1.4Outer Measures

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In this section, we define a tool used to construct measures called the outer measure. To motivate the idea, we consider how area is defined for a region $E \subseteq \bR^2$ . We can partition $\bR^2$ into many rectangles (whose area is known immediately), and we use these rectangles to approximate the area for $E$ . Indeed, we can define a lower bound on the area of $E$ by summing the area of the rectangles that are a subset of $E$ . Conversely, we may define an upper bound on the area of $E$ by summing the area of the rectangles that intersect $E$ . As we take these rectangles to be smaller and smaller, we produce what is known as the inner area and outer area of $E$ . If these coincide, then we simply have the area of $E$ .

We will generalize the notion of the outer area (since, for any bounding rectangle $R$ of $E$ , the inner area of $E$ is simply the area of $R$ minus the outer area of $R \setminus E$ ). We replace the outer area with the outer measure, and we replace the rectangles with “simple sets”, sets that we already know the area of.

[1.4.1]Definition(Outer Measure)#

Let $X$ be any set. An outer measure is a function $\mu^*: \P(X) \to [0, \infty]$ satisfying the following properties.

Nullity. $\mu^*(\emptyset) = 0$ .
Monotonicity. For any subsets $A, B$ of $X$ , $\mu^*(A) \le \mu^*(B)$ if $A \subseteq B$ .
Countable Subadditivity. For any countable sequence $A_1, A_2, \dots$ of subsets of $X$ , we have that $\mu^*(\bigcup_{k = 1}^\infty A_k) \le \sum_{k = 1}^\infty \mu^*(A_k)$ .

Give $X$ some collection $\mathcal{E}$ called “elementary” or “simple” sets. Like rectangles, these are sets with some elementary notion of measure. We want $\mathcal{E}$ to contain $\emptyset$ and $X$ , of course. Using this, we build an outer measure on $X$ by using the elementary sets as follows.

[1.4.2]Theorem(Construction of Outer Measure)#

Let $X$ be any set. Define some collection of elementary sets $\mathcal{E}$ , such that $\emptyset, X \in \mathcal{E}$ , and a notion of measure $\rho: \mathcal{E} \to [0, \infty]$ , such that $\mu(\emptyset) = 0$ . For each $A \subseteq X$ , define

\mu^*(A) = \inf\left\{ \sum_{k = 1}^\infty \rho(E_k): E_k \in \mathcal{E} \text{ and } A \subseteq \bigcup_{k = 1}^\infty E_k \right\}.

Then $\mu^*: \P(X) \to [0, \infty]$ is an outer measure on $X$ .

Proof.

Immediately, we have that $\mu^*(\emptyset) = 0$ . For any subsets $A \subseteq B$ of $X$ , note that any countable cover of $B$ also contains $A$ , and so $\mu^*(A) \le \mu^*(B)$ is also immediate. Finally, we show $\mu^*$ is countable subadditive. Fix $\epsilon > 0$ . Consider a sequence $A_1, A_2, \dots$ of subsets of $X$ . For each $j \in \bZ_+$ , consider $A_j$ . Then, by definition of the outer measure, there is a sequence of sets $\{E_{j, k}\}_{k = 1}^\infty \subseteq \mathcal{E}$ such that

A_j \subseteq \bigcup_{k = 1}^\infty E_{j, k} \quad \sum_{k = 1}^\infty \rho(E_{j, k}) \le \mu^*(A_j) + \frac{\epsilon}{2^j}.

Then note

\mu^*\left(\bigcup_{j = 1}^\infty A_j\right) \le \sum_{j = 1}^\infty \sum_{k = 1}^\infty \rho(E_{j, k}) \le \sum_{j = 1}^\infty \left[\mu^*(A_j) + \frac{\epsilon}{2^j}\right] = \sum_{j = 1}^\infty \mu^*(A_j) + \epsilon \sum_{j = 1}^\infty \frac{1}{2^j} = \sum_{j = 1}^\infty \mu^*(A_j) + \epsilon.

