Approximation by Open/Compact Sets

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Locally finite Borel Measures on are completely characterized[1] 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.

Theorem Statement

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


Then, and

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.


  1. Gerald Folland Real Analysis: Modern Techniques and their Applications