Measures and $\sigma$-algebras

An outer measure such as $\mu^*$ is a rather crude object since, even if the $A_i$ are disjoint, there is generally strict inequality in (1.14). It turns out to be unreasonable to expect equality in (1.14), for disjoint unions, for a function defined on all subsets of $X$. We therefore restrict attention to smaller collections of subsets.

For a general outer measure $\mu^*$ we define the notion of $\mu^*$-measurability of a set.

Proof. Suppose $E$ is $\mu^*$-measurable, then $E^C$ is $\mu^*$-measurable by the symmetry of (2.1).

Suppose $A$, $E$ and $F$ are any three sets. Then

$$A \cap (E \cup F) = (A \cap E \cap F) \cup (A \cap E \cap F^C) \cup (A \cap E^C \cap F)$$

$$A \cap (E \cup F)^C = A \cap E^C \cap F^C .$$

From the subadditivity of $\mu^*$

$$\begin{multline*} \mu^* (A \cap (E \cup F)) + \mu^* (A \cap (E \cup F)^C)\\ \le... ...) \\ + \mu^* (A \cap E^C \cap F) + \mu^* (A \cap E^C \cap F^C). \end{multline*}$$

Now, if $E$ and $F$ are $\mu^*$-measurable then applying the definition twice, for any $A$,

$$\begin{multline*} \mu^* (A) = \mu^* (A \cap E \cap F) + \mu^* (A \cap E \cap F... ...q \mu^* (A \cap (E \cup F)) + \mu^* (A \cap (E \cup F)^C) . \end{multline*}$$

The reverse inequality follows from the subadditivity of $\mu^*$, so $E \cup F$ is also $\mu^*$-measurable.

If $\left\{ E_i \right\}^{\infty}_{i=1}$ is a sequence of disjoint $\mu^*$-measurable sets, set $F_n= \bigcup^n_{i=1} E_i$ and $F = \bigcup^{\infty}_{i=1} E_i$. Then for any $A$,

$$\mu^* (A \cap F_n) = \mu^* (A \cap F_n \cap E_n) + \mu^* (A \cap F_n \cap E^C_n)$$

$$= \mu^* (A \cap E_n) + \mu^* (A \cap F_{n-1}) .$$

Iterating this shows that

$$\mu^* (A \cap F_n) = \sum^n_{i=1} \mu^* (A \cap E_j) .$$

From the $\mu^*$-measurability of $F_n$ and the subadditivity of $\mu^*$,

$$\mu^* (A) = \mu^* (A \cap F_n) + \mu^* (A \cap F^C_n)$$

$$\geq \sum^n_{i=1} \mu^* (A \cap E_j) + \mu^* (A \cap F^C) .$$

Taking the limit as $n \to \infty$ and using subadditivity,

$$\mu^* (A) \geq \sum^n_{i=1} \mu^* (A \cap E_j) + \mu^* (A \cap F^C)$$ (13)

$$\geq \mu^* (A \cap F) + \mu^*(A \cap F^C) \ge \mu^* (A)$$

proves that inequalities are equalities, so $F$ is also $\mu^*$-measurable.

In general, for any countable union of $\mu^*$-measurable sets,

$$\bigcup^{\infty}_{j=1} A_j = \bigcup^{\infty}_{j=1} \widetilde{A}_j ,$$

$$\widetilde{A}_j = A_j \backslash \bigcup^{j-1}_{i=1} A_i = A_j \cap \left( \bigcup^{j-1}_{i=1} A_i \right)^C$$

is $\mu^*$-measurable since the $\widetilde{A}_j$ are disjoint. $\qedsymbol$

A measure (sometimes called a positive measure) is an extended function defined on the elements of a $\sigma$-algebra $\mathcal{M}$:

$$\mu : \mathcal{M} \to [0, \infty ]$$

such that

$$\mu(\emptyset ) = 0$$ (14)

$$\begin{split}\mu \left( \bigcup^{\infty}_{i=1} A_i \right) = ... ...cal{M} \hbox{ and } A_i \cap A_j = \phi &i \neq j.\end{split}$$ (15)

The elements of $\mathcal{M}$ with measure zero, i.e., $E \in \mathcal{M}$, $\mu (E) =0$, are supposed to be `ignorable'. The measure $\mu$ is said to be complete if

$$E \subset X \hbox{ and } \exists F \in \mathcal{M} , \mu (F) =0 , E \subset F \Rightarrow E \in \mathcal{M} .$$ (16)

See Problem 4.

