Lebesgue Theory Part 3

Previously, we defined a measure m*(I) to be the infimum of the sum of the components (covers) of the measure (i.e. due to the measure being additive). We talked about moving forward and refining our example. In some texts, they continue with denoting a measure using m(I). However, most texts move forward with a different notation and an additional exposition of properties regarding the set functions and how they play in with being a measure.

Now, in the last post we defined A=\bigcup\limits_{i=1}^n(I_i). We can now cash-out why that is significant.  Since we have defined A as the finite union of a finite union of intervals, A is said to be elementary. Let’s talk a little about elementary sets:

Q1: What the is an elementary set?

A1: To describe what it is, let’s take a look at its properties:

p1. Every bounded interval is an elementary set.

p2. The intersection of two elementary sets is an elementary set.

p3. The difference of two elementary sets is also an elementary set.

p4.  Set E is an elementary set IFF it is the union of finitely many intervals.

Now, ask yourself, do these look familiar? These are the same properties that we described, sans p4, for a ring. Notice, however, that an elementary set, denoted by \mathscr{E}, is not a \sigma-ring! Why? Well, because we said that a set is elementary IFF it can be expressed as a finite union of intervals. A \sigma-ring contains the property if being closed under countable (possibly infinite) union.

Now that we have the notion of elementary sets down, we have some definitions that need evaluating.

Definition 1: A nonnegative additive set function \phi defined on \mathscr{E} is said to be regular if the following is true: to every A\in\mathscr{E} and to every \epsilon>0, there exists F\in\mathscr{E} and G\in\mathscr{E} where F is compact (and closed) and G is open, so that : F\subset A \subset G and

\phi(G)-\epsilon\leq \phi(A) \leq \phi(F) + \epsilon

If A=I_1\cup I_2\cup I_3\cup \cdots \cup I_n and if these intervals are pairwise disjoint, then (think additivity) m(A)=m(I_1)+\cdots+m(I_n);

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