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    Outer measure and the construction

    Contents

    Let \((X,\mathcal M,\mu)\) be a measure space. Countable additivity gives continuity of measure: if \(E_1\subseteq E_2\subseteq\cdots\) then \(\mu\qty{\bigcup_n E_n} = \lim_n\mu(E_n)\), and if \(E_1\supseteq E_2\supseteq\cdots\) with \(\mu(E_1)<\infty\) then \(\mu\qty{\bigcap_n E_n}=\lim_n\mu(E_n)\). For arbitrary \(E_n\in\mathcal M\), the sets \(F_n\coloneqq E_n\setminus\bigcup_{k<n}E_k\) are disjoint with the same union, so countable additivity gives countable subadditivity \(\mu\qty{\bigcup_nE_n}=\sum_n\mu(F_n)\leq\sum_n\mu(E_n)\).

    Measure on a measurable space

    Let \((X, \mcm)\) be a measurable space. A measure on \((X,\mcm)\) is a function \(\mu\colon \mcm \to [0,\infty]\) such that

    1. \(\mu(\emptyset) = 0\), and

    2. \(\mu\) is countably additive: if \((E_k)_{k\geq 1}\) is a sequence of pairwise disjoint sets in \(\mcm\), then \[ \mu\qty{\bigcup_{k\geq 1} E_k} = \sum_{k\geq 1} \mu(E_k). \]

    Countable additivity is also called \(\sigma\)-additivity. A function \(\mu\colon\mcm\to[0,\infty]\) with \(\mu(\emptyset)=0\) for which the identity in (2) is required only for finite families of pairwise disjoint sets is called finitely additive.

    Subtraction of measures

    Removing a finite-measure subset subtracts its measure.Suggest a slogan

    Let \((X,\mcm,m)\) be a measure space and let \(A,C\in\mcm\) with \(C\subseteq A\) and \(m(C) < \infty\). Then \[ m(A\setminus C) = m(A) - m(C) \] [@Fol13].

    The hypothesis \(m(C)<\infty\) cannot be dropped: for Lebesgue measure with \(A = C = \RR\), the right-hand side \(\infty-\infty\) is undefined.

    Monotonicity and countable subadditivity of measures

    Measures grow with inclusion and never exceed the sum of a countable cover.Suggest a slogan

    Let \((X, \mcm, \mu)\) be a measure space.

    1. (Monotonicity) If \(E,F\in\mcm\) and \(E \subseteq F\), then \(\mu(E) \leq \mu(F)\).

    2. (Countable subadditivity) If \(E_k\in\mcm\) for \(k\geq1\), then \[ \mu\qty{\bigcup_{k\geq 1} E_k} \leq \sum_{k\geq 1} \mu(E_k). \]

    Replacing a sequence of sets by a sequence of disjoint sets

    Let \((E_j)_{j\in \NN}\) be a sequence of subsets of a set \(X\), and put \[ F_j\coloneqq E_j \setminus \bigcup_{k<j} E_k . \] The sets \(F_j\) are pairwise disjoint, \(F_j\subseteq E_j\), and \(\bigcup_{j\le n} F_j=\bigcup_{j\le n} E_j\) for every \(n\), so \(\bigcup_j E_j = \coprod_j F_j\). If each \(E_j\) belongs to a \(\sigma\)-algebra \(\mcm\) on \(X\), then so does each \(F_j\).

    Continuity of measure from below and from above

    For increasing $E_n$, $\mu(E_n)\to\mu(\bigcup E_n)$; for decreasing $E_n$ with $\mu(E_1)<\infty$, $\mu(E_n)\to\mu(\bigcap E_n)$.Suggest a slogan

    Let \((X,\mcm,\mu)\) be a measure space and let \(E_n\in\mcm\) for \(n\geq1\) [@Fol13].

    1. If \(E_1\subseteq E_2\subseteq\cdots\) and \(E\coloneqq\bigcup_{n\geq1}E_n\), then \(\mu(E_n)\to\mu(E)\).

    2. If \(E_1\supseteq E_2\supseteq\cdots\), \(\mu(E_1)<\infty\), and \(E\coloneqq\bigcap_{n\geq1}E_n\), then \(\mu(E_n)\to\mu(E)\).