Since $\epsilon > 0$ , we have shown countable subadditivity. Thus $\mu^*$ is an outer measure.

Note that an outer measure is weaker than a measure. At the cost of being defined on all subsets of $X$ , it gives up an important property: countable additivity. A measure on a $\sigma$ -algebra acts additively on a countable collection of pairwise disjoint sets, while an outer measure can only hope to be countably subadditive. The rest of this section will be dedicated to trying to take different notions of measure (outer measure, and in the future, a premeasure) and find a way to canonically induce a $\sigma$ -algebra and a measure that faithfully represents the original construction.

Let $X$ be a set and $\mu^*$ an outer measure on $X$ . How do we represent $\mu^*$ faithfully by a measure? Intuitively, we should identify the largest collection of subsets of $X$ on which $\mu^*$ behaves like a genuine measure, show that this collection forms a $\sigma$ -algebra, and then restrict $\mu^*$ to it. To determine which sets belong to this $\sigma$ -algebra, we therefore seek the sets across which $\mu^*$ behaves additively. This brings us to the concept of $\mu^*$ -measurability (also called Caratheodory's measurability criterion).

[1.4.3]Definition(Caratheodory Measurability Criterion)#

Let $X$ be a set together with an outer measure $\mu^*: \P(X) \to [0, \infty]$ . We say some $A \subseteq X$ is $\mu^*$ -measurable if

\mu^*(E) = \mu^*(E \cap A) \cup \mu^*(E \cap A^c)

for every $E \subseteq X$ .

Let $A \subseteq X$ be fixed. For any $E \subseteq X$ , note we have that

E = E \cap X = E \cap (A \cup A^c) = (E \cap A) \cup (E \cap A^c).

Note that $E \cap A$ and $E \cap A^c$ are disjoint, and so $E$ is actually the disjoint union of $E \cap A$ and $E \cap A^c$ . By saying $A$ is $\mu^*$ -measurable, we are saying that for any $E \subseteq X$ , we have that

\mu^*(E) = \mu^*(E \cap X) = \mu^*(E \cap (A \cup A^c)) = \mu^*((E \cap A) \sqcup (E \cap A^c)) = \mu^*(E \cap A) + \mu^*(E \cap A^c)

for each $E \subseteq X$ . The last equality is exactly the behavior we want: a measurable set $A$ should partition every $E \subseteq X$ into two disjoint pieces without altering its total outer measure. In other words, $A$ is a “legal cut” across which $\mu^*$ behaves additively. If we have that $A_1, A_2, \dots, A_n$ are $\mu^*$ -measurable, then by induction we have that

\mu^*(E) = \bigsqcup_{k = 1}^n \mu^*(E \cap A_k)

for every $E \subseteq X$ . Intuitively, we seem to have found the sets we want to keep in our $\sigma$ -algebra, and indeed we have.

As an aside, note that $\mu^*(E) \le \mu^*(E \cap A) \cup \mu^*(E \cap A^c)$ is true for all $A, E \subseteq X$ . Thus $A$ is $\mu^*$ -measurable if and only if $\mu^*(E) \ge \mu^*(E \cap A) \cup \mu^*(E \cap A^c)$ .

[1.4.4]Theorem(Caratheodory Extension Theorem)#

Let $X$ be a set together with an outer measure $\mu^*: \P(X) \to [0, \infty]$ . Then the set of all $\mu^*$ -measurable sets, call it $\M$ , forms a $\sigma$ -algebra. Then the restriction $\mu^*|_\M$ forms a complete measure on $\M$ .

We will first use Caratheodory's theorem for extending premeasures (which we define) on algebras to $\sigma$ -algebras.

[1.4.5]Definition(Premeasure)#

Let $X$ be a set together with an algebra $\A \subseteq \P(X)$ . Then a function $\mu_0: \A \to [0, \infty]$ is said to be a premeasure if it satisfies the following properties.

