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 :math:\mathbb R^m. .. index:: Measure .. prf:definition:: Measure :label: def:measure A *measure* :math:\mu on a set :math:X is a function .. math:: \mu \colon \{A : A\subset X\} \to [0,\infty] such that #. :math:\mu(\emptyset)=0; #. :math:\mu(A)\subset \mu(B) whenever :math:A\subset B \subset X (:math:\mu is *monotonic*); #. :math:\mu(\bigcup_{i\in\mathbb N}A_i) \leq \sum_{i\in\mathbb N} \mu(A_i) whenever :math:A_1,A_2,\ldots \subset X (:math:\mu is *countably subadditive*). .. index:: Measurable set .. prf:definition:: Measurable set :label: def:measurable-set Let :math:\mu be a measure on a set :math:X. A set :math:A\subset X is :math:\mu-measurable if, for every :math:E\subset X, .. math:: :label: measurable-set \mu(E)= \mu(E \cap A) + \mu(E \setminus A). .. prf:remark:: #. Since a measure is countably sub-additive, it is sufficient to check the :math:\geq inequality in :eq:measurable-set. #. In particular, it suffices to check :eq:measurable-set for :math:E\subset X with :math:\mu(E)<\infty. #. If :math:A is :math:\mu-measurable then so is :math:X\setminus A. #. If :math:\mu(A)=0 then :math:A is :math:\mu-measurable. .. prf:definition:: Restriction If :math:\mu is a measure on a set :math:X and :math:S\subset X, the *restriction* of :math:\mu to :math:A is defined as .. math:: \mu|_S(A) := \mu(S\cap A). .. prf:lemma:: Let :math:\mu be a measure on a set :math:X and :math:S\subset X. Then :math:\mu|_S is a measure on :math:X and any :math:\mu-measurable set is also :math:\mu|_S-measurable. .. prf:proof:: The fact that :math:\mu|_S is a measure follows immediately from the fact that :math:\mu is a measure. If :math:A\subset X is :math:\mu-measurable, then for any :math:E\subset X, .. math:: \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), as required. .. prf:theorem:: :label: measurable-properties Let :math:\mu be a measure on a set :math:X and let :math:\mathcal M be the set of :math:\mu-measurable subsets of :math:X. #. If :math:A_1,A_2,\ldots \in \mathcal M then :math:\bigcup_{i\in\mathbb N}A_i \in\mathcal M and :math:\bigcap_{i\in\mathbb N} A_i \in \mathcal M. #. :math:\mu is countably additive on :math:\mathcal M. That is, if :math:A_1,A_2\ldots \in\mathcal M are disjoint then .. math:: \mu\left(\bigcup_{i\in\mathbb N}A_i\right) = \sum_{i\in\mathbb N}\mu(A_i). #. If :math:A_1\subset A_2 \subset \ldots \in \mathcal M then .. math:: \mu\left(\bigcup_{i\in\mathbb N}A_i\right) = \lim_{i\to\infty} \mu(A_i). #. If :math:A_1\supset A_2 \supset \ldots \in \mathcal M and :math:\mu(A_1)<\infty then .. math:: \mu\left(\bigcap_{i\in\mathbb N}A_i\right) = \lim_{i\to\infty} \mu(A_i). .. prf:proof:: We first prove the first point for finite unions and intersections. If :math:A,B\in\mathcal M then for every :math:E\subset X, .. math:: \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)) by sub-additivity. Thus :math:A\cup B is :math:\mu-measurable and induction gives finite unions. Taking complements gives finite intersections. To prove the second point, note that the inequality :math:\leq is given by sub-additivity. For the other inequality, for each :math:i\in\mathbb N let :math:A_i\in\mathcal M be disjoint and for each :math:j\in\mathcal M let .. math:: B_j = \bigcup_{i=1}^j A_i, which is measurable by the first conclusion. Note that .. math:: B_j = B_{j-1}\cup A_j and that this