    The hypothesis \(\mu(E_1)<\infty\) in (b) is needed: for Lebesgue measure \(m\) on \(\RR\) and \(E_n\coloneqq[n,\infty)\), \(m(E_n)=\infty\) for all \(n\) but \(\bigcap_n E_n=\emptyset\).

    State continuity of measure from above and from below.

    Outer measure

    An outer measure \(\mu^*\) on \(X\) is defined on every subset of \(X\), with \(\mu^*(\varnothing)=0\), monotone, and countably subadditive. A set \(A\) is Carathéodory measurable if \(\mu^*(E) = \mu^*(E\cap A)+\mu^*(E\setminus A)\) for every \(E\subseteq X\). By Carathéodory’s theorem, the Carathéodory measurable sets form a \(\sigma\)-algebra on which \(\mu^*\) is countably additive, hence a measure.

    Properties of Lebesgue outer measure

    Lebesgue outer measure is monotone, countably subadditive, outer regular, and additive on sets at positive distance.Suggest a slogan

    Let \(m_*\) be Lebesgue outer measure on \(\RR^n\) [@SS05].

    1. If \(E\subseteq F\subseteq\RR^n\), then \(m_*(E) \leq m_*(F)\).

    2. If \(E_i\subseteq\RR^n\) for \(i\geq1\), then \(m_*\big(\bigcup_{i\geq1} E_{i}\big) \leq \sum_{i\geq1} m_*(E_{i})\).

    3. For every \(E\subseteq\RR^n\) and \(\varepsilon>0\) there exists an open set \(G\supseteq E\) with \(m_*(G) \leq m_*(E) + \varepsilon\).

    4. If \(A,B\subseteq\RR^n\) satisfy \(\dist(A,B)\coloneqq\inf\theset{\abs{a-b} : a\in A,\ b\in B}>0\), then \(m_*(A\cup B) = m_*(A) + m_*(B)\).

    In (d), positive distance is sufficient but not necessary: for \(A\coloneqq[0,1)\) and \(B\coloneqq[1,2]\) in \(\RR\), \(\dist(A,B)=0\) and \(m_*(A\cup B)=2=m_*(A)+m_*(B)\). For disjoint sets in general, additivity fails: there are disjoint \(A,B\subseteq\RR\) with \(m_*(A\cup B)<m_*(A)+m_*(B)\), constructed from a non-measurable Vitali set.

    Lebesgue measurability

    Let \(m_*\) be the outer measure on \(\RR^n\). A set \(E\subseteq\RR^n\) is measurable if for every \(\varepsilon>0\) there exists an open set \(G\supseteq E\) with \(m_*(G\setminus E) < \varepsilon\).

    Carathéodory's criterion for Lebesgue measurability

    A set $E$ is measurable exactly when, for every set $A$, the outer measure of $A$ is additive over $A\cap E$ and $A\cap E^c$.Suggest a slogan

    Let \(m_*\) be the outer measure on \(\RR^n\). A set \(E\subseteq \RR^n\) is Lebesgue measurable if and only if for every \(A\subseteq \RR^n\), \[ m_*(A) = m_*(A\cap E) + m_*(A\cap E^c). \]

    State the Caratheodory characterization of measurability.

    Carathéodory criterion for Lebesgue measurability

    A set $E$ is measurable exactly when, for every set $A$, the outer measure of $A$ is additive over $A\cap E$ and $A\setminus E$.Suggest a slogan

    Let \(m_*\) be the outer measure on \(\RR^n\). A set \(E\subseteq \RR^n\) is Lebesgue measurable if and only if for every \(A\subseteq \RR^n\), \[ m_*(A) = m_*(A\cap E) + m_*(A\setminus E). \]

    What is the Carathéodory criterion for Lebesgue measurability?

    Measures on \(\RR^d\)

    Lebesgue measure \(m\) on \(\RR^d\) is translation invariant, satisfies \(m(aE) = \abs a^d m(E)\) for \(a\in\RR\), and assigns to a rectangle the product of its side lengths. It is complete, and for every Lebesgue measurable \(E\) and \(\varepsilon>0\) there are an open \(U\supseteq E\) and a closed \(F\subseteq E\) with \(m(U\setminus E)<\varepsilon\) and \(m(E\setminus F)<\varepsilon\). For measurable \(E_n\), the set \(\limsup_n E_n\) of points lying in infinitely many \(E_n\) is measurable, and by the Borel–Cantelli lemma it is null when \(\sum_n m(E_n)<\infty\).