Nullity. $\mu_0(\emptyset) = 0$ .
Countable Additivity. If $\{A_k\}_{k = 1}^\infty \subseteq \A$ is a sequence of sets such that $\bigcup_{k = 1}^\infty A_k \in \A$ , then
$\mu_0\left( \bigcup_{k = 1}^\infty A_k \right) = \sum_{k = 1}^\infty \mu_0(A_k).$

A premeasure can be realized as a measure defined on an algebra instead of a $\sigma$ -algebra by simply ignoring cases where it fails (like when a countable sequence of sets escapes the algebra and is thus not measurable). We essentially pretend that our algebra is a $\sigma$ -algebra—unless it isn't, in which case we just ignore the failure. In this vein, we may also define finite, $\sigma$ -finite, and semifinite premeasures readily.

We can consider an algebra $\A \subseteq \P(X)$ as a collection of simple sets, together with the simple measure being the premeasure $\mu_0$ . In this fashion, it should make sense to define an outer measure with respect to the premeasure as per usual:

\mu^*(A) = \inf\left\{ \sum_{k = 1}^\infty \mu_0(A_k): A_k \in \A, A \subseteq \bigcup_{k = 1}^\infty A_k \right\}.

This construction works, but we also seek to show that the outer measure constructed faithfully represents the premeasure it is constructed from.

[1.4.6]Theorem(Caratheodory Outer Measure Construction)#

Let $X$ be a set together with some algebra $\A \subseteq \P(X)$ and a premeasure $\mu_0: \A \to [0, \infty]$ . Then there exists an outer measure $\mu^*: \P(X) \to [0, \infty]$ defined by

\mu^*(A) = \inf\left\{ \sum_{k = 1}^\infty \mu_0(A_k): A_k \in \A, A \subseteq \bigcup_{k = 1}^\infty A_k \right\}

satisfying the following conditions.

$\mu^*|_\A = \mu_0$ .
Every set in $\A$ is $\mu^*$ -measurable.

Proof.

Fix some $A \in \A$ . Immediately, consider the sequence of sets $(E_k)_{k = 1}^\infty$ with $E_k = A$ for $k = 1$ and $E_k = \emptyset$ otherwise, and note
$\mu^*(A) = \inf\left\{ \sum_{k = 1}^\infty \mu_0(A_k) : A_k \in \A, A \subseteq \bigcup_{k = 1}^\infty A_k \right\} \le \sum_{k = 1}^\infty \mu_0(E_k) = \mu_0(A),$
and so $\mu^*(A) \le \mu_0(A)$ . We show that $\mu_0(A) \le \mu^*(A)$ , proving equality. For any sequence of sets $A_k$ such that $A \subseteq \bigcup_{k = 1}^\infty A_k$ , we may produce a sequence of disjoint sets
$B_n = A \cap \left[A_n \setminus \bigcup_{k = 1}^{n - 1} A_{k} \right]$
for each $n \in \bZ_+$ . Then $B_n$ is a disjoint sequence of sets such that $A = \bigcup_{k = 1}^\infty B_n$ . Thus
$\mu_0(A) = \mu_0\left( \bigcup_{k = 1}^\infty B_k \right) = \sum_{k = 1}^\infty \mu_0(B_k) \le \sum_{k = 1}^\infty \mu_0(A_k),$
with the last inequality since $B_k \subseteq A_k$ for each $k$ . Since $\mu_0(A)$ is less than the sum of the premeasure of every single sequence of sets that covers $A$ , we have that $\mu_0(A) \le \mu^*(A)$ as desired, and so we have reached equality.
Fix $A \in \A$ . For $A$ to be $\mu^*$ -measurable, we need that for any $E \subseteq X$ , we have that
$\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c).$
We already have that $\mu^*(E) \le \mu^*(E \cap A) + \mu^*(E \cap A^c)$ by countable subadditivity, so we prove the other direction. Fix $\epsilon > 0$ . Then there is a sequence of sets $\{B_k\}_{k = 1}^\infty \subseteq \A$ such that $E \subseteq \bigcup_{k = 1}^\infty B_k$ and $\sum_{k = 1}^\infty \mu_0(B_k) \le \mu^*(E) + \epsilon$ . Since $\mu_0$ is additive on sets in $\A$ , note that
$\mu^*(E) + \epsilon \ge \sum_{k = 1}^\infty \mu_0(B_k) = \sum_{k = 1}^\infty \mu_0((B_k \cap A) \cup (B_k \cap A^c)) = \sum_{k = 1}^\infty \mu_0(B_k \cap A) + \sum_{k = 1}^\infty \mu_0(B_k \cap A^c) \ge \mu^*(E \cap A) + \mu^*(E \cap A^c).$
Since $\epsilon > 0$ , we have completed the proof.

