Measures#

We wish to assign a value to the size of subsets of some given space, such as the length, area or volume of subsets of \(\mathbb R^m\).

Definition 1 (Measure)

A measure \(\mu\) on a set \(X\) is a function

\[\mu \colon \{A : A\subset X\} \to [0,\infty]\]

such that

  1. \(\mu(\emptyset)=0\);

  2. \(\mu(A)\subset \mu(B)\) whenever \(A\subset B \subset X\) (\(\mu\) is monotonic);

  3. \(\mu(\bigcup_{i\in\mathbb N}A_i) \leq \sum_{i\in\mathbb N} \mu(A_i)\) whenever \(A_1,A_2,\ldots \subset X\) (\(\mu\) is countably subadditive).

Definition 2 (Measurable set)

Let \(\mu\) be a measure on a set \(X\). A set \(A\subset X\) is \(\mu\)-measurable if, for every \(E\subset X\),

(1)#\[ \mu(E)= \mu(E \cap A) + \mu(E \setminus A).\]

Remark 1

  1. Since a measure is countably sub-additive, it is sufficient to check the \(\geq\) inequality in (1).

  2. In particular, it suffices to check (1) for \(E\subset X\) with \(\mu(E)<\infty\).

  3. If \(A\) is \(\mu\)-measurable then so is \(X\setminus A\).

  4. If \(\mu(A)=0\) then \(A\) is \(\mu\)-measurable.

Definition 3 (Restriction)

If \(\mu\) is a measure on a set \(X\) and \(S\subset X\), the restriction of \(\mu\) to \(A\) is defined as

\[\mu|_S(A) := \mu(S\cap A).\]

Lemma 1

Let \(\mu\) be a measure on a set \(X\) and \(S\subset X\). Then \(\mu|_S\) is a measure on \(X\) and any \(\mu\)-measurable set is also \(\mu|_S\)-measurable.

Proof. The fact that \(\mu|_S\) is a measure follows immediately from the fact that \(\mu\) is a measure. If \(A\subset X\) is \(\mu\)-measurable, then for any \(E\subset X\),

\[\begin{split}\mu|_S(E)=\mu(E\cap S) &= \mu(E\cap S \cap A)+\mu(E\cap S\setminus A)\\ &=\mu|_S(E\cap A) +\mu|_S(E\setminus A),\end{split}\]

as required.

Theorem 1

Let \(\mu\) be a measure on a set \(X\) and let \(\mathcal M\) be the set of \(\mu\)-measurable subsets of \(X\).

  1. If \(A_1,A_2,\ldots \in \mathcal M\) then \(\bigcup_{i\in\mathbb N}A_i \in\mathcal M\) and \(\bigcap_{i\in\mathbb N} A_i \in \mathcal M\).*

  2. \(\mu\) is countably additive on \(\mathcal M\). That is, if \(A_1,A_2\ldots \in\mathcal M\) are disjoint then

    \[\mu\left(\bigcup_{i\in\mathbb N}A_i\right) = \sum_{i\in\mathbb N}\mu(A_i).\]
  3. If \(A_1\subset A_2 \subset \ldots \in \mathcal M\) then

    \[\mu\left(\bigcup_{i\in\mathbb N}A_i\right) = \lim_{i\to\infty} \mu(A_i).\]
  4. If \(A_1\supset A_2 \supset \ldots \in \mathcal M\) and \(\mu(A_1)<\infty\) then

    \[\mu\left(\bigcap_{i\in\mathbb N}A_i\right) = \lim_{i\to\infty} \mu(A_i).\]

Proof. We first prove the first point for finite unions and intersections. If \(A,B\in\mathcal M\) then for every \(E\subset X\),

\[\begin{split}\mu(E) &= \mu(E\cap A) +\mu(E\setminus A)\\ &= \mu(E\cap A) + \mu((E\setminus A)\cap B) + \mu(E\setminus (A\cup B))\\ &\geq \mu(E\cap (A\cup B)) +\mu(E\setminus (A\cup B))\end{split}\]

by sub-additivity. Thus \(A\cup B\) is \(\mu\)-measurable and induction gives finite unions. Taking complements gives finite intersections.

