The ideal: for every , a function on the sets , (or ), so that is the -dimensional measure of , with given by the usual integrals, for example.
Naively, we wish for these to hold for :
(i) If (finite, or a sequence ) are all disjoint, then .
(ii) If ( for congruent here), i.e. can be transformed into by translations, rotations and reflections, then .
(iii) , where is the unit cube: .
These, it turns out, are mutually inconsistent. Let . Define on such that iff . Let be such that contains precisely one member of each equivalence class (by AOC). Let , and for every let
(0) ()
(1)
(2) (measure invariance under congruence)
(3) for disjoint
iff ;, so define , with elements of . is uncountable.
Let have 1 and only 1 point of each equivalence class (: choose one point from within ). So .
Lemma 1.1.1 (Claim). If , then either or .
Proof. Assume . Let the intersection be denoted . Take , so with , and with . Ah! So
since , so , BUT then ! So is equivalent to , and therefore .∎
Now consider (the sum the union of sets whose pairwise intersections are ).
Lemma 1.1.2 (Claim (2))..
and . Here , disjoint union.
So then
Note. Note that , since can be written .
Now , where , is such that .
So . Thus
Ah! But . So then
But there are infinitely many terms in this sum, one for each (and itself is clearly uncountable), and since they are all the same, it increases at a constant rate. Thus is either , or it must be since it needs to be (and is bounded by a finite number).
But is , so
Folland writes this as: . “is” (one representative per class), so for every , the class is represented in ., and for every we let
So every is in one and only one . If is such that , i.e. , then where if , or if . If , then (or ) and (or ) would be distinct elements of belonging to the same equivalence class, which cannot be.
Suppose satisfies (i), (ii), (iii). Then
Since is countable, . But by (iii), , and since , is either (if ) or (if ). Thus no such can exist.
Theorem 1.1.3 (Banach–Tarski Paradox). Let be open bounded sets in ,. There exist and subsets ,, of such that:
(a) the ’s are disjoint and their union is ;
(b) the ’s are disjoint and ;
(c) (congruent) for all .
Thus these measure axioms do not hold well in the “real world” (the of the parts is the whole, and the parts of each are congruent to another’s decomposition, thus for all ,). Now say with the same axioms as before. Let be a ball such that , and let be two disjoint translates of (i.e. disjoint from , and disjoint from and , with ).
Then
which can only be if is or ; contradiction, since we assume .
Moral: is so strangely put together that it’s impossible to define a geometrically reasonable notion of measure.
So instead of trying defined on ALL subsets of , we try on a class of subsets of . Such a class should be such that it includes all sets one is likely to meet in practice.
Let’s think more abstractly/generally! (ii) and (iii) relate to Euclidean geometry, but (i) doesn’t, so a set function satisfying (i) we may call a measure.
Example 1.1.4 (Examples). in a physics problem represents the total mass in the region ; in probability theory, consider a set which represents the possible outcomes of an experiment, and for , is the probability that the outcome lies in .