proof of Carathéodory’s extension theorem

The first step is to extend the set function ${\mu }_0$ to the power set $P(X)$. For any subset $S\subseteq X$ the value of ${\mu }^{*}(S)$ is defined by taking sequences $S_i$ in $A$ which cover $S$,

$${\mu }^{*}(S)\equiv inf\{\sum _{i=1}^{\mathrm{\infty }}{\mu }_0(S_i):S_i\in A,S\subseteq \bigcup _{i=1}^{\mathrm{\infty }}{S}_i\}.$$(1)

We show that this is an outer measure (http://planetmath.org/OuterMeasure2). First, it is clearly non-negative. Secondly, if $S=\mathrm{\varnothing }$ then we can take $S_i=\mathrm{\varnothing }$ in (1) to obtain ${\mu }^{*}(S)\le {\sum }_i{\mu }_0(\mathrm{\varnothing })=0$, giving ${\mu }^{*}(\mathrm{\varnothing })$=0. It is also clear that ${\mu }^{*}$ is increasing, so that if $S\subseteq T$ then ${\mu }^{*}(S)\le {\mu }^{*}(T)$. The only remaining property to be proven is subadditivity. That is, if $S_i$ is a sequence in $P(X)$ then

$${\mu }^{*}\left(\bigcup _i{S}_i\right)\le \sum _i{\mu }^{*}(S_i).$$(2)

To prove this inequality, choose any $ϵ>0$ and, by the definition (1) of ${\mu }^{*}$, for each $i$ there exists a sequence $S_{i,j}\in A$ such that $S_i\subseteq {\bigcup }_j{S}_{i,j}$ and,

$$\sum _{j=1}^{\mathrm{\infty }}{\mu }_0(S_{i,j})\le {\mu }^{*}(S_i)+2^{-i}ϵ.$$

As ${\bigcup }_i{S}_i\subseteq {\bigcup }_{i,j}{S}_{i,j}$, equation (1) defining ${\mu }^{*}$ gives

$${\mu }^{*}\left(\bigcup _i{S}_i\right)\le \sum _{i,j}{\mu }_0(S_{i,j})=\sum _i\sum _j{\mu }_0(S_{i,j})\le \sum _i({\mu }^{*}(S_i)+2^{-i}ϵ)=\sum _i{\mu }^{*}(S_i)+ϵ.$$

As $ϵ>0$ is arbitrary, this proves subadditivity (2). So, ${\mu }^{*}$ is indeed an outer measure.

The next step is to show that ${\mu }^{*}$ agrees with ${\mu }_0$ on $A$. So, choose any $S\in A$. The inequality ${\mu }^{*}(S)\le {\mu }_0(S)$ follows from taking $S_1=S$ and $S_i=\mathrm{\varnothing }$ in (1), and it remains to prove the reverse inequality. So, let $S_i$ be a sequence in $A$ covering $S$, and set

$$S_i^{\prime }=(S\cap S_i)\setminus \bigcup _{j=1}^{i-1}{S}_j\in A.$$

Then, $S_i^{\prime }$ are disjoint sets satisfying ${\bigcup }_{j=1}^i{S}_j^{\prime }=S\cap {\bigcup }_{j=1}^i{S}_j$ and, therefore, ${\bigcup }_i{S}_i^{\prime }=S$. By the countable additivity of ${\mu }_0$,

$$\sum _i{\mu }_0(S_i)=\sum _i({\mu }_0(S_i^{\prime })+{\mu }_0(S_i\setminus S_i^{\prime }))\ge \sum _i{\mu }_0(S_i^{\prime })={\mu }_0(S).$$

As this inequality hold for any sequence $S_i\in A$ covering $S$, equation (1) gives ${\mu }^{*}(S)\ge {\mu }_0(S)$ and, by combining with the reverse inequality, shows that ${\mu }^{*}$ does indeed agree with ${\mu }_0$ on $A$.

We have shown that ${\mu }_0$ extends to an outer measure ${\mu }^{*}$ on the power set of $X$. The final step is to apply Carathéodory’s lemma on the restriction of outer measures. A set $S\subseteq X$ is said to be ${\mu }^{*}$-measurable if the inequality

$${\mu }^{*}(E)\ge {\mu }^{*}(E\cap S)+{\mu }^{*}(E\cap S^c)$$(3)

is satisfied for all subsets $E$ of $X$. Carathéodory’s lemma then states that the collection $ℱ$ of ${\mu }^{*}$-measurable sets is a $\sigma$-algebra (http://planetmath.org/SigmaAlgebra) and that the restriction of ${\mu }^{*}$ to $ℱ$ is a measure.To complete the proof of the theorem it only remains to be shown that every set in $A$ is ${\mu }^{*}$-measurable, as it will then follow that $ℱ$ contains $𝒜=\sigma (A)$ and the restriction of ${\mu }^{*}$ to $𝒜$ is a measure.

So, choosing any $S\in A$ and $E\subseteq X$, the proof will be complete once it is shown that (3) is satisfied.Given any $ϵ>0$, equation (1) says that there is a sequence $E_i$ in $A$ such that $E\subseteq {\bigcup }_i{E}_i$ and

$$\sum _i{\mu }_0(E_i)\le {\mu }^{*}(E)+ϵ.$$

As $E\cap S\subseteq {\bigcup }_i(E_i\cap S)$ and $E\cap S^c\subseteq {\bigcup }_i(E_i\cap S^c)$,

$${\mu }^{*}(E\cap S)+{\mu }^{*}(E\cap S^c)\le \sum _i{\mu }_0(E_i\cap S)+\sum _i{\mu }_0(E_i\cap S^c)=\sum _i{\mu }_0(E_i)\le {\mu }^{*}(E)+ϵ.$$

Since $ϵ$ is arbitrary, this shows that (3) is satisfied and $S$ is ${\mu }^{*}$-measurable.