Wikipedia (and our measure theory books) tell us the following:
The modern construction of the Lebesgue measure, based on outer measures, is due to Carathéodory. It proceeds as follows:
For any subset B of Rn, we can define an outer measure λ * by:
, and is a countable union of products of intervals .
Here, vol(M) is sum of the product of the lengths of the involved intervals. We then define the set A to be Lebesgue measurable if
for all sets B. These Lebesgue measurable sets form a σ-algebra, and the Lebesgue measure is defined by λ(A) = λ*(A) for any Lebesgue measurable set A.
It seems like magic that this strange characterization of a Lebesgue measurable set would be a sigma-algebra, it is not at all obvious that the property of Lebesgue measurability has anything to do with what our intuition would expect it to be, and the fact that this specific definition of the outer measure would have any properties we would want it to have (sigma-additivity). Also, even if the definition of the outer measure is somehow intuitive, why not define it as the supremum of the contained boxes instead of the infimum of the containing boxes?
My suggestion is to start fresh, and try to define the measure on R naively, and see how it works out. Maybe by doing this we'll learn why we need Cartheodory's crazy magic.
How would we want to define a measure function on
? It might not be defined on all subsets of , since the Vitali Set shows that if you assume the Axiom of Choice then you may not define a translation-invariant (meaning ) measure on all subsets of .Here are some intuitive properties we would expect out naive measure function:
It should be defined on all segments, and It should be translation-invariant.
It should be sigma-additive.
For any set A for which the measure is defined, it should also be defined for the complement of A, and would be equal to... ...what?There seems to be a problem here - we can't really define the measure this way, since
has infinite measure. But if we were to do the same thing and try to define a measure on subsets of [0,1], or rather, [0,1), we could use this definition to define a measure on all of . If is a measure defined on a certain class of subsets of [0,1), and we want our new measure on to be translation invariant, we can define it as following -- split any set A into its intersection with all the segments of the form (n, n+1). On each one of them, use to calculate the measure of that "chunk" of A, and then add them all together to find the meausure of A. Formally,
So let's try to define as a measure on [0,1), and then get back to the measure defined on all of . Now we can go back to the properties we expect to have:
It should be defined on all segments, and It should be translation-invariant.
It should be sigma-additive.
For any set A for which the measure is defined, it should also be defined for the complement of A, and (this is actually implied by sigma-additivity).In fact, the class of sets for which this measure will be defined is precisely the Borel Sets, which is nice because indeed we would "expect" these sets to be measurable (would we expect other sets to be measurable?). This definition of the measure in fact gives us a direct way to (recursively) compute the measure of any Borel Set, which is also very nice -- and it is all naively intuitive.
BUT! Who says this definition is consistent. Meaning, if two Borel sets can be represented in two different ways using countable unions and complements, how do we know we'll get the same value for the measure of the set? For example, consider the set A = (0, 1/2]. By property 1, its measure should be 1/2. But we can also represent A as the union of all sets of the form
. In this case, we can prove that these two representations of A have the same measure if we use our recursive definition. But how would we prove that this is always the case?
Is it possible that we need Cartheodory's crazy constructions in order to prove this? Would this be a proper introduction to the study of the Cartheodory system?