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.