To prove the second point, note that the inequality \(\leq\) is given by sub-additivity. For the other inequality, for each \(i\in\mathbb N\) let \(A_i\in\mathcal M\) be disjoint and for each \(j\in\mathcal M\) let

\[B_j = \bigcup_{i=1}^j A_i,\]

which is measurable by the first conclusion. Note that

\[B_j = B_{j-1}\cup A_j\]

and that this union is disjoint. Therefore, since \(A_j\) is \(\mu\)-measurable,

\[\begin{split}\mu(B_j) &= \mu(B_j\cap A_j) + \mu(B_j \setminus A_j)\\ &= \mu(A_j) + \mu(B_{j-1}),\end{split}\]

since the \(A_i\) are all disjoint. Therefore, by induction, \(\mu(B_j)= \sum_{i=1}^j \mu(A_i)\) for each \(j\in\mathbb N\). Finally, for each \(j\in\mathbb N\), since \(\mu\) is monotonic,

\[\mu\left(\bigcup_{i\in\mathbb N}A_i\right) \geq \mu(B_j) = \sum_{i=1}^j \mu(A_i)\]

and so letting \(j\to\infty\) gives the second conclusion.

The third conclusion follows by applying the second conclusion to the disjoint measurable sets \(B_j = A_j \setminus A_{j-1}\).

The fourth conclusion follows from the third by setting \(B_j=A_1\setminus A_j\), so that

\[A_1 =\bigcap_{i\in\mathbb N}A_i \cup \bigcup_{i\in\mathbb N}B_i\]

and the \(B_j\) increase. By sub-additivity,

\[\begin{split}\mu(A_1) &\leq \mu\left(\bigcap_{i\in\mathbb N}A_i\right) + \lim_{j\to\infty}\mu(B_j)\\ &= \mu\left(\bigcap_{i\in\mathbb N}A_i\right) + \lim_{j\to\infty} \mu(A_1)-\mu(A_j),\end{split}\]

by applying the first conclusion for finite unions. Since \(\mu(A_1)<\infty\), the fourth point follows.

Finally, to prove the first point for countable unions, for each \(j\in\mathbb N\) let

\[B_j = \bigcup_{i=1}^j A_i,\]

an increasing sequence, and let \(E\subset X\) with \(\mu(E)<\infty\). Since the \(B_j\) are \(\mu\)-measurable,

\[\begin{split}\mu(E) &= \lim_{j\to\infty}\mu(E\cap B_j) + \lim_{j\to\infty}\mu(E\setminus B_j)\\ &= \mu\left(E\cap \bigcup_{i\in\mathbb N}B_i\right) +\mu\left(E\setminus \bigcup_{i\in\mathbb N}B_i\right)\\ &=\mu\left(E\cap \bigcup_{i\in\mathbb N}A_i\right) +\mu\left(E\setminus \bigcup_{i\in\mathbb N}A_i\right),\\\end{split}\]

using the fact that the \(B_j\) are \(\mu|_E\)-measurable in the second equality. Taking complements shows that countable intersections of measurable sets are measurable.

Definition 4 (Sigma algebra)

A collection \(\Sigma\) of subsets of a set \(X\) is a \(\sigma\)-algebra if

  1. \(\emptyset\in \Sigma\);

  2. \(A\in\Sigma \Rightarrow X\setminus A \in \Sigma\);

  3. \(A_1,A_2,\ldots \in \Sigma \Rightarrow \bigcup_{i\in\mathbb N}A_i \in \Sigma\).

Theorem 1 shows that the set of \(\mu\)-measurable sets is a \(\sigma\)-algebra.

