Home > Uncategorized > Lecture 3: The outer measure generated by a measure

## Lecture 3: The outer measure generated by a measure

I’ll post lecture 1 and 2 soon.

In lecture 1 and 2, I defined a measure $\mu$ on a semiring $S$ and an outer measure on the power set of $X$. Our goal is that we want to define a measurable set $E\subseteq X.$ As we recall, a measurable set $E\subseteq X$ satifies the following condition: $\mu(A) =\mu(A\cap E) + \mu(A\cap E^c)$ for any $A\subseteq X.$ Notice that $\mu$ here is an outer measure.

In this lecture, a measure $\mu$ is extended to an outer measure $\mu^*.$ As proved before, the collection $\Lambda$ of measurable sets with respect to this outer measure is a $\sigma-algebra.$ We will show that $S\subseteq\Lambda.$ Thus, semirings are a basic collections of sets which a measure theory can be built on.

Let a measure space $(X,S,\mu)$ be fixed. The outer measure $\mu^*$ is defined as follows:

$\mu^*(A) = \inf \{\sum_{i=1}^{\infty} \mu(A_i): A_i \in S \text{ and } A\subseteq \bigcup_{i=1}^\infty A_i \}.$

Theorem: $\mu^*$ is an outer measure.

Proof:

i) Let $A_i=\emptyset$ for $i\geq 1.$ Then $\mu^*(\bigcup_{i=1}^\infty A_i)=\mu^*(\emptyset) \leq \sum_{i=1}^\infty \mu(\emptyset)=0.$ Thus $\mu*(\emptyset)=0.$

ii) Let $A,B\in X$ and $A\subseteq B.$ We consider a sequence $\{A_i\}$ of sets in $S$ such that $B\subseteq \bigcup_{i=1}^\infty A_i.$ Then clearly, $A\subseteq\bigcup_{i=1}^\infty A_i.$ Thus, $\mu^*(A)\leq\sum_{i=1}^\infty \mu(A_i)$ Further, we have $\mu^*(B)=\inf \{\sum_{i=1}^{\infty} \mu(A_i): A_i \in S \text{ and } B\subseteq \bigcup_{i=1}^\infty A_i \}.$ This follows that $\mu^*(A)\leq \mu^*(B).$ Hence, $\mu^*$ is monotone.

iii) Let $\{A_i\}$ be a sequence of pairwise disjoint sets in $X.$ If $\sum_{i=1}^\infty \mu^*(A_i)=\infty$. Then clearly, $\mu^*(\bigcup_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu^*(A_i).$

Thus we can assume $\sum_{i=1}^\infty \mu^*(A_i)<\infty.$ Since each $\mu^*(A_i) <\infty,$ there exists a sequence $\{A_i\}^{n}$ of sets in $S$ such that $A_i\subseteq \bigcup_{n=1}^\infty A_i^n$ and $\sum_{n=1}^\infty \mu(A_i^n)\le \mu^* (A_i)+2^{-i}\epsilon.$ Hence, $\bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty \bigcup_{n=1}^\infty A_i^n.$ Furthermore,

$\mu^* (\bigcup_{i=1}^\infty A_i)\leq \sum_{i=1}^\infty\sum_{n=1}^\infty\mu(A_i^n) 0,$ we have $\mu^* (\bigcup_{i=1}^\infty A_i) \leq \sum_{i=1}^\infty \mu^*(A_i) .$ This proves the $\sigma$-subadditivity of $\mu^*. \square$

The next theorem will show that $\mu^*$ is indeed an extension of $\mu$ from $S$ to $\mathcal{P}(X).$

Theorem: If $E\in S,$ then $\mu^*(E)=\mu(E).$

Proof:

Let $A_1=E$ and $A_i=\emptyset$ for $i>1.$ Then $E\subseteq \bigcup_{i=1}^\infty A_i.$ Thus, $\mu^*(E)\leq \mu(\bigcup_{i=1}^\infty A_i)=\mu(E).$

Consider a sequence $\{A_i\}$ of sets in $S$ satisfying $A\subseteq \bigcup_{i=1}^\infty A_i.$ Since $\mu$ is a measure on semiring $S.$ It is $\sigma$-subadditive on $S.$ Therefore, we have $\mu(A)\leq\sum_{i=1}^\infty \mu(A_i).$ Thus, $\mu(A)\leq\mu^*(A). \square$

The outer measure generated by a measure have some nice characteristics which can be shown by the following theorem.

Theorem: Let $E\subseteq X.$ The followings are equivalent:

(1) $E$ is $\mu^*$-measurable.

(2) $\mu(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)$ with any $A\in S$ and $\mu(A)<\infty.$

(3) $\mu(A)\geq \mu^*(A\cap E)+\mu^*(A\cap E^c)$ with any $A\in S$ and $\mu(A)<\infty.$

(4) $\mu^*(A)\geq \mu^*(A\cap E)+\mu^*(A\cap E^c)$ with any $A\subseteq X.$

Proof: