4 Outer Measure and Premeasure

outer measure

Let $X$ be a set. An outer measure is a function ${\mu }^{\ast }:\mathcal{P}(X)\to$ $[0,\infty ]$ satisfying

• ${\mu }^{\ast }(\varnothing )=0$;
• if $A\subset B$, then ${\mu }^{\ast }(A)⩽{\mu }^{\ast }(B)$;
• ${\mu }^{\ast }\left({\cup }_{i=1}^{\infty }{A}_i\right)⩽{\sum }_{i=1}^{\infty }{\mu }^{\ast }\left(A_i\right)$ whenever $A_1,A_2,\dots$ are subsets of $X$.

Proposition

Suppose $\mathcal{C}$ is a collection of subsets of $X$ such that $\varnothing \in \mathcal{C}$ and there exist $D_1,D_2,\dots$ in $\mathcal{C}$ such that $X={\cup }_{i=1}^{\infty }{D}_i$. Suppose $l:\mathcal{C}\to [0,\infty ]$ with $l(\varnothing )=0$. Define

$${\mu }^{\ast }(E)=\inf \left\{\sum _{i=1}^{\infty }l\left(A_i\right):A_i\in \mathcal{C}\text{ for each }i\text{ and }E\subset \bigcup _{i=1}^{\infty }{A}_i\right\}$$

Then ${\mu }^{\ast }$ is an outer measure.

\begin{proof} Note that (i), (ii) is easy. So we only need to check (iii). Let $\{A_i\}\subseteq \mathcal{P}(X)$. If ${\mu }^{\ast }(A_i)=\infty$, it is OK. Assume that ${\mu }^{\ast }(A_i)<\infty$ for all $i\in \mathbb{N}$. For any each $i$, any $ϵ>0$, there exists $\{C_{ij}{\}}_{j=1}^{\infty }\subseteq \mathcal{C}$ such that ${\sum }_{i=1}^{\infty }l(c_{ij})⩽{\mu }^{\ast }(A_i)+\frac{ϵ}{2^i}$. Then ${\cup }_{i=1}^{\infty }{A}_i\subseteq {\cup }_{i,j}{C}_{ij}$ and ${\mu }^{\ast }({\cup }_{i=1}^{\infty }{A}_i)⩽{\sum }_{i,j}l(C_{ij})⩽{\sum }_{i=1}^{\infty }{\mu }^{\ast }(A_i)+ϵ$. Take $ϵ\to 0$, then we finish the proof. \end{proof}

Definition

Let ${\mu }^{\ast }$ be an outer measure. A set $A\subset X$ is ${\mu }^{\ast }$-measurable if

$${\mu }^{\ast }(E)={\mu }^{\ast }(E\cap A)+{\mu }^{\ast }\left(E\cap A^c\right)$$

for all $E\subset X$.

Note. Since ${\mu }^{\ast }(E)⩽{\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\cap A^c)$ by definition, to verify $A$ is ${\mu }^{\ast }$-measurable, it suffices to show ${\mu }^{\ast }(E)⩾{\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\cap A^c)$ with ${\mu }^{\ast }(E)$ finite.

Theorem

If ${\mu }^{\ast }$ is an outer measure on $X$, then:

• The collection $\mathcal{A}$ of ${\mu }^{\ast }$-measurable sets is a $\sigma$-algebra.
• If $\mu$ is the restriction of ${\mu }^{\ast }$ to $\mathcal{A}$, then $\mu$ is a complete measure.

\begin{proof} i) Note that $X,\varnothing \in \mathcal{A}$, and $A\in \mathcal{A}$ iff $A^c\in \mathcal{A}$. Let $A,B\in \mathcal{A}$. Aim to show $A\cap B\in \mathcal{A}$. That is, ${\mu }^{\ast }(E)⩾{\mu }^{\ast }(E\cap A\cap B)+{\mu }^{\ast }(E\cap (A\cap B{)}^c)$. Since

$$\begin{array}{rl}{\mu }^{\ast }(E)& ⩾{\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\cap A^c)\\ & ⩾{\mu }^{\ast }(E\cap A\cap B)+{\mu }^{\ast }(E\cap A\cap B^c)+{\mu }^{\ast }(E\cap A^c)\\ & ⩾{\mu }^{\ast }(E\cap A\cap B)+{\mu }^{\ast }(E\cap (A^c\cup B^c))\\ & ⩾{\mu }^{\ast }(E\cap A\cap B)+{\mu }^{\ast }(E\cap (A\cap B{)}^c),\end{array}$$

then $A\cap B\in \mathcal{A}$. It remains to show that if $\{A_i\}\subseteq \mathcal{A}$ with $A_i\cap A_j=\varnothing$, then $\cup A_i\in \mathcal{A}$. Let $B_n={\cup }_{i=1}^n{A}_i$. Then ${\cup }_{n=1}^{\infty }{B}_n={\cup }_{i=1}^{\infty }{A}_i=A$. We want to show ${\mu }^{\ast }(E)⩾{\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\cap A^c)$. Since

$$\begin{array}{rl}{\mu }^{\ast }(E\cap B_n)& ⩾{\mu }^{\ast }(E\cap B_n\cap A_n)+{\mu }^{\ast }(E\cap B_n\cap A_n^c)\\ & ={\mu }^{\ast }(E\cap A_n)+{\mu }^{\ast }(E\cap B_{n-1})\\ & =\sum _{i=1}^n{\mu }^{\ast }(E\cap A_i).\end{array}$$

Then there is

$${\mu }^{\ast }(E)⩾{\mu }^{\ast }(E\cap B_n)+{\mu }^{\ast }(E\cap B_n^c)⩾\sum _{i=1}^n{\mu }^{\ast }(E\cap A_i)+{\mu }^{\ast }(E\cap A^c).$$