union is disjoint. Therefore, since :math:A_j is :math:\mu-measurable, .. math:: \mu(B_j) &= \mu(B_j\cap A_j) + \mu(B_j \setminus A_j)\\ &= \mu(A_j) + \mu(B_{j-1}), since the :math:A_i are all disjoint. Therefore, by induction, :math:\mu(B_j)= \sum_{i=1}^j \mu(A_i) for each :math:j\in\mathbb N. Finally, for each :math:j\in\mathbb N, since :math:\mu is monotonic, .. math:: \mu\left(\bigcup_{i\in\mathbb N}A_i\right) \geq \mu(B_j) = \sum_{i=1}^j \mu(A_i) and so letting :math:j\to\infty gives the second conclusion. The third conclusion follows by applying the second conclusion to the disjoint measurable sets :math:B_j = A_j \setminus A_{j-1}. The fourth conclusion follows from the third by setting :math:B_j=A_1\setminus A_j, so that .. math:: A_1 =\bigcap_{i\in\mathbb N}A_i \cup \bigcup_{i\in\mathbb N}B_i and the :math:B_j increase. By sub-additivity, .. math:: \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), by applying the first conclusion for finite unions. Since :math:\mu(A_1)<\infty, the fourth point follows. Finally, to prove the first point for countable unions, for each :math:j\in\mathbb N let .. math:: B_j = \bigcup_{i=1}^j A_i, an increasing sequence, and let :math:E\subset X with :math:\mu(E)<\infty. Since the :math:B_j are :math:\mu-measurable, .. math:: \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),\\ using the fact that the :math:B_j are :math:\mu|_E-measurable in the second equality. Taking complements shows that countable intersections of measurable sets are measurable. .. index:: Sigma algebra .. prf:definition:: Sigma algebra :label: def-sigm-algebra A collection :math:\Sigma of subsets of a set :math:X is a :math:\sigma-algebra if #. :math:\emptyset\in \Sigma; #. :math:A\in\Sigma \Rightarrow X\setminus A \in \Sigma; #. :math:A_1,A_2,\ldots \in \Sigma \Rightarrow \bigcup_{i\in\mathbb N}A_i \in \Sigma. :prf:ref:measurable-properties shows that the set of :math:\mu-measurable sets is a :math:\sigma-algebra. For :math:\Omega a set of subset of a set :math:X, the :math:\sigma-algebra generated by :math:\Omega is .. math:: \Sigma(\Omega):= \bigcap\{\Sigma': \Sigma'\supset \Omega,\ \Sigma' \text{ a } \sigma\text{-algebra}\}. By :prf:ref:sigma-alg-generated, it is a :math:\sigma-algebra. The Borel :math:\sigma-algebra of a topological space :math:X is the :math:\sigma-algebra generated by the open (respectively closed) subsets of :math:X. It will be denoted by :math:\mathcal B(X) and its elements called the *Borel* subsets of :math:X. A measure for which all Borel sets are measurable is a *Borel measure*. It is *Borel regular* if for every :math:A\subset X there exists a Borel :math:B\supset A with :math:\mu(B)=\mu(A). .. index:: Carathéodory criterion .. prf:theorem:: Carathéodory criterion :label: caradtheodory-criterion Let :math:(X,d) be a metric space and :math:\mu a measure on :math:X which is additive on separated sets. That is, whenever :math:A,B\subset X with .. math:: \inf\{d(x,y): x\in A,\ y\in B\}>0, we have .. math:: \mu(A\cup B)=\mu(A) + \mu(B). Then :math:\mu is a Borel measure. .. prf:proof:: Let :math:C\subset X be closed and :math:E\subset X with :math:\mu(E)<\infty. We need to show .. math:: \mu(E) \geq \mu(E\cap C)+ \mu(E\setminus C). For each :math:j\in \mathbb N let .. math:: E_j=\{x\in E: \frac{1}{j+1}< \operatorname{dist}(x,C) \leq \frac{1}{j}\} and .. math:: E_0=\{x\in E: \operatorname{dist}(x,C)>1\}. Since :math:C is closed, .. math:: E\setminus C = E_0\cup \bigcup_{j\in\mathbb N}E_j. Moreover, the :math:E_j with :math:j odd are pairwise separated so .. math:: \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 .. math:: \sum_{j\geq n}\mu(E_j)\to 0 \quad \text{as } n\to\infty. Therefore .. math:: \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), using the additivity on separated sets for the equality and countable sub-additivity for the second inequality. .. index:: Carathéodory construction .. prf:definition:: Carathéodory construction :label: caradtheodory-construction Let :math:(X,d) be a metric space, :math:F a set of subsets of :math:X and :math:\zeta \colon F\to [0,\infty]. For each :math:\delta>0 and :math:A\subset X define .. math:: \psi_\delta(A)= \inf \sum_{S\in G}\zeta(S), where the infimum is taken over all countable .. math:: G\subset \{S\in F: \operatorname{diam}(S)\}<\delta such that .. math:: A \subset \bigcup_{S\in G}S. Finally, define :math:\psi(A)=\sup_{\delta>0}\psi_\delta(A). For any :math:\delta>0, :math:\psi_\delta is a measure, as is :math:\psi. Theorem 8 <#caradtheodory-criterion>__ shows that :math:\psi is a Borel measure on :math:X. Indeed, if :math:\operatorname{dist}(A,B)>\delta then .. math:: \psi_\delta(A\cup B) \geq \psi_\delta(A) +\psi_\delta(B). If :math:F consists only of Borel sets then :math:\psi is Borel regular. .. prf:remark:: The fact that :math:\psi_{\delta'}\leq \psi_{\delta} whenever :math:\delta'\geq \delta implies that .. math:: \psi(A)=\lim_{\delta\to 0}\psi_\delta(A). .. index:: Measure; Sigma finite .. index:: Measure; Borel regular .. index:: Measure; Radon .. prf:definition:: We define some properties of a measure :math:\mu on a topological space :math:X. #. :math:\mu is locally finite if every point in :math:X has a neighbourhood of finite measure. #. :math:\mu is :math:\sigma-finite if there exist measurable :math:X_i\subset X with :math:\mu(X_i)<\infty and :math:X=\bigcup_{i\in\mathbb N} X_i. #. :math:\mu is finite if :math:\mu(X)<\infty. #. A Borel regular measure :math:\mu is a Radon measure if #. :math:\mu(K)<\infty for all compact :math:K\subset X, #. :math:\mu(A)= \sup\{\mu(K): K\subset A \text{ compact}\} for all Borel :math:A\subset X. #. :math:\mu(A) = \inf\{\mu(U): U\supset A \text{ open}\} for all Borel :math:A\subset X. .. index:: Almost everywhere .. prf:definition:: Almost everywhere :label: almost-everywhere Let :math:\mu be a measure on a set :math:X. A property of points in :math:X holds :math:\mu almost everywhere (or :math:\mu-a.e.) if the set of points for which the property doesn’t hold has :math:\mu measure zero. .. index:: Measure; Push forward .. prf:definition:: Push forward Let :math:X,Y be sets, :math:\mu be a measure on :math:X and let :math:f\colon X \to Y. The *push forward* of :math:\mu under :math:f, written :math:f_{\#}\mu is defined by .. math:: f_{\#}\mu(S) = \mu(f^{-1}(S)). .. index:: Measure; Lebesgue .. prf:definition:: Lebesgue measure :label: lebesgue-defn The *Lebesgue measure* on :math:\mathbb R^n, denoted :math:\mathcal L^n, is defined using the Carathéodory construction with :math:F the set of cubes and :math:\zeta(Q)= \operatorname{vol}(Q). Its measurable sets are called the *Lebesgue measurable* subsets of :math:\mathbb R^n. The following lemma is very useful. .. prf:lemma:: :label: exhaustion Let :math:\mu be a finite measure