The previous theorem shows that the outer measure induced by the premeasure is “reasonable” (it generalizes the premeasure and declares sets in the algebra as Caratheodory-measurable).

[1.4.7]Theorem(Caratheodory Extension Theorem)#

Let $X$ be a set together with an algebra $\A \subseteq \P(X)$ and a premeasure $\mu_0: \A \to [0, \infty]$ . Let $\mu^*$ be the outer measure induced by $\mu_0$ , and let $\M$ be the $\sigma$ -algebra generated by $\A$ . Then the following hold:

Existence. The restriction $\mu = \mu^*|_\M$ is a measure on $\M$ that extends $\mu_0$ (meaning $\mu(A) = \mu_0(A)$ for all $A \in \A$ ).
Bounding Alternative Extensions. If $\nu$ is any other measure on $\M$ that extends $\mu_0$ , then $\nu(E) \le \mu(E)$ for all $E \in \M$ . Furthermore, if $\mu(E) < \infty$ , then $\nu(E) = \mu(E)$ .
Uniqueness. If $\mu_0$ is $\sigma$ -finite, then $\mu$ is the unique measure on $\M$ that extends $\mu_0$ .

Proof.

Recall that all sets in $\A$ are $\mu^*$ -measurable. Then note that the $\sigma$ -algebra generated by $\A$ , say $\M$ , is a subset of all $\mu^*$ -measurable sets. By Caratheodory's theorem, there is a measure $\mu: \M \to [0, \infty]$ such that $\mu|_\M = \mu^* = \mu_0$ .

We have blackboxed certain mechanical proofs, such as the Caratheodory theorem for outer measures.

1.4.1Exercises#

[1.4.8]Problem#

Let $X$ be a set together with an algebra $\A \subseteq \P(X)$ . Let $\A_\sigma$ be the collection of countable unions of sets from $\A$ , and $\A_{\sigma \delta}$ the collection of countable intersections of sets from $\A_\sigma$ . Let $\mu_0: \mathcal{A} \to [0, \infty]$ be a premeasure and $\mu^*$ the induced premeasure.

Show that, for any $E \subseteq X$ and $\epsilon > 0$ , there is some $A \in \A_\sigma$ such that $E \subseteq A$ and $\mu^*(A) \le \mu^*(E) + \epsilon$ .

Proof.

Fix $E \subseteq X$ and $\epsilon > 0$ . If $\mu^*(E) = \infty$ , then any $A \in \A_\sigma$ satisfies this problem, so suppose $E$ has finite measure. Since $\mu^*(E) + \epsilon > \mu^*(E)$ , there exists a sequence of sets $\{A_k\}_{k = 1}^\infty \subseteq \A$ such that $E \subseteq \bigcup_{k = 1}^\infty A_k$ and
$\mu^*(E) \le \mu^*\left(\bigcup_{k = 1}^\infty A_k \right) \le \sum_{k = 1}^\infty \mu_0(A_k) \le \mu^*(E) + \epsilon,$
by the definition of the infimum. Then let $A = \bigcup_{k = 1}^\infty A_k$ , and note $A \in \A_\sigma$ .

[1.4.9]Definition(Borel Measure)#

A Borel measure is a measure whose domain is a Borel $\sigma$ -algebra for the topological space $(X, \Tau)$ .