For \(\Omega\) a set of subset of a set \(X\), the \(\sigma\)-algebra generated by \(\Omega\) is

\[\Sigma(\Omega):= \bigcap\{\Sigma': \Sigma'\supset \Omega,\ \Sigma' \text{ a } \sigma\text{-algebra}\}.\]

By Example 2, it is a \(\sigma\)-algebra.

The Borel \(\sigma\)-algebra* of a topological space \(X\) is the \(\sigma\)-algebra generated by the open (respectively closed) subsets of \(X\). It will be denoted by \(\mathcal B(X)\) and its elements called the Borel subsets of \(X\).

A measure for which all Borel sets are measurable is a Borel measure. It is Borel regular if for every \(A\subset X\) there exists a Borel \(B\supset A\) with \(\mu(B)=\mu(A)\).

Theorem 2 (Carathéodory criterion)

Let \((X,d)\) be a metric space and \(\mu\) a measure on \(X\) which is additive on separated sets. That is, whenever \(A,B\subset X\) with

\[\inf\{d(x,y): x\in A,\ y\in B\}>0,\]

we have

\[\mu(A\cup B)=\mu(A) + \mu(B).\]

Then \(\mu\) is a Borel measure.

Proof. Let \(C\subset X\) be closed and \(E\subset X\) with \(\mu(E)<\infty\). We need to show

\[\mu(E) \geq \mu(E\cap C)+ \mu(E\setminus C).\]

For each \(j\in \mathbb N\) let

\[E_j=\{x\in E: \frac{1}{j+1}< \operatorname{dist}(x,C) \leq \frac{1}{j}\}\]

and

\[E_0=\{x\in E: \operatorname{dist}(x,C)>1\}.\]

Since \(C\) is closed,

\[E\setminus C = E_0\cup \bigcup_{j\in\mathbb N}E_j.\]

Moreover, the \(E_j\) with \(j\) odd are pairwise separated so

\[\mu(E)\geq \mu(\bigcup_{j \text{ odd}}E_j)=\sum_{j\text{ odd}}\mu(E_j)\]

and so the sum is convergent. Similarly the sum over even indices is convergent and so

\[\sum_{j\geq n}\mu(E_j)\to 0 \quad \text{as } n\to\infty.\]

Therefore

\[\begin{split}\mu(E) &\geq \mu\left(E\cap C \cup \bigcup_{j=0}^n E_j\right)\\ &= \mu(E\cap C) + \mu\left(\bigcup_{j=0}^n E_j\right)\\ &\geq \mu(E\cap C) +\mu(E\setminus C) - \sum_{j>n} \mu(E_j)\\ &\to \mu(E\cap C) + \mu(E\setminus C),\end{split}\]

using the additivity on separated sets for the equality and countable sub-additivity for the second inequality.

Definition 5 (Carathéodory construction)

Let \((X,d)\) be a metric space, \(F\) a set of subsets of \(X\) and \(\zeta \colon F\to [0,\infty]\). For each \(\delta>0\) and \(A\subset X\) define

\[\psi_\delta(A)= \inf \sum_{S\in G}\zeta(S),\]

where the infimum is taken over all countable

\[G\subset \{S\in F: \operatorname{diam}(S)\}<\delta\]

such that

\[A \subset \bigcup_{S\in G}S.\]

Finally, define \(\psi(A)=\sup_{\delta>0}\psi_\delta(A)\).

For any \(\delta>0\), \(\psi_\delta\) is a measure, as is \(\psi\). Theorem 8 shows that \(\psi\) is a Borel measure on \(X\). Indeed, if \(\operatorname{dist}(A,B)>\delta\) then

\[\psi_\delta(A\cup B) \geq \psi_\delta(A) +\psi_\delta(B).\]

If \(F\) consists only of Borel sets then \(\psi\) is Borel regular.

Remark 2

The fact that \(\psi_{\delta'}\leq \psi_{\delta}\) whenever \(\delta'\geq \delta\) implies that

\[\psi(A)=\lim_{\delta\to 0}\psi_\delta(A).\]

Definition 6

We define some properties of a measure \(\mu\) on a topological space \(X\).

