebra, be a Dynkin system, and be two collections of subsets of . Then
The next proposition is a useful tool for checking the equivalence of two measures on a given measurable space.
Proposition 5. (Uniqueness of Measures) Assume that is a measurable space and that is generated by a family such that
- There exists an exhausting sequence with
Then any two measures that coincide on and satisfy are equal on (i.e. ).
Proof. Let be two measures that coincide on and satisfy for each . For each , define
I claim that is a Dynkin system. Clearly, , and if , then by additivity,
If is pairwise disjoint, then using that ,
Obviously by hypothesis that is -stable and that agree on . Hence, for each . Since is -stable, by Proposition 3, . Hence, . Since , for every ,
We denote the Lebesgue measure on by .
Theorem 6. (Lebesgue Translation-Invariance)
The -dimensional Lebesgue measure is invariant under translations:
Every measure on which is invariant under translations and satisfies is a real multiple of the Lebesgue measure: .
Proof. We first verify that is indeed a Borel set. Define
I claim that is a -algebra. It is clear that , and is closed under countable union. For closure under relative complement, I claim that . Indeed,
Since translation is a homeomorphism, we have that , where is the collection of open (in the Euclidean topology) subsets of . By definition, , which implies that .
For , set for fixed. It is clear that is a measure on . Let denote the collection of half-open rectangles in . Let and observe that
which shows that . Since is -stable, generates , and has an exhausting sequence , we have by Proposition 5 that .
Let be a measure on which satisfies and is translation invariant. Observe that, if is a positive integer, then by additivity of , we see that
which implies that . If are positive integers, then
where we use the additivity of measures. It follows from the translation-invariance of that, for half-open rectangle as above, but with for , the formula
holds. By Proposition 5, .