Let $n\to \infty$. Then

$${\mu }^{\ast }(E)⩾\sum _{i=1}^{\infty }{\mu }^{\ast }(E\cap A_i)+{\mu }^{\ast }(E\cap A^c)⩾{\mu }^{\ast }(E\cap A)+{\mu }^{\ast }(E\cap A^c)⩾{\mu }^{\ast }(E).$$

So ${\cup }_{i=1}^{\infty }{A}_i\in \mathcal{A}$. In addition, since ${\sum }_{i=1}^{\infty }{\mu }^{\ast }(E\cap A_i)={\mu }^{\ast }(E\cap A)$ for all $E\subseteq X$. Then if $E=A=\cup A_i$, we have that ${\sum }_{i=1}^{\infty }{\mu }^{\ast }(A_i)={\mu }^{\ast }(A)$. Thus, ${\mu }^{\ast }$ is a measure on $\mathcal{A}$.

Note that if $A\subseteq X$ with ${\mu }^{\ast }(A)=0$, then for any $E\subseteq X$ there is ${\mu }^{\ast }(A\cap E)=0$ and so ${\mu }^{\ast }(E)⩾{\mu }^{\ast }(A^c\cap E)+{\mu }^{\ast }(A\cap E)$ and so $A\in \mathcal{A}$. Thus $\mathcal{A}$ is a complete measure. \end{proof}

Example. Let $X=\mathbb{R}$ and let $\mathcal{C}$ be the collection of intervals of the form $(a,b]$. Let $l(I)=b-a$ if $I=(a,b]$. Let $\mathcal{L}$ be the set of ${\mu }^{\ast }$-measurable set, then it is a $\sigma$-algebra, called the Lebesgue $\sigma$-algebra. However, there are subsets in $\mathbb{R}$ which are non-measurable with respect to ${\mu }^{\ast }$ and so ${\mu }^{\ast }$ is not a measure on $\mathcal{P}(\mathbb{R})$. See here.

premeasure

Let ${\mathcal{A}}_0$ be an algebra. Saying $\mu :{\mathcal{A}}_0\to [0,\infty ]$ is a measure on ${\mathcal{A}}_0$ means the following: $\mu (\varnothing )=0$ and if $A_1,A_2,\dots$ are pairwise disjoint elements of ${\mathcal{A}}_0$ and also ${\cup }_{i=1}^{\infty }{A}_i\in {\mathcal{A}}_0$, then $\mu \left({\cup }_{i=1}^{\infty }{A}_i\right)={\sum }_{i=1}^{\infty }\mu \left(A_i\right)$. Sometimes one calls a measure on an algebra a premeasure .

Caratheodory-Hahn

Suppose ${\mathcal{A}}_0$ is an algebra on $X$ and $l$ : ${\mathcal{A}}_0\to [0,\infty ]$ is a measure. For $E\in \mathcal{P}(X)$, define

$${\mu }^{\ast }(E)=\inf \left\{\sum _{i=1}^{\infty }l\left(A_i\right):A_i\in {\mathcal{A}}_0\text{ for each }i\text{ and }E\subset \bigcup _{i=1}^{\infty }{A}_i\right\}$$

Then

• ${\mu }^{\ast }$ is an outer measure;
• ${\mu }^{\ast }(A)=l(A)$ if $A\in {\mathcal{A}}_0$;
• every set in ${\mathcal{A}}_0$ is ${\mu }^{\ast }$-measurable;
• if $l$ is $\sigma$-finite, then ${\mu }^{\ast }$ is a measure on $\sigma \left({\mathcal{A}}_0\right)$ which is the unique extension of l to $\sigma \left({\mathcal{A}}_0\right)$.

\begin{proof} i)-iii) are easy. See here.

iv) We firstly consider the case of finite measure and then consider the case of $\sigma$-finite.

Case 1. Assume that $l$ is finite. Let $\nu$ is a measure on $\sigma ({\mathcal{A}}_0)$. For any $A\in \sigma ({\mathcal{A}}_0)$, there is

$$l(X)=\nu (X)=\nu (A)+\nu (A^c)⩽{\mu }^{\ast }(A)+{\mu }^{\ast }(A^c)={\mu }^{\ast }(X)=l(X)$$

and so $\nu (A)={\mu }^{\ast }(A)$.

Case 2. Assume that $l$ is $\sigma$-finite. Let $\{C_i\}\subseteq {\mathcal{A}}_0$ such that $C_i\uparrow X$ and $l(C_i)<\infty$. Consider $l_i:{\mathcal{A}}_0\to [0,\infty ),l_i(A)=l(A\cap C_i)$. Let $\nu ,\mu$ be measures on $\sigma ({\mathcal{A}}_0)$ with $\nu =\mu$ on ${\mathcal{A}}_0$. Define ${\mu }_i,{\nu }_i$ as ${\nu }_i(A)=\nu (A\cap C_i)$ and ${\mu }_i(A)=\mu (A\cap C_i)$. By Case 1 we have ${\nu }_i={\mu }_i$ for all $A\in \sigma ({\mathcal{A}}_0)$. Thus $\nu (A)=\lim \nu (A\cap C_i)=\lim \mu (A\cap C_i)=\mu (A)$.
\end{proof}

Remark. Note that $\mathcal{A}$ defined in ^036d19 is a $\sigma$-algebra and $\mathcal{A}\supseteq {\mathcal{A}}_0$. Thus $\sigma ({\mathcal{A}}_0)\subseteq \mathcal{A}$.