Locally finite Borel Measures on are completely characterized by the Lebesque-Stieltjes outer Measure associated to a monotone increasing right continuous functions by defining the measure of a Borel set to be
where the infimum is taken over all coverings of by half open intervals. This is a fairly unruly albeit useful definition so it is valuable to have alternatives at hand to approximate the measure of Borel sets. The Approximation by Open and Compact sets is a great way to do so, as open and compact sets are well studied and understood, and as the following theorem will show, they approximate measurable sets arbitrarily well.
Let be a Lebesque-Stieltjes measure for a right continuous increasing function . By Caratheodory's Theorem, we know that the algebra of measurable sets makes into a measure space. Then, if we have that
To prove this statement, it is useful to state and prove the following lemma.
Lemma 1: In the notation of the Theorem Statement, we find that
proof. Let us denote the value on the right as . To show that , we will first show that and then that to conclude with equality.
Let . Then, we let be a sequence of real numbers so that and . Then, let . We find that can be expressed as the disjoint union so that by disjoint additivity,
given the definition of . Since this holds for all covers of by open sets, the claim follows.
Let Then, there exists a collection of half open intervals with so that
By the right continuity of , we know that for each , there exists some so that
proving by the definition of , that , completing the proof of the lemma.
proof. By the aforementioned lemma, we find that if , then there exist a sequence of intervals covering so that Taking then gives the following by countable subadditivity and monotonicity of .
which proves that Next, to show equality for compact sets. We will devolve into cases: is bounded and when it is unbounded.
Case 1: is bounded.
Let and let be open so that . Let . Then, is closed and bounded by construction. Then, since is measurable, we find that so that
which proves the result for this case.
Case 2: is unbounded
Let and . Then, is bounded so we can apply the previous case to find a compact set so that . Then, is compact with Then, since for all , we can apply continuity of to find that
from which the desired result follows. This proves the theorem.
- ↑ Gerald Folland Real Analysis: Modern Techniques and their Applications