  1. \(\mu\) is locally finite if every point in \(X\) has a neighbourhood of finite measure.

  2. \(\mu\) is \(\sigma\)-finite if there exist measurable \(X_i\subset X\) with \(\mu(X_i)<\infty\) and \(X=\bigcup_{i\in\mathbb N} X_i\).

  3. \(\mu\) is finite if \(\mu(X)<\infty\).

  4. A Borel regular measure \(\mu\) is a Radon measure if

    1. \(\mu(K)<\infty\) for all compact \(K\subset X\),

    2. \(\mu(A)= \sup\{\mu(K): K\subset A \text{ compact}\}\) for all Borel \(A\subset X\).

    3. \(\mu(A) = \inf\{\mu(U): U\supset A \text{ open}\}\) for all Borel \(A\subset X\).

Definition 7 (Almost everywhere)

Let \(\mu\) be a measure on a set \(X\). A property of points in \(X\) holds \(\mu\) almost everywhere (or \(\mu\)-a.e.) if the set of points for which the property doesn’t hold has \(\mu\) measure zero.

Definition 8 (Push forward)

Let \(X,Y\) be sets, \(\mu\) be a measure on \(X\) and let \(f\colon X \to Y\). The push forward of \(\mu\) under \(f\), written \(f_{\#}\mu\) is defined by

\[f_{\#}\mu(S) = \mu(f^{-1}(S)).\]

Definition 9 (Lebesgue measure)

The Lebesgue measure on \(\mathbb R^n\), denoted \(\mathcal L^n\), is defined using the Carathéodory construction with \(F\) the set of cubes and \(\zeta(Q)= \operatorname{vol}(Q)\). Its measurable sets are called the Lebesgue measurable subsets of \(\mathbb R^n\).

The following lemma is very useful.

Lemma 2

Let \(\mu\) be a finite measure on a set \(X\) and let \(\mathcal S\) be a set of \(\mu\)-measurable subsets of \(X\). There exists disjoint \(S_i\in\mathcal S\) such that any \(S\in \mathcal S\) with

\[S\subset X\setminus \bigcup_{i\in\mathbb N} S_i\]

satisfies \(\mu(S)=0\).

In particular, if each \(\mu\)-measurable subset of \(X\) of positive measure contains an element of \(\mathcal S\) of positive measure then we can decompose almost all of \(X\) into countably many disjoint elements of \(\mathcal S\).

Proof. We find the \(S_i\) by induction. First let \(\mathcal S^1\subset\mathcal S\) be countable and disjoint such that

\[\mu(\cup \mathcal S^1) \geq \sup\{\mu(\cup \mathcal S'): \mathcal S' \subset \mathcal S \text{ countable and disjoint}\} -1/1.\]

Now let \(\mathcal M_2\) be the set of all \(\mathcal S'\subset \mathcal S\) that are countable, disjoint and disjoint from \(\cup \mathcal S^1\). Let \(\mathcal S^2\in \mathcal M_2\) be such that

\[\mu(\cup \mathcal S^2) \geq \sup\{\mu(\cup\mathcal S'): \mathcal S'\in\mathcal M_2\} -1/2.\]

Inductively, given countable, disjoint \(\mathcal S^1,\ldots,\mathcal S^{i-1}\) such that each \(\mathcal S^j\) and \(\mathcal S^k\) are disjoint for \(k<j\), let \(\mathcal M_i\) be the set of all \(\mathcal S'\subset \mathcal S\) that are countable, disjoint and disjoint from \(\mathcal S^1\cup \ldots \cup \mathcal S^{i-1}\). Let \(\mathcal S^i\in \mathcal M_i\) be such that

\[\mu(\cup \mathcal S^i) \geq \sup\{\mu(\cup\mathcal S'): \mathcal S'\in\mathcal M_i\} -1/i.\]

We claim that any \(S\in \mathcal S\) with

\[S\subset X\setminus \bigcup_{i\in\mathbb N}\bigcup \mathcal S^i\]

satisfies \(\mu(S)=0\). If not, let \(i\in \mathbb N\) be such that \(1/i<\mu(S)\). Then \(\mathcal T:=\mathcal S^i\cup \{S\}\in \mathcal M_i\) and

\[\mu(\cup \mathcal T) > \sup\{\mu(\cup\mathcal S'): \mathcal S'\in\mathcal M_i\} -1/i +1/i,\]

a contradiction.

