Outer Measures

by AKG
Tags: measures, outer
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Oct3-06, 02:18 AM
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If X is a set, an algebra A on X is a non-empty collection of subsets of X which is closed under complements with respect to X, and finite unions.

Given an algebra A, a premeasure on A is a function [itex]p\, :\, A \to [0,\, \infty][/itex] such that:
a) [itex]p(\emptyset ) = 0[/itex]
b) If B is a countable collection of disjoint elements of A whose union is in A, then [itex]p(\bigcup _{S \in B}S) = \sum _{S \in B}p(S)[/itex] (additivity)

A premeasure p on an algebra A of a set X is [itex]\mathbf{\sigma -finite}[/itex] iff there is a countable collection of elements of A, each with finite p-premeasure, whose union is X.

Given a set X, an outer measure w is a function from the power set of X to the extended non-negative reals which satisfies:
a) [itex]w(\emptyset ) = 0[/itex]
b) [itex]A \subset B \rightarrow w(A) \leq w(B)[/itex] (monotonicity)
c) [itex]w(\bigcup _{i=1}^{\infty} A_i) \leq \sum_{i=1}^{\infty}w(A_i)[/itex] (subadditivity)

A [itex]\mathbf{\sigma -algebra}[/itex] is an algebra which is closed under countable union.

A measure m on a [itex]\sigma -algebra[/itex] M is a function [itex]m\, :\, M \to [0\, ,\, \infty ][/itex] satisfying the same properties as a premeasure. Note, however, that since M is closed under countable union, the condition in property b) that the union be contained in the domain is redundant.

If f is a premeasure or a measure, and has domain D, then the induced outer measure, wf is defined by:

[tex]w_f(E) = \inf \left \{ \sum_{i=1} ^{\infty}f(A_i)\ :\ A_i \in D,\ E \subseteq \bigcup _{i=1} ^{\infty}A_i \right \}[/tex]

A measure is finite iff it doesn't map anything to infinity.

If E is a subset of a set X, and w is an outer measure on the power set of X, then E is w-measurable iff for every subset F of X:

[tex]w(F) = w(F-E) + w(F\cap E)[/tex]


1. Let M be a [itex]\sigma -algebra[/itex] on a set X, and let m be a measure with domain M. Let w be the induced outer measure by m. Suppose E is a subset of X satisfying w(E) = w(X) (E need not be in M). Prove that if A and B are in M, and [itex]A \cap E = B \cap E[/itex], then m(A) = m(B)

2. A is an algebra on X, [itex]A_{\sigma}[/itex] is the collection of all countable unions of sets in A, and [itex]A_{\sigma \delta}[/itex] is the collection of countable intersections of sets in [itex]A_{\sigma }[/itex]. Let p be a premeasure on A and w the induced outer measure. Prove that if w(E) is finite, then E is w-measurable iff there is a B in [itex]A_{\sigma \delta }[/itex] containing E such that w(B-E) = 0. Moreover, prove that if p is [itex]\sigma -finite[/itex], the restriction that w(E) be finite is superfluous.

For question 1, I've tried just playing around with taking outer measures of sets, applying the definition of measurability, using monotonicity and subadditivity, etc. but haven't gotten anywhere.

For question 2, I've actually proved most of it. It's just the last part, proving that if p is [itex]\sigma -finite[/itex] that the restriction that w(E) be finite is superfluous, which I can't get. The only time in proving the previous part where I used the restriction is when proving that if E is w-measurable, then there exists a B such that ... What happened was I find an appropriate B in [itex]A_{\sigma \delta }[/itex] containing E such that w(B) = w(E). After easily proving that B is w-measurable, I got:

[itex]w(B) = w(B-E) + w(B\cap E)[/itex]
w(E) = w(B-E) + w(E)

Finiteness of w(E) allowed me to cancel it from both sides, concluding w(B-E) = 0 as desired. But it seems that if I replace the assumption that w(E) is finite for the assumption that p is [itex]\sigma -finite[/itex], then I have to take a whole new approach to the proof. Any hints?

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