    Open sets are countable unions of disjoint intervals or almost disjoint cubes

    Open sets decompose countably into disjoint intervals in one dimension and almost disjoint closed cubes in higher dimensions.Suggest a slogan
    1. Every open set \(U\subseteq\RR\) is a union of countably many pairwise disjoint open intervals, possibly unbounded, and these intervals are uniquely determined by \(U\): they are the connected components of \(U\).

    2. For \(n\geq1\), every open set \(U\subseteq\RR^n\) is a union of countably many pairwise almost disjoint closed cubes [@SS05].

    The decomposition in (b) is not unique: replacing one cube of such a family by the \(2^n\) closed cubes of half its side length obtained by bisecting each of its edges gives another countable family of pairwise almost disjoint closed cubes with union \(U\).

    Translation and dilation of Lebesgue measurable sets

    Lebesgue measure is translation invariant and scales by $\delta^n$ under dilation.Suggest a slogan

    Let \(m\) be Lebesgue measure on \(\RR^n\), let \(E\subseteq\RR^n\) be Lebesgue measurable, let \(h\in\RR^n\), and let \(\delta>0\). Then \(E+h\coloneqq\theset{x+h\suchthat x\in E}\) and \(\delta E\coloneqq\theset{\delta x\suchthat x\in E}\) are Lebesgue measurable, and \[ m(E+h)=m(E), \qquad m(\delta E)=\delta^n\,m(E). \]

    Translation invariance and dilation of the Lebesgue integral

    Let \(f\colon\RR^d\to[0,\infty]\) be Lebesgue measurable, or let \(f\in L^1(\RR^d)\), and let \(h\in\RR^d\) and \(\lambda\in\RR\setminus\theset{0}\). Then \(x\mapsto f(x+h)\) and \(x\mapsto f(\lambda x)\) are measurable, and \[ \int_{\RR^d} f(x+h)\dx = \int_{\RR^d} f(x)\dx, \qquad \int_{\RR^d} f(\lambda x)\dx = \abs{\lambda}^{-d}\int_{\RR^d} f(x)\dx. \]

    We use that for a Lebesgue measurable \(E \subseteq \RR^d\) the sets \(E + h \coloneqq \theset{x + h \suchthat x \in E}\) and \(\lambda E\coloneqq\theset{\lambda x\suchthat x\in E}\) are measurable, with \(m(E + h) = m(E)\) and \(m(\lambda E) = \abs{\lambda}^d m(E)\). Put \(\tau f(x)\coloneqq f(x+h)\) and \(\delta f(x)\coloneqq f(\lambda x)\). For a Borel set \(B\), \(\inverseof{(\tau f)}(B) = \inverseof{f}(B) - h\) and \(\inverseof{(\delta f)}(B) = \lambda^{-1}\inverseof{f}(B)\), so \(\tau f\) and \(\delta f\) are measurable.

    Characteristic functions. For measurable \(E\), \(\tau\chi_E = \chi_{E-h}\) and \(\delta\chi_E = \chi_{\lambda^{-1}E}\), so \[ \int\tau\chi_E = m(E - h) = m(E) = \int\chi_E, \qquad \int\delta\chi_E = m(\lambda^{-1}E) = \abs{\lambda}^{-d}\int\chi_E. \]

    Simple functions. A nonnegative simple function is a finite sum \(\phi = \sum_i c_i \chi_{E_i}\) with \(c_i\geq 0\) and \(E_i\) measurable, and \(\tau\phi = \sum_i c_i\tau\chi_{E_i}\) and \(\delta\phi = \sum_i c_i\delta\chi_{E_i}\). By the case of characteristic functions, \(\int\tau\phi = \int\phi\) and \(\int\delta\phi = \abs{\lambda}^{-d}\int\phi\).

    Nonnegative measurable functions. By the definition of the integral, \(\int f\) is the supremum of \(\int\phi\) over simple \(0 \leq \phi \leq f\). The maps \(\phi\mapsto\tau\phi\) and \(\phi\mapsto\delta\phi\) are bijections from the simple functions \(0\leq\phi\leq f\) onto the simple functions \(0\leq\psi\leq\tau f\) and \(0\leq\psi\leq\delta f\) respectively, with inverses given by translation by \(-h\) and dilation by \(\lambda^{-1}\). They multiply integrals by \(1\) and by \(\abs{\lambda}^{-d}\), so the suprema satisfy \(\int\tau f = \int f\) and \(\int\delta f = \abs{\lambda}^{-d}\int f\).