As such, a Borel measure on $\bR$ is simply a measure whose domain is the Borel $\sigma$ -algebra generated by the standard topology on $\bR$ . We will motivate the section by first introducing what is known as a cumulative distribution function. If $\mu$ is a finite Borel measure on $\bR$ , then we may define the cumulative distribution $F(x) = \mu((-\infty, x])$ . $F(x)$ accumulates the value of the measure $\mu$ at all points along the real line, similar to how a cumulative distribution function in probability accumulates the values of a probability measure across an event space. Notably, $F$ starts from $0$ , is increasing, and is right continuous as follows.

If $x < y$ , then $(-\infty, x] \subseteq (-\infty, y]$ , which implies $F(x) = \mu((-\infty, x]) \le \mu((-\infty, y]) = F(y)$ , and so $F$ is monotonically increasing. Moreover, $F$ is right continuous. Indeed, if $x_n \searrow x$ is a monotonically decreasing sequence converging to $x$ , then note that the intervals $(-\infty, x_n]$ form a decreasing sequence of measurable sets whose intersection is exactly $(-\infty, x]$ . That is, $\bigcap_{n=1}^\infty (-\infty, x_n] = (-\infty, x]$ . Because $\mu$ is a finite measure, we can invoke continuity from above to conclude that

\lim_{n \to \infty} F(x_n) = \lim_{n \to \infty} \mu((-\infty, x_n]) = \mu\left(\bigcap_{n=1}^\infty (-\infty, x_n]\right) = \mu((-\infty, x]) = F(x).

Since $\lim_{n \to \infty} F(x_n) = F(x)$ holds for any sequence $x_n \searrow x$ , it follows that $F$ is right continuous.

Moreover, if $b > a$ , then note $(-\infty, b] = (-\infty, a] \cup (a, b]$ , and by countable additivity, we have that

\mu((-\infty, b]) = \mu((-\infty, a]) + \mu((a, b]),

and so $\mu((a, b]) = F(b) - F(a)$ .

Noting all of this, we work backwards. Given a function $F: \bR \to \bR$ that is increasing and right-continuous, we seek a Borel measure on $\bR$ generated by $F$ , say $\mu_F$ , satisfying $\mu_F((a, b]) = F(b) - F(a)$ as from before. This is called the Lebesgue-Stieltjes construction. Eventually, we will show this construction works in reverse as well—that is, for an arbitrary Borel measure $\mu$ , there is some function $F: \bR \to \bR$ such that $\mu = \mu_F$ . This is called the Lebesgue-Stieltjes correspondence.

The correspondence isn't necessarily unique, however; indeed, two functions may generate the same Borel measure under the Lebesgue-Stieltjes construction, and accordingly, an arbitrary Borel measure may give rise to multiple functions that induce it. Of course, this would be problematic if the various functions generating a Borel measure were completely different (and vice versa), as there isn't even much of a correspondence if there is no determinism. Thankfully, the “various functions” differ by only an additive constant, as we will state more formally later.

In any case, the Lebesgue-Stieltjes construction and correspondence will lead to a rich theory of assigning length to Borel sets in $\bR$ . As a specific case, if our function is given by $F(x) = x$ , the measure we derive is the Lebesgue measure, the standard way to assign length to Borel sets of $\bR$ .

We develop the theory of the Lebesgue-Stieltjes construction by considering the half-open intervals $(a, b]$ and rays $(a, \infty)$ and $(-\infty, a)$ . More specifically, we consider the collection of all half-open intervals and rays, and close this set under complementation and finite unions. A general element of this algebra is then simply a finite disjoint union of half-open intervals and rays. Given a function $F: \bR \to \bR$ , we define a premeasure on this algebra that extends to a measure whose distribution function is $F$ , up to an additive constant.

[1.4.10]Theorem(Increasing Right-Continuous Function Induces a Premeasure)#

Let $F: \bR \to \bR$ be increasing and right-continuous. Let $\mathcal A$ be the algebra of finite disjoint unions of half-open intervals and rays. For a standard half-open interval $I = (a, b]$ , define

\mu_0(I) = F(b) - F(a).

