Measure Theory, Measure Space

Extended Reals

Extend the reals by plus and minus infinity. Reals can be added as usual, but ∞+x = ∞, and -∞+x = -∞. It's not clear what ∞-∞ would be, so call it 0. Multiplication can also be defined, with ∞ trumping all the finite real numbers, other than 0. (We stillw want 0 times anything to be 0.)

Note that this structure is not a ring, since addition is not associative. Consider 1+(∞-∞) and (1+∞)-∞. The results are 1 and 0 respectively. However, the nonnegative extended reals do form a commutative semiring. Verify the properties of a ring, other than additive inverse. Remember to consider all the finite and infinite combinations. I'll leave the details to you.

How do we extend our linear topology? The real line is open, and belongs to σ, hence the complement, plus and minus infinity, also belongs to σ. Every set that was in σ before, can be augmented with ±∞, or not, as you please. Nothing separates +∞ and -∞, unless we put one of these points specifically in σ. Of course, only +∞ is relevant if we are considering the nonnegative extended reals.

Note that the sum or product of measurable functions into the extended reals remains measurable. Use the same proof as before. The quotient is measurable too, if we can avoid the ambiguities of 0 and infinity in the denominator.

Measure Space

A measure space (X,σ,μ) is a measurable space (X,σ), along with a function μ, called a measure, from σ into the extended reals, having the following properties.

  1. μ(∅) = 0.
  2. μ(E) ≥ 0 for every E in σ.
  3. If E is the countable union of pairwise disjoint sets Ej, them μ(E) is the sum over μ(Ej). Since all values are nonnegative, the series is absolutely convergent, and the order of the sets Ej is not significant.

The measure μ assigns a value to each set in σ. Property (1) assigns 0 to the empty set, and (2) is sometimes called positivity, since it prohibits negative values. Without positivity, μ is called a charge, rather than a measure. Finally, (3) combines disjoint sets together and adds their measures.

Finite unions are implied by (3). Pad the rest of the series with repeated instances of the empty set. These are (technically) pairwise disjoint, so we're ok. And they all contribute zeros to the infinite sum. Thus μ of a finite disjoint union is the sum of μ of each set.

If a countable collection of sets is not disjoint, then the measure of the union is less than or equal to the sum of the individual measures. Let's prove this now.

Let F be a proper subset of E, with both F and E in σ. Let G = E-F. In other words, G is the part of E that is not in F. Remember that the complement of F is in σ, so intersect this with E to show G is in σ. Now μ(E) = μ(F) + μ(G), and μ(F) ≤ μ(E). I will call this the subset lemma.

If μ(F) is finite, then μ(G) = μ(E)-μ(F). Show that E has a finite measure iff G does, then verify the relationship for μ(G) finite and μ(G) infinite.

Now, let Ej be a sequence of sets that may not be disjoint. Let Gj be the extra stuff in Ej that has not been brought in by E1 through Ej-1. Since G1 = E1, they have the same measure. Bring in E2 and G2. Since G2 is a subset of E2, its measure is no larger than that of E2 (subset lemma). Thus the measure of G1∪G2 is no larger than μ(E1) + μ(E2). Yet G1∪G2 = E1∪E2. Thus the measure of the union is no larger than the sum of the two measures. This extends, by induction, to all finite unions. Take the limit and find the same result for the countable union over Ej, which has a measure no larger then the infinite sum of the individual measures.

Counting Measure

Let μ(E) be the size of E. Infinite cardinalities map to the generic value ∞. Verify the properties of a measure space. In fact, μ remains a measure when σ is the discrete algebra on X, or any weaker subalgebra thereof. That is why I call this a counting measure, rather than specifying a particular measure space σ.

Lebesgue Measure

Let X be the power set of the reals. Certain sigma algebras on X are called Lebesgue algebras. I will postpone the definition of a Lebesgue algebra for now. Let's just say they exist. Now, such an algebra admits a Lebesgue measure μ, where μ on the closed interval [a,b] is b-a. Thus μ of a single point is 0. This definition holds whether the interval is closed, open, or half open.

Expanding / Contracting

The sequence of sets En is expanding if En is included in En+1 - and contracting if En contains En+1. If En is expanding, let U be the union over En. The measure of U is the limit of the measure of En. Let Gn equal En - En-1. This is the extra stuff brought in by En. Now U is the disjoint union over Gn, and its measure is the sum of the measures of Gn. This is, by definition, the limit of the partial sums. Yet the nth partial sum is the measure of En. Therefore μ(U) is the limit of μ(En).

Since a series does not typically have terms equal to infinity, verify the theorem when En is infinite. By the subset lemma, each larger set has infinite measure, and the same holds for U. In our extended arithmetic, an infinite term leads to an infinite sum; hence the theorem still holds.

A similar result holds for a contracting sequence, with one additional constraint. At least one set in the sequence must have a finite measure. Without this, one could construct a sequence of descending sets, each infinite, with an empty intersection. The open intervals from 0 to 1/n under the counting measure, for instance. Since 0 is not the limit of infinity, this doesn't work. Therefore, let's assume at least one term has finite measure, whence all the terms beyond have finite measure. We may as well start the sequence at this point, so that all measures are finite. Let I be the intersection, and let μ(I) = c. Let Gn = En - En+1. Now E1 is the disjoint union over Gn, along with I. Let G0 = I. Now the sum of the measures of Gn, starting with n = 0, is equal to μ(E1. Subtract the measures from both sides, starting at G1, and c becomes the limit of the partial sums, which is the limit of μ(En). That completes the expanding/contracting theorem.