    Integrable functions. For real-valued \(f\in L^1(\RR^d)\), write \(f = f^+ - f^-\) with \(f^\pm \geq 0\); then \(\tau f = \tau f^+ - \tau f^-\) and \(\delta f = \delta f^+ - \delta f^-\), and the nonnegative case applies to each part. For complex-valued \(f\), apply the real case to \(\operatorname{Re} f\) and \(\operatorname{Im} f\).

    Existence of a non-measurable subset of $\RR$

    Not every subset of the real line is Lebesgue measurable.Suggest a slogan

    There exists a set \(A\subseteq \RR\) that is not Lebesgue measurable.

    Limits superior and inferior of measurable sets are measurable

    Countable unions and intersections keep set limsup and liminf measurable.Suggest a slogan

    Let \((X,\mcm)\) be a measurable space and let \(A_n\in\mcm\) for \(n\geq1\). Then \(\limsup_n A_n\) and \(\liminf_n A_n\), as defined by \(\bigcap_{N\geq1}\bigcup_{n\geq N}A_n\) and \(\bigcup_{N\geq1}\bigcap_{n\geq N}A_n\), belong to \(\mcm\).

    Borel--Cantelli lemma

    If $\sum_k m(E_k)<\infty$, then almost every point lies in only finitely many of the sets $E_k$.Suggest a slogan

    Let \((E_k)_{k\geq1}\) be a sequence of Lebesgue measurable subsets of \(\RR^n\) with \[ \sum_{k=1}^\infty m(E_k) < \infty . \] Then \(m\qty{\limsup_{k\to\infty} E_k} = 0\), where \(\limsup_k E_k\) is the limit superior of the sets \(E_k\); equivalently, almost every \(x\in\RR^n\) lies in only finitely many of the sets \(E_k\).

    The same statement holds for a sequence of measurable sets in any measure space. For a probability measure \(P\) and events \(E_k\) with \(\sum_k P(E_k)<\infty\), it says that with probability \(0\) infinitely many of the events \(E_k\) occur.

    Borel--Cantelli lemma

    Summable event probabilities give finitely many occurrences almost surely; under independence, divergent sums give infinitely many almost surely.Suggest a slogan

    Let \((\Omega,\mcf,P)\) be a probability space, let \((E_n)_{n\geq 1}\) be a sequence of events, and let \(\limsup_n E_n\) be its limit superior.

    1. If \(\sum_{n\geq 1} P(E_n) < \infty\), then \(P(\limsup_n E_n) = 0\).

    2. If the events \(E_n\) are independent and \(\sum_{n\geq 1} P(E_n) = \infty\), then \(P(\limsup_n E_n) = 1\).

    For (1), \(\limsup_n E_n\subseteq\bigcup_{n\geq N}E_n\) for every \(N\), so \(P(\limsup_n E_n)\leq\sum_{n\geq N}P(E_n)\to 0\) as \(N\to\infty\).

    For (2), the complement of \(\limsup_n E_n\) is \(\bigcup_{N\geq 1}\bigcap_{n\geq N}E_n^c\). For \(M\geq N\), independence and \(1-t\leq e^{-t}\) give \[ P\qty{\bigcap_{n=N}^{M}E_n^c} = \prod_{n=N}^{M}(1-P(E_n))\leq\exp\qty{-\sum_{n=N}^{M}P(E_n)}\to 0 \quad (M\to\infty), \] so \(P\qty{\bigcap_{n\geq N}E_n^c} = 0\) for every \(N\), and the countable union of these null events is null.

    What is the Borel-Cantelli lemma?

    First Borel--Cantelli lemma for measures

    Let \((X,\mcm,\mu)\) be a measure space, let \((E_j)_{j\geq 1}\) be a sequence in \(\mcm\) with \(\sum_{j\geq 1}\mu(E_j) < \infty\), and let \(E\coloneqq\limsup_j E_j\) be its limit superior. Then \(E\in\mcm\) and \(\mu(E) = 0\).