on a set :math:X and let :math:\mathcal S be a set of :math:\mu-measurable subsets of :math:X. There exists disjoint :math:S_i\in\mathcal S such that any :math:S\in \mathcal S with .. math:: S\subset X\setminus \bigcup_{i\in\mathbb N} S_i satisfies :math:\mu(S)=0. In particular, if each :math:\mu-measurable subset of :math:X of positive measure contains an element of :math:\mathcal S of positive measure then we can decompose almost all of :math:X into countably many disjoint elements of :math:\mathcal S. .. prf:proof:: We find the :math:S_i by induction. First let :math:\mathcal S^1\subset\mathcal S be countable and disjoint such that .. math:: \mu(\cup \mathcal S^1) \geq \sup\{\mu(\cup \mathcal S'): \mathcal S' \subset \mathcal S \text{ countable and disjoint}\} -1/1. Now let :math:\mathcal M_2 be the set of all :math:\mathcal S'\subset \mathcal S that are countable, disjoint and disjoint from :math:\cup \mathcal S^1. Let :math:\mathcal S^2\in \mathcal M_2 be such that .. math:: \mu(\cup \mathcal S^2) \geq \sup\{\mu(\cup\mathcal S'): \mathcal S'\in\mathcal M_2\} -1/2. Inductively, given countable, disjoint :math:\mathcal S^1,\ldots,\mathcal S^{i-1} such that each :math:\mathcal S^j and :math:\mathcal S^k are disjoint for :math:k \sup\{\mu(\cup\mathcal S'): \mathcal S'\in\mathcal M_i\} -1/i +1/i, a contradiction. Exercises ^^^^^^^^^ .. prf:example:: :label: extend-outer-meas Usually in measure theory, a measure is defined as a countably additive function defined on a :math:\sigma-algebra. However, using our definition is simply a convenience rather than a restriction. Indeed, suppose :math:\mu is a countably additive function defined on a :math:\sigma-algebra :math:\Sigma of :math:X with :math:\mu(\emptyset)=0. Show that it can be extended to the power set of :math:X by .. math:: \overline\mu(A)=\inf\{\mu(B): A\subset B\in \Sigma\} and that any :math:B\in\Sigma is :math:\mu-measurable. What about .. math:: \underline\mu(A)=\sup\{\mu(B): A\supset B\in \Sigma\}? Conversely, any measure is countably additive when restricted to any :math:\sigma-algebra of measurable sets. .. prf:example:: :label: sigma-alg-generated Let :math:\Omega be a set of subsets of a set :math:X. Show that :math:\Sigma(\Omega) is a :math:\sigma-algebra. Note that it is the smallest :math:\sigma-algebra of :math:X containing :math:\Omega. .. prf:example:: Show that the following sets are Borel subsets of :math:\mathbb R: :math:\mathbb Q, :math:[0,1), the set of points in :math:[0,1] whose first decimal is even. Let :math:f\colon [0,1]\to [0,1]. Show that the set of points where :math:f is continuous is a Borel set. What about the set of points where :math:f is differentiable? .. index:: Measure; Dirac .. index:: Measure; Counting .. prf:example:: Let :math:X be a set and :math:x\in X. The Dirac measure at :math:x is defined as :math:\delta_x(A)= 1 if :math:x\in A, :math:\delta_x(A)=0 otherwise. Show that :math:\delta_x is a measure on :math:X. What are its measurable sets? Define the *counting measure* on :math:X to be the cardinality (finite or :math:\infty) of any subset of :math:X. Show that this is a measure. What are its measurable sets? .. index:: Hausdorff measure .. prf:example:: :label: hausdorff-defn For :math:(X,d) a metric space and :math:s \geq 0 the :math:s-dimensional Hausdorff measure on :math:X, denoted :math:\mathcal H^s, is defined using the Carathéodory