The first part of the following important result due to Caratheodory was shown above.

Proof. We have already shown that the collection of $\mu^*$-measurable subsets of $X$ is a $\sigma$-algebra. To see the second part, observe that taking $A=F$ in (2.2) gives

$$\mu^* (F) = \sum_j \mu^* (E_j) \hbox{ if } F = \bigcup^{\infty}_{i=1} E_j$$

and the $E_j$ are disjoint elements of $\mathcal{M}$. This is (2.4).

Similarly if $\mu^*(E) =0$ and $F \subset E$ then $\mu^* (F)=0$. Thus it is enough to show that for any subset $E\subset X,$ $\mu^*(E) =0$ implies $E \in \mathcal{M}$. For any $A \subset X,$ using the fact that $\mu^*(A\cap E)=0,$ and the `increasing' property of $\mu^*$

$$\mu^* (A) \leq \mu^* (A \cap E) + \mu^* (A \cap E^C)$$

$$=\mu^* (A \cap E^C) \leq \mu^* (A)$$

shows that these must always be equalities, so $E \in \mathcal{M}$ (i.e., is $\mu^*$-measurable). $\qedsymbol$

Going back to our primary concern, recall that we constructed the outer measure $\mu^*$ from $0 \leq u \in (\mathcal{C}_0 (X))'$ using (1.11) and (1.12). For the measure whose existence follows from Caratheodory's theorem to be much use we need

Proof. Let $U\subset X$ be open. We only need to prove (2.1) for all $A \subset X$ with $\mu^* (A) < \infty$.²

Suppose first that $A \subset X$ is open and $\mu^* (A) < \infty$. Then $A \cap U$ is open, so given $\epsilon >0$ there exists $f \in C(X)$ $\operatorname{supp}(f) \Subset A \cap U$ with $0 \leq f \leq 1$ and

$$\mu^*(A \cap U)=\mu (A \cap U) \leq u (f)+ \epsilon .$$

Now, $A \backslash \operatorname{supp} (f)$ is also open, so we can find $g \in C(X) , 0 \leq g \leq 1 , \operatorname{supp} (g) \Subset A \backslash \operatorname{supp} (f)$ with

$$\mu^*(A \backslash\operatorname{supp} (f))=\mu(A\backslash\operatorname{supp} (f)) \leq u(g) + \epsilon .$$

Since

$$A \backslash \operatorname{supp} (f) \supset A \cap U^C , 0 \leq f+g \leq 1 , \operatorname{supp} (f+g) \Subset A ,$$

$$\mu (A) \geq u(f+g) =u(f)+u(g)$$

$$> \mu^* (A \cap U) + \mu^* (A \cap U^C) - 2\epsilon$$

$$\geq \mu^* (A) - 2 \epsilon$$

using subadditivity of $\mu^*.$ Letting $\epsilon \downarrow 0$ we conclude that

$$\mu^* (A) \leq \mu^* (A \cap U) + \mu^* (A \cap U^C) \leq \mu^* (A) = \mu (A) .$$

This gives (2.1) when $A$ is open.

