construction of outer measures

The following theorem is used in measure theory to construct outer measures (http://planetmath.org/OuterMeasure2) on a set $X$, starting with a non-negative function on a collection of subsets of $X$. For example, if we take $X$ to be the real numbers, $𝒞$ to be the collection of bounded open intervals of $ℝ$ and define $p$ by $p((a,b))=b-a$ for real numbers , then the Lebesgue outer measure is obtained.

Theorem.

Let $X$ be a set, $\mathrm{C}$ be a family of subsets of $X$ containing the empty set and $p\mathrm{:}\mathrm{C}\mathrm{\to }\mathrm{R}\mathrm{\cup }\mathrm{\{}\mathrm{\infty }\mathrm{\}}$ be a function satisfying $p\mathit{}\mathrm{(}\mathrm{\varnothing }\mathrm{)}\mathrm{=}\mathrm{0}$.Then the function ${\mu }^{\mathrm{*}}\mathrm{:}\mathrm{P}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{\to }\mathrm{R}\mathrm{\cup }\mathrm{\{}\mathrm{\infty }\mathrm{\}}$ defined by

$${\mu }^{*}(A)=inf\{\sum _{i=1}^{\mathrm{\infty }}p(A_i):A_i\in 𝒞,A\subseteq \bigcup _{i=1}^{\mathrm{\infty }}{A}_i\}$$(1)

is an outer measure.

Proof.

The definition of ${\mu }^{*}$ immediately gives ${\mu }^{*}(A)\le {\mu }^{*}(B)$ for sets $A\subseteq B$, and if $A=\mathrm{\varnothing }$ then we can take $A_i=\mathrm{\varnothing }$ in (1) to obtain ${\mu }^{*}(\mathrm{\varnothing })\le {\sum }_{i}p(\mathrm{\varnothing })=0$, giving ${\mu }^{*}(\mathrm{\varnothing })=0$. Only the countable subadditivity of ${\mu }^{*}$ remains to be shown.That is, if $A_i$ is a sequence in $𝒫(X)$ then

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

To prove this inequality, we may restrict to the case where for each $i$ so that, choosing any $ϵ>0$, equation (1) says that there exists a sequence $A_{i,j}\in 𝒞$ such that $A_i\subseteq {\bigcup }_j{A}_{i,j}$ and,

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

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

$${\mu }^{*}\left(\bigcup _i{A}_i\right)\le \sum _{i,j}p(A_{i,j})=\sum _i\sum _{j}p(A_{i,j})\le \sum _i({\mu }^{*}(A_i)+2^{-i}ϵ)=\sum _i{\mu }^{*}(A_i)+ϵ.$$

As $ϵ>0$ is arbitrary, this proves subadditivity (2).∎

Although this result is rather general, placing few restrictions on the function $p$, there is no guarantee that the outer measure ${\mu }^{*}$ will agree with $p$ for the sets in $𝒞$ nor that $𝒞$ will consist of ${\mu }^{*}$-measurable (http://planetmath.org/CaratheodorysLemma) sets.

For example, if $X=ℝ$, $𝒞$ consists of the bounded open intervals, and $p((a,b))={(b-a)}^2$ for real numbers , then ${\mu }^{*}((a,b))=0\ne p((a,b))$.

Alternatively if $p((a,b))=\sqrt{b-a}$ for all then it follows that ${\mu }^{*}((a,b))=\sqrt{b-a}$ so

$${\mu }^{*}((0,1))+{\mu }^{*}([1,2))=1+1\ne {\mu }^{*}((0,2))=\sqrt{2},$$

and $(0,1)$ is not ${\mu }^{*}$-measurable.