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Lecture 3: The outer measure generated by a measure

September 17, 2011 Leave a comment

I’ll post lecture 1 and 2 soon.

In lecture 1 and 2, I defined a measure \mu on a semiring S and an outer measure on the power set of X. Our goal is that we want to define a measurable set E\subseteq X. As we recall, a measurable set E\subseteq X satifies the following condition: \mu(A) =\mu(A\cap E) + \mu(A\cap E^c) for any A\subseteq X. Notice that \mu here is an outer measure.

In this lecture, a measure \mu is extended to an outer measure \mu^*. As proved before, the collection \Lambda of measurable sets with respect to this outer measure is a \sigma-algebra. We will show that S\subseteq\Lambda. Thus, semirings are a basic collections of sets which a measure theory can be built on.

Let a measure space (X,S,\mu) be fixed. The outer measure \mu^* is defined as follows:

\mu^*(A) = \inf \{\sum_{i=1}^{\infty} \mu(A_i): A_i \in S \text{ and } A\subseteq \bigcup_{i=1}^\infty A_i \}.

Theorem: \mu^* is an outer measure.

Proof:

i) Let A_i=\emptyset for i\geq 1. Then \mu^*(\bigcup_{i=1}^\infty A_i)=\mu^*(\emptyset) \leq \sum_{i=1}^\infty \mu(\emptyset)=0. Thus \mu*(\emptyset)=0.

ii) Let A,B\in X and A\subseteq B. We consider a sequence \{A_i\} of sets in S such that B\subseteq \bigcup_{i=1}^\infty A_i. Then clearly, A\subseteq\bigcup_{i=1}^\infty A_i. Thus, \mu^*(A)\leq\sum_{i=1}^\infty \mu(A_i) Further, we have \mu^*(B)=\inf \{\sum_{i=1}^{\infty} \mu(A_i): A_i \in S \text{ and } B\subseteq \bigcup_{i=1}^\infty A_i \}. This follows that \mu^*(A)\leq \mu^*(B). Hence, \mu^* is monotone.

iii) Let \{A_i\} be a sequence of pairwise disjoint sets in X. If \sum_{i=1}^\infty \mu^*(A_i)=\infty. Then clearly, \mu^*(\bigcup_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu^*(A_i).

Thus we can assume \sum_{i=1}^\infty \mu^*(A_i)<\infty. Since each \mu^*(A_i) <\infty, there exists a sequence \{A_i\}^{n} of sets in S such that A_i\subseteq \bigcup_{n=1}^\infty A_i^n and \sum_{n=1}^\infty \mu(A_i^n)\le \mu^* (A_i)+2^{-i}\epsilon. Hence, \bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty \bigcup_{n=1}^\infty A_i^n. Furthermore,

\mu^* (\bigcup_{i=1}^\infty A_i)\leq \sum_{i=1}^\infty\sum_{n=1}^\infty\mu(A_i^n) 0, we have \mu^* (\bigcup_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu^*(A_i) . This proves the \sigma-subadditivity of \mu^*. \square

The next theorem will show that \mu^* is indeed an extension of \mu from S to \mathcal{P}(X).

Theorem: If E\in S, then \mu^*(E)=\mu(E).

Proof:

Let A_1=E and A_i=\emptyset for i>1. Then E\subseteq \bigcup_{i=1}^\infty A_i. Thus, \mu^*(E)\leq \mu(\bigcup_{i=1}^\infty A_i)=\mu(E).

Consider a sequence \{A_i\} of sets in S satisfying A\subseteq \bigcup_{i=1}^\infty A_i. Since \mu is a measure on semiring S. It is \sigma-subadditive on S. Therefore, we have \mu(A)\leq\sum_{i=1}^\infty \mu(A_i). Thus, \mu(A)\leq\mu^*(A). \square

The outer measure generated by a measure have some nice characteristics which can be shown by the following theorem.

Theorem: Let E\subseteq X. The followings are equivalent:

(1) E is \mu^*-measurable.

(2) \mu(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c) with any A\in S and \mu(A)<\infty.

(3) \mu(A)\geq \mu^*(A\cap E)+\mu^*(A\cap E^c) with any A\in S and \mu(A)<\infty.

(4) \mu^*(A)\geq \mu^*(A\cap E)+\mu^*(A\cap E^c) with any A\subseteq X.

Proof:

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