Definition. An outer measure on a nonempty set X is a function \mu^*: \mathcal P(X)\to[0, \infty] that satisfies

1) \mu^*(\emptyset)=0,

2) \mu^*(A)\le\mu^*(B) if $A\subset B$,

3) \mu^*(\cup_1^\infty A_j)\le \sum_1^\infty\mu^*(A_j).

 

The most common way to obtain outer measure is to start with a family \mathcal E of “elementary sets” on which a notion of measure is defined (such as rectangles in the plane) and then to approximate arbitrary sets “from the outside” by countable unions of members of \mathcal E. The precise construction is as follows.

 

Proposition. Let \mathcal E\subset\mathcal P(X) and \rho: \mathcal E\to[0, \infty] be such that \emptyset\in\mathcal E, X\in\mathcal E and \rho(\emptyset)=0. For any A\subset X, define

\mu^*(A)=\inf\{\sum_1^\infty\rho(E_j): E_j\in\mathcal E and A\subset \cup_1^\infty E_j\}.

Then \mu^* is an outer measure.

Proof. For any  A\subset X, \exists \{E_j\}_1^\infty\subset\mathcal E such that A\subset\cup_1^\infty E_j (take E_j=X for all j) so the definition of \mu^* makes sense. Obviously \mu^*(\emptyset)=0 (take E_j=\emptyset for all j), and \mu^*(A)\le\mu^*(B) for A\subset B because the set over which the infimum is taken in the definition of \mu^*(A) includes the corresponding set in the definition of \mu^*(B). To prove countable subadditivity, suppose \{A_j\}_1^\infty\subset\mathcal P(X) and \epsilon>0. For each j there exists \{E_j^k\}_{k=1}^\infty\subset\mathcal E such that A_j\subset\cup_{k=1}^\infty E_j^k and \sum_{k=1}^\infty\rho(E_j^k)\le\mu^*(A_j)+\epsilon 2^{-j}. But then if A=\cup_1^\infty A_j, we have A\subset\cup_{j, k=1}^\infty E_j^k and \sum_{j, k}\rho(E_j^k)\le\sum_j\mu^*(A_j)+\epsilon, whence \mu^*(A)+\epsilon. Since \epsilon is arbitrary, we are done. \Box

 

Definition. If \mu^* is an outer measure on X, a set A\subset X is called \mu^*-measurable if

\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\cap A^c) for all E\subset X.

The inequality \mu^*(E)\le\mu^*(E\cap A)+\mu^*(E\cap A^c) holds for any A and E, so to prove that A is \mu^*-measurable, it suffices to prove that the reverse inequality. The latter is trivial if \mu^*(E)=\infty, so we see that A is \mu^*-measurable iff

\mu^*(E)\ge\mu^*(E\cap A)+\mu^*(E\cap A^c) for all E\subset X such that \mu^*(E)<\infty.

If E is a “well-behaved” set such that E\supset A, the equation \mu^*(E)=\mu^*(E\cap A)+\mu^*(E\cap A^c) says that the outer measure of A, \mu^*(A) is equal to the “inner measure” of A, \mu^*(E)-\mu^*(E\cap A^c).

Caratheodory’s Theorem. If \mu^* is an outer measure on X, the collection \mathcal M of \mu^*-measurable sets is a \sigma-algebra, and the restriction of \mu^* to \mathcal M is a complete measure.

Proof. First, we observe that \mathcal M is closed under complements since the definition of \mu^*-measurability of A is symmetric in A and A^c. Next, if A, B\in\mathcal M and E\subset X,

\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\cap A^c)=\mu^*(E\cap A\cap B)+\mu^*(E\cap A\cap B^c)+\mu^*(E\cap A^c\cap B)+\mu^*(E\cap A^c\cap B^c).

But (A\cup B)=(A\cap B)\cup(A\cap B^c)\cup(A^c\cap B), so by subadditivity,

\mu^*(E\cap A\cap B)+\mu^*(E\cap A\cap B^c)+\mu^*(E\cap A^c\cap B)\ge\mu^*(E\cap(A\cup B)),

and hence

\mu^*(E)\ge\mu^*(E\cap(A\cup B))+\mu^*(E\cap(A\cup B)^c).

It follows that A\cup B=\mathcal M, so \mathcal M is an algebra. Moreover, if A, B\in\mathcal M and A\cap B=\emptyset,

\mu^*(A\cup B)=\mu^*((A\cup B)\cap A)+\mu^*((A\cup B)\cap A^c)=\mu^*(A)+\mu^*(B),

so \mu^* is finitely additive on \mathcal M.

To show that \mathcal M is a \sigma-algebra, it will suffice to show that \mathcal M is closed under countable disjoint unions. If \{A_j\}_1^\infty is a sequence of disjoint sets in \mathcal M, let B_n=\cup_1^n A_j and B=\cup_1^\infty A_j. Then for any E\subset X,

\mu^*(E\cap B_n)=\mu^*(E\cap B_n\cap A_n)+\mu^*(E\cap B_n\cap A_n^c)=\mu^*(E\cap A_n)+\mu^*(E\cap B_{n-1}),

so a simple induction shows that \mu^*(E\cap B_n)=\sum_1^n\mu^*(E\cap A_j). Therefore,

\mu^*(E)=\mu^*(E\cap B_n)+\mu^*(E\cap B_n^c)\ge\sum_1^n\mu^*(E\cap A_j)+\mu(E\cap B^c),

and letting n\to\infty we obtain

\mu^*(E)\ge\sum_1^\infty (E\cap A_j)+\mu^*(E\cap B^c)\ge\mu^*(\cup_1^\infty(E\cap A_j))+\mu^*(E\cap B^c)=\mu^*(E\cap B)+\mu^*(E\cap B^c)\ge \mu^*(E).

All the inequalities in the last calculation are thus equalities. It follows that B\in\mathcal M and taking E=B that \mu^*(B)=\sum_1^\infty\mu^*(A_j), so \mu^* is countably additive on \mathcal M. Finally, if \mu^*(A)=0, for any E\subset X we have

\mu^*(E)\le\mu^*(E\cap A)+\mu^*(E\cap A^c)=\mu^*(E\cap A^c)\le\mu(E),

so that A\in\mathcal M. Therefore \mu^*|\mathcal M is a complete measure. \Box

Let go of all thoughts of limitations about how you should or should not behave based on how old or young you are. Age is nothing but a number, “an issue of mind over matter. If you don’t mind, it doesn’t matter.” ~ Mark Twain

References:

[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 28-30.

[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.