The measure of a ray follows inevitably from the countable additivity required of any premeasure on disjoint sets whose union remains in the algebra. By partitioning a ray into a countable union of half-open intervals—for instance, $(a, \infty) = \bigsqcup_{k=1}^\infty (x_{k-1}, x_k]$ where $a = x_0 < x_1 < x_2 < \dots$ and $x_n \to \infty$ —the measure evaluates as a telescoping sum:

\mu_0((a, \infty)) = \sum_{k=1}^\infty \left[ F(x_k) - F(x_{k-1}) \right] = F(\infty) - F(a).

By symmetry, $\mu_0((-\infty, b]) = F(b) - F(-\infty)$ . For the empty set, $\mu_0(\emptyset) = 0$ .

Let $A \in \mathcal{A}$ be arbitrary, and write $A = \bigcup_{k = 1}^n A_k$ , with $\{A_k\}_{k = 1}^n$ a collection of pairwise disjoint half-open intervals or rays. Define

\mu_0(A) = \sum_{k = 1}^n \mu_0(A_k).

Then $\mu_0$ is a well-defined premeasure on $\mathcal A$ .

Note that, for an increasing function $F$ , we write

F(-\infty) = \lim_{x \to -\infty} F(x) \quad F(\infty) = \lim_{x \to \infty} F(x),

with the value possibly being $\pm \infty$ if the function diverges at the tails.

[1.4.11]Theorem(Lebesgue-Stieltjes Correspondence)#

Suppose $F: \bR \to \bR$ is increasing and right-continuous. Then there exists a unique Borel measure $\mu_F$ on $\bR$ such that $\mu_F((a, b]) = F(b) - F(a)$ for all $a, b \in \bR$ . If $G$ is another such function, we have that $\mu_F = \mu_G$ if and only if $F - G \equiv c$ for some constant $c \in \bR$ .

Conversely, if $\mu$ is some Borel measure on $\bR$ that is finite on all bounded Borel sets, we can define

F_k(x) = \begin{cases} \mu((k, x]) & x > k, \\ 0 & x = k, \\ -\mu((x, k]) & x < k \end{cases}

for any $k \in \bR$ . Then each $F_k$ is increasing, right-continuous, and such that $\mu = \mu_{F_k}$ .

Crucially, this correspondence only exists for Borel measures that are finite on all bounded sets; without this condition, the construction of a real-valued $F$ would immediately collapse. Even with this guarantee, one might naturally attempt to define the generating function globally as simply $F(x) = \mu((-\infty, x])$ . However, this strictly requires the measure to be finite on the unbounded ray $(-\infty, x]$ , which is not guaranteed for general boundedly-finite Borel measures (such as the standard Lebesgue measure, which is infinite on all such rays). By introducing a finite anchor point $k \in \bR$ , we restrict our evaluations entirely to bounded intervals. Because the measure is finite on bounded sets, this ensures $F_k(x)$ safely remains finite and well-defined everywhere.

Consequently, the correspondence very explicitly associates a measure with a family of functions that all differ by an additive constant. The theorem already dictates that any two functions generating the same Borel measure via the Lebesgue-Stieltjes construction must differ by a constant. Also, the constructed family of functions $F_k$ generating an arbitrary Borel measure differ by a constant as well. Indeed, without loss of generality, let us compare $F_k$ against the function $F_0$ anchored at $0$ . For any $x > k > 0$ , finite additivity yields

F_0(x) = \mu((0, x]) = \mu((0, k]) + \mu((k, x]) = F_0(k) + F_k(x).

Rearranging this gives $F_k(x) = F_0(x) - F_0(k)$ . It is quick to show this exact identity holds for all relative orderings of $x, k,$ and $0$ . Thus, we can readily recover one function from another; modifying the anchor point to $k$ simply translates the function $F_0$ vertically by exactly $F_0(k)$ .