    We have \(E = \bigcap_{k\geq 1}\bigcup_{j\geq k} E_j\), a countable intersection of countable unions of sets in \(\mcm\), so \(E\in\mcm\). For every \(k\geq 1\), \(E\subseteq\bigcup_{j\geq k}E_j\), so by countable subadditivity \[ \mu(E)\leq\sum_{j\geq k}\mu(E_j). \] The right side is the tail of a convergent series, so it tends to \(0\) as \(k\to\infty\), and \(\mu(E) = 0\).

    Closure properties of Lebesgue measurable functions

    Measurability is preserved by absolute values, countable suprema, infima, and limits, finite sums and differences, extension to a product space, invertible linear changes of variable, and $(x,y)\mapsto f(x-y)$.Suggest a slogan

    All functions below are Lebesgue measurable functions with values in \([-\infty,\infty]\) unless stated otherwise.

    • For a Lebesgue measurable set \(E\subseteq\RR^d\), the characteristic function \(\chi_E\) is measurable.

    • If \(f_n\colon\RR^d\to[-\infty,\infty]\) are measurable for \(n\geq1\), then \(\abs{f_n}\), \(\sup_n f_n\), \(\inf_n f_n\), \(\limsup_n f_n\), and \(\liminf_n f_n\) are measurable, and so is \(\lim_n f_n\) wherever the sequence converges everywhere.

    • If \(f,g\colon\RR^d\to\RR\) are measurable and finite-valued, then \(f+g\) and \(f-g\) are measurable.

    • If \(f\colon\RR^{d_1}\to[-\infty,\infty]\) is measurable, then \(F\colon\RR^{d_1}\times\RR^{d_2}\to[-\infty,\infty]\), \(F(x,y)\coloneqq f(x)\), is measurable.

    • If \(f\colon\RR^d\to[-\infty,\infty]\) is measurable and \(T\colon\RR^d\to\RR^d\) is an invertible linear map, then \(f\circ T\) is measurable.

    • If \(f\colon\RR^d\to[-\infty,\infty]\) is measurable, then \((x,y)\mapsto f(x-y)\) is measurable on \(\RR^d\times\RR^d\).

    $\sigma$-finite measure spaces

    A measure space \((X, \mcm, \mu)\) is \(\sigma\)-finite if there exist sets \(E_1,E_2,\ldots\in\mcm\) with \(X=\bigcup_{k=1}^\infty E_k\) and \(\mu(E_k)<\infty\) for every \(k\).

    Borel measures finite on balls are outer regular by open sets and inner regular by closed sets

    Borel measures finite on balls can approximate measurable sets from outside by open sets and inside by closed sets.Suggest a slogan

    Let \(X\) be a metric space with Borel \(\sigma\)-algebra \(\mcb\), and let \(\mu\) be a measure on \(\mcb\) with \(\mu(B)<\infty\) for every open ball \(B\) of finite radius. Then for every \(E \in \mcb\) and every \(\varepsilon>0\):

    • there exists an open set \(O\subseteq X\) with \(E \subseteq O\) and \(\mu(O\setminus E) < \varepsilon\);

    • there exists a closed set \(F\subseteq X\) with \(F\subseteq E\) and \(\mu(E\setminus F) < \varepsilon\).

    A \(\sigma\)-finite measure space is a countable union of sets of finite measure, so theorems proved for finite measures, such as the Tonelli and Fubini theorems for product measures, extend to \(\sigma\)-finite measures by exhaustion. For Lebesgue measure, every measurable set of finite measure \(E\) also contains a compact \(K\) with \(m(E\setminus K)<\varepsilon\).

    Volume of a rectangle is additive over almost disjoint decompositions and subadditive over covers

    Rectangle volume is additive over almost disjoint decompositions and subadditive over covers by rectangles.Suggest a slogan

    A closed rectangle in \(\RR^n\) is a product \(R=\prod_{i=1}^n[a_i,b_i]\) with \(a_i\leq b_i\), with volume \(\abs{R}\coloneqq\prod_{i=1}^n(b_i-a_i)\). Let \(R\) and \(R_1,R_2,\ldots\) be closed rectangles in \(\RR^n\), finitely or countably many.

    1. If \(R=\bigcup_j R_j\) and the \(R_j\) are pairwise almost disjoint, then \(\abs{R}=\sum_j\abs{R_j}\).

    2. If \(R\subseteq\bigcup_j R_j\), then \(\abs{R}\leq\sum_j\abs{R_j}\).

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