construction with :math:F the set of all sets and :math:\zeta(S)=\operatorname{diam}(S)^s. #. Show that :math:\mathcal L^n and :math:\mathcal H^n are non-zero, translation invariant and :math:n-homogenous measures. That is, for any :math:A\subset \mathbb R^n, :math:x\in\mathbb R^n and :math:t>0, :math:\mathcal L^n(A+x)=\mathcal L^n(A) and :math:\mathcal L^n(tA)=t^n\mathcal L^n(A) (and similarly for :math:\mathcal H^n). #. On :math:\mathbb R^n show that there exists a :math:C>0 such that :math:\mathcal H^n/C \leq \mathcal L^n \leq C\mathcal H^n. #. Let :math:f\colon X\to Y be an :math:L-Lipschitz function between two metric spaces. Show that for any :math:s\geq 0 and :math:A\subset X, .. math:: \mathcal H^s(f(A)) \leq L^s \mathcal H^s(A). #. For any metric space :math:X, show that :math:\mathcal H^0 is the counting measure on :math:X. #. For :math:0\leq s < t <\infty, suppose that :math:\mathcal H^{s}(A)<\infty. Show that :math:\mathcal H^t(A)=0. Hence there exists a single :math:0\leq s \leq \infty for which :math:\mathcal H^{t}(A)=0 for all :math:t>s and :math:\mathcal H^{t}(A)=\infty for all :math:t__, 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 :math:\sim on :math:\mathbb R defined by :math:x \sim y iff :math:x-y\in\mathbb Q. By the density of :math:\mathbb Q in :math:\mathbb R, each equivalence class :math:V_x intersects :math:[0,1]. Therefore, by the axiom of choice(!), we may construct a set :math:\mathcal V\subset [0,1] consisting of exactly one member of each equivalence class. Show: #. If :math:p\neq q are rational then :math:p+\mathcal V and :math:q+\mathcal V are disjoint. #. :math:[0,1] \subset \bigcup\{q+\mathcal V: q\in\mathbb Q \cap [-1,1]\} \subset [-1,2]. #. Show that :math:\mathcal L^1(\mathcal V) \neq 0. #. Deduce that :math:\mathcal V is not Lebesgue measurable. .. prf:example:: :label: ext_meas_not_unique Show that the two extensions given in :prf:ref:extend-outer-meas may not agree. For example, after extending Lebesgue measure (restricted to the Borel sets), what are the values of a Vitali set? .. prf:example:: :label: borel-is-regular Let :math:\mu be a finite Borel measure on a metric space :math:X. Prove that for every Borel :math:B\subset X, .. math:: :label: inner-reg-closed \mu(B)=\sup\{\mu(C): C\subset B \text{ closed}\} and .. math:: :label: outer-reg-open \mu(B)=\inf\{\mu(U): U\supset B \text{ open}\}. Property :eq:inner-reg-closed is called *inner regularity by closed sets* and :eq:outer-reg-open is called *outer regularity by open sets*. Hint: observe that it suffices to show that all Borel sets satisfy :eq:inner-reg-closed. Show that the set .. math:: \{B\subset X: B \text{ and } X\setminus B \text{ are inner regular by closed sets} \} is a :math:\sigma-algebra that contains all closed subsets of :math:X. Show that a :math:\sigma-finite :math:\mu is inner regular by closed sets. Show that a :math:\sigma-finite :math:\mu is outer regular by open sets if there exist open sets :math:U_i\subset X with :math:\mu(U_i)<\infty for all :math:i\in \mathbb N and :math:X=\bigcup_{i\in \mathbb N}U_i. Give an example of a :math:\sigma-finite :math:\mu that is *not* outer regular by open sets. .. prf:example:: Let :math:X be a complete and separable metric space. Show that any finite Borel measure on :math:X` is a Radon measure. Hint: a metric space is compact if and only if it is complete and totally bounded.