In general, if $E \subset X$ and $\mu^* (E) < \infty$ then given $\epsilon >0$ there exists $A \subset X$ open with $\mu^* (E) > \mu^* (A) - \epsilon$. Thus,

$$\mu^* (E) \geq \mu^* (A \cap U) + \mu^* (A \cap U^C) -\epsilon$$

$$\geq \mu^* (E \cap U) + \mu^* (E \cap U^C) -\epsilon$$

$$\geq \mu^* (E) - \epsilon .$$

This shows that (2.1) always holds, so $U$ is $\mu^*$-measurable if it is open. We have already observed that $\mu (U) = \mu^* (U)$ if $U$ is open. $\qedsymbol$

Thus we have shown that the $\sigma$-algebra given by Caratheodory's theorem contains all open sets. You showed in Problem 3 that the intersection of any collection of $\sigma$-algebras on a given set is a $\sigma$-algebra. Since $\mathcal{P} (X)$ is always a $\sigma$-algebra it follows that for any collection $\mathcal{E} \subset \mathcal{P} (X)$ there is always a smallest $\sigma$-algebra containing $\mathcal{E}$, namely

$$\mathcal{M}_{\mathcal{E}} = \bigcap \left\{ \mathcal{M} \supset \... ...box{ is a $\sigma$-algebra }, \mathcal{M} \subset \mathcal{P} (X) \right\} .$$

The elements of the smallest $\sigma$-algebra containing the open sets are called `Borel sets'. A measure defined on the $\sigma$-algebra of all Borel sets is called a Borel measure. This we have shown:

Proof. This is what Proposition 2.5 says! See how easy proofs are. $\qedsymbol$

We can even continue in the same vein. A Borel measure is said to be outer regular on $E \subset X$ if

$$\mu (E) = \inf \left\{ \mu (U) ; U \supset E , U \hbox{ open} \right\} .$$ (17)

Thus the measure constructed in Proposition 2.5 is outer regular on all Borel sets! A Borel measure is inner regular on $E$ if

$$\mu (E) = \sup \left\{ \mu (K) ; K \subset E , K \hbox{ compact} \right\} .$$ (18)

Here we need to know that compact sets are Borel measurable. This is Problem 5.

Proof. Suppose $K \subset X$ is compact. Let $\chi_K$ be the characteristic function of $K , \chi_K =1$ on $K , \chi_K =0$ on $K^C$. Suppose $f \in \mathcal{C}_0 (X) , \operatorname{supp} (f) \Subset X$ and $f \geq \chi_K$. Set

$$U_{\epsilon } = \left\{ x \in X ; f (x) > 1-\epsilon \right\}$$

where $\epsilon >0$ is small. Thus $U_{\epsilon }$ is open, by the continuity of $f$ and contains $K$. Moreover, we can choose $g \in C (X) , \operatorname{supp} (g) \Subset U_{\epsilon } , 0 \leq g \leq 1$ with $g=1$ near³ $K$. Thus, $g \leq (1-\epsilon )^{-1} f$ and hence

$$\mu^* (K) \leq u (g) = (1-\epsilon)^{-1} u (f) .$$

Letting $\epsilon \downarrow 0$, and using the measurability of $K$,

$$\mu (K) \leq u (f)$$

$$\Rightarrow \mu (K) = \inf \left\{ u (f) ; f \in C (X) , \operatorname{supp} (f) \Subset X , f \geq \chi_K \right\} .$$

In particular this implies that $\mu (K) < \infty$ if $K \Subset X$, but is also proves (2.7). $\qedsymbol$

Let me now review a little of what we have done. We used the positive functional $u$ to define an outer measure $\mu^*$, hence a measure $\mu$ and then checked the properties of the latter.

This is a pretty nice scheme; getting ahead of myself a little, let me suggest that we try it on something else.

Let us say that $Q \subset \mathbb{R}^n$ is `rectangular' if it is a product of finite intervals (open, closed or half-open)

$$Q= \prod^n_{i=1} (\hbox{or}[a_i , b_i]\hbox{or}) a_i \leq b_i$$ (19)

we all agree on its standard volume:

$$v(Q) = \prod^n_{i=1} (b_i - a_i ) \in [0, \infty) .$$ (20)

Clearly if we have two such sets, $Q_1 \subset Q_2$, then $v(Q_1) \le v(Q_2)$. Let us try to define an outer measure on subsets of $\mathbb{R}^n$ by

$$v^* (A) = \inf \left\{ \sum^{\infty}_{i=1} v(Q_i) ; A \subset \bigcup^{\infty}_{i=1} Q_i, Q_i\hbox{ rectangular}\right\} .$$ (21)