Exercises#

Example 1

Usually in measure theory, a measure is defined as a countably additive function defined on a \(\sigma\)-algebra. However, using our definition is simply a convenience rather than a restriction.

Indeed, suppose \(\mu\) is a countably additive function defined on a \(\sigma\)-algebra \(\Sigma\) of \(X\) with \(\mu(\emptyset)=0\). Show that it can be extended to the power set of \(X\) by

\[\overline\mu(A)=\inf\{\mu(B): A\subset B\in \Sigma\}\]

and that any \(B\in\Sigma\) is \(\mu\)-measurable. What about

\[\underline\mu(A)=\sup\{\mu(B): A\supset B\in \Sigma\}?\]

Conversely, any measure is countably additive when restricted to any \(\sigma\)-algebra of measurable sets.

Example 2

Let \(\Omega\) be a set of subsets of a set \(X\). Show that \(\Sigma(\Omega)\) is a \(\sigma\)-algebra. Note that it is the smallest \(\sigma\)-algebra of \(X\) containing \(\Omega\).

Example 3

Show that the following sets are Borel subsets of \(\mathbb R\): \(\mathbb Q\), \([0,1)\), the set of points in \([0,1]\) whose first decimal is even.

Let \(f\colon [0,1]\to [0,1]\). Show that the set of points where \(f\) is continuous is a Borel set. What about the set of points where \(f\) is differentiable?

Example 4

Let \(X\) be a set and \(x\in X\). The Dirac measure at \(x\) is defined as \(\delta_x(A)= 1\) if \(x\in A\), \(\delta_x(A)=0\) otherwise. Show that \(\delta_x\) is a measure on \(X\). What are its measurable sets?

Define the counting measure on \(X\) to be the cardinality (finite or \(\infty\)) of any subset of \(X\). Show that this is a measure. What are its measurable sets?

Example 5

For \((X,d)\) a metric space and \(s \geq 0\) the \(s\)-dimensional Hausdorff measure on \(X\), denoted \(\mathcal H^s\), is defined using the Carathéodory construction with \(F\) the set of all sets and \(\zeta(S)=\operatorname{diam}(S)^s\).

  1. Show that \(\mathcal L^n\) and \(\mathcal H^n\) are non-zero, translation invariant and \(n\)-homogenous measures. That is, for any \(A\subset \mathbb R^n\), \(x\in\mathbb R^n\) and \(t>0\), \(\mathcal L^n(A+x)=\mathcal L^n(A)\) and \(\mathcal L^n(tA)=t^n\mathcal L^n(A)\) (and similarly for \(\mathcal H^n\)).

  2. On \(\mathbb R^n\) show that there exists a \(C>0\) such that \(\mathcal H^n/C \leq \mathcal L^n \leq C\mathcal H^n\).

  3. Let \(f\colon X\to Y\) be an \(L\)-Lipschitz function between two metric spaces. Show that for any \(s\geq 0\) and \(A\subset X\),

    \[\mathcal H^s(f(A)) \leq L^s \mathcal H^s(A).\]
  4. For any metric space \(X\), show that \(\mathcal H^0\) is the counting measure on \(X\).

  5. For \(0\leq s < t <\infty\), suppose that \(\mathcal H^{s}(A)<\infty\). Show that \(\mathcal H^t(A)=0\). Hence there exists a single \(0\leq s \leq \infty\) for which \(\mathcal H^{t}(A)=0\) for all \(t>s\) and \(\mathcal H^{t}(A)=\infty\) for all \(t<s\). This \(t\) is called the Hausdorff dimension of \(A\), denoted \(\dim_{\mathrm H} A\).