We want to show that (2.10) does define an outer measure. This is pretty easy; certainly $v(\emptyset) =0$. Similarly if $\left\{ A_i \right\}^{\infty}_{i=1}$ are (disjoint) sets and $\left\{ Q_{ij} \right\}^{\infty}_{i=1}$ is a covering of $A_i$ by open rectangles then all the $Q_{ij}$ together cover $A= \bigcup_i A_i$ and

$$v^* (A) \leq \sum_i \sum_j v(Q_{ij})$$

$$\Rightarrow v^* (A) \leq \sum_i v^* (A_i) .$$

So we have an outer measure. We also want

Assuming this, the measure defined from $v^*$ using Caratheodory's theorem is called Lebesgue measure.

To prove this we just need to show that (open) rectangular sets are $v^*$-measurable.

Suppose that $\mathcal{M}$ is a $\sigma$-algebra on a set $X$⁴ and $\mathcal{N}$ is a $\sigma$-algebra on another set $Y$. A map $f: X \to Y$ is said to be measurable with respect to these given $\sigma$-algebras on $X$ and $Y$ if

$$f^{-1} (E) \in \mathcal{M} \forall E \in \mathcal{N} .$$ (22)

Notice how similar this is to one of the characterizations of continuity for maps between metric spaces in terms of open sets. Indeed this analogy yields a useful result.

Proof. The main point to note here is that $f^{-1}$ as a map on power sets, is very well behaved for any map. That is if $f: X \to Y$ then $f^{-1}: \mathcal{P}(Y) \to \mathcal{P}(X)$ satisfies:

$$\begin{gathered}f^{-1} (E^C) = (f ^{-1} (E) )^C \ [1ex] f^{-... ... f^{-1} (\phi) = \phi , f^{-1} (Y) = X . \end{gathered}$$ (23)

Putting these things together one sees that if $\mathcal{M}$ is any $\sigma$-algebra on $X$ then

$$\left\{ E \subset Y ; f^{-1} (E) \in \mathcal{M} \right\}$$ (24)

is always a $\sigma$-algebra on $Y$.

Going back to the proof of the proposition, the continuity of $f$ shows that $f^{-1}(E) \subset X$ is open if $E \subset Y$ is open. Thus the $\sigma$-algebra on $Y$ defined by (2.13) from the Borel $\sigma$-algebra on $X$ contains all open sets. It therefore contains the Borel $\sigma$-algebra on $Y$, i.e.,

$$f^{-1} (\mathcal{B} (Y)) \subset\mathcal{B}(X) .$$

$\qedsymbol$

We are mainly interested in functions on $X$. If $\mathcal{M}$ is a $\sigma$-algebra on $X$ then $f:X \to \mathbb{R}$ is measurable if it is measurable with respect to the Borel $\sigma$-algebra on $\mathbb{R}$ and $\mathcal{M}$ on $X$. More generally, for an extended function $f: X \to [ - \infty , \infty ]$ we take as the `Borel' $\sigma$-algebra in $[- \infty , \infty]$ the smallest $\sigma$-algebra containing all open subsets of $\mathbb{R}$ and all sets $(a, \infty ]$ and $[- \infty , b);$ in fact it is generated by the sets $(a,\infty].$ (See Problem 6.)

Our main task is to define the integral of a measurable function: we start with simple functions. Observe that the characteristic function of a set

$$\chi_E = \left\{ \begin{array}{cc} 1 & x \in E \ 0 & x \notin E \end{array} \right.$$

is measurable if and only if $E \in \mathcal{M}$. More generally a simple function,

$$f = \sum^N_{i=1} a_i \chi_{E_i}$$ (25)

is measurable if the $E_i$ are measurable. The presentation, (2.14), of a simple function is not unique. We can make it so, getting the minimal presentation, by insisting that all the $a_i$ are non-zero and

$$E_i = \left\{ x \in E ; f(x) = a_i \right\}$$

then $f$ in (2.14) is measurable iff all the $E_i$ are measurable.