Example 6 (Cantor set)

The Cantor set \(K\subset [0,1]\) is defined as follows. Let \(K_0=[0,1]\) and for each \(i\in \mathbb N\) let \(K_i\) be obtained from deleting the “middle third” open interval from each of the intervals in \(K_{i-1}\). That is, \(K_1 = [0,1/3] \cup [2/3,1]\),

\[K_2 = [0,1/9] \cup [2/9,1/3] \cup [2/3,7/9] \cup [8/9,1],\]

etc. Define \(K= \bigcap_{i\in\mathbb N}K_i\). Note that \(K\) is compact and hence Borel.

  1. Show that \(K\) is uncountable.

  2. Let \(s=\log 2/\log 3\). Show that \(0<\mathcal H^s(K)<\infty\).

In particular, \(K\) is an uncountable subset of \(\mathbb R\) with \(\mathcal L^1(K)=0\).

Example 7 (Restriction measure)

Give an examples of \(S\subset \mathbb R^2\) with \(\dim_{\mathrm H}S =1\) for which \(\mathcal H^1|_S\) is:

  1. finite,

  2. \(\sigma\)-finite but not finite,

  3. not \(\sigma\)-finite.

Example 8 (Vitali set)

The fundamental properties of measures are those given in Theorem 6, in particular countable additivity. It is necessary for us to only require this to be true for measurable sets, as can be seen from the existence of non-measurable sets.

Define a Vitali set as follows. Consider the equivalence relation \(\sim\) on \(\mathbb R\) defined by \(x \sim y\) iff \(x-y\in\mathbb Q\). By the density of \(\mathbb Q\) in \(\mathbb R\), each equivalence class \(V_x\) intersects \([0,1]\). Therefore, by the axiom of choice(!), we may construct a set \(\mathcal V\subset [0,1]\) consisting of exactly one member of each equivalence class.

Show:

  1. If \(p\neq q\) are rational then \(p+\mathcal V\) and \(q+\mathcal V\) are disjoint.

  2. \([0,1] \subset \bigcup\{q+\mathcal V: q\in\mathbb Q \cap [-1,1]\} \subset [-1,2]\).

  3. Show that \(\mathcal L^1(\mathcal V) \neq 0\).

  4. Deduce that \(\mathcal V\) is not Lebesgue measurable.

Example 9

Show that the two extensions given in Example 1 may not agree. For example, after extending Lebesgue measure (restricted to the Borel sets), what are the values of a Vitali set?

Example 10

Let \(\mu\) be a finite Borel measure on a metric space \(X\). Prove that for every Borel \(B\subset X\),

(2)#\[\mu(B)=\sup\{\mu(C): C\subset B \text{ closed}\}\]

and

(3)#\[\mu(B)=\inf\{\mu(U): U\supset B \text{ open}\}.\]

Property (2) is called inner regularity by closed sets and (3) is called outer regularity by open sets.

Hint: observe that it suffices to show that all Borel sets satisfy (2). Show that the set

\[\{B\subset X: B \text{ and } X\setminus B \text{ are inner regular by closed sets} \}\]

is a \(\sigma\)-algebra that contains all closed subsets of \(X\).

Show that a \(\sigma\)-finite \(\mu\) is inner regular by closed sets. Show that a \(\sigma\)-finite \(\mu\) is outer regular by open sets if there exist open sets \(U_i\subset X\) with \(\mu(U_i)<\infty\) for all \(i\in \mathbb N\) and \(X=\bigcup_{i\in \mathbb N}U_i\). Give an example of a \(\sigma\)-finite \(\mu\) that is not outer regular by open sets.

Example 11

Let \(X\) be a complete and separable metric space. Show that any finite Borel measure on \(X\) is a Radon measure.

Hint: a metric space is compact if and only if it is complete and totally bounded.