1 Measure Theory
1.1 Preliminaries
The open ball in \(\mathbb{R}^d\) centered at \(x\) and of radius \(r\) is defined by \[ B_r(x)=\{y\in \mathbb{R}^d :|y−x|<r\}. \] A subset \(E\subset\mathbb{R}^d\) is open if for every \(x\in E\) there exists \(r>0\) with \(B_r(x)\subset E\). By definition, a set is closed if its complement is open.
We note that any (not necessarily countable) union of open sets is open, while in general the intersection of only finitely many open sets is open. A similar statement holds for the class of closed sets, if one interchanges the roles of unions and intersections.
A set \(E\) is bounded if it is contained in some ball of finite radius. A bounded set is compact if it is also closed. Compact sets enjoy the Heine-Borel covering property:
- Assume \(E\) is compact, \(E\subset \bigcup_{\alpha}\cal{O}_{\alpha}\), and each \(\cal{O}_{\alpha}\) is open. Then there are finitely many of the open sets, \(\cal{O}_{\alpha_1},{O}_{\alpha_2},\dots,{O}_{\alpha_N}\), such that \(E\subset \bigcup_{j=1}^n\cal{O}_{\alpha_j}\).
In words, any covering of a compact set by a collection of open sets contains a finite subcovering.
A point \(x\in \mathbb{R}^d\) is a limit point of the set \(E\) if for every \(r>0\),the ball \(B_r(x)\) contains points of \(E\). This means that there are points in \(E\) which are arbitrarily close to \(x\). An isolated point of \(E\) is a point \(x\in E\) such that there exists an \(r > 0\) where \(B_r(x)\cap E\) is equal to \(\{x\}\).
A point \(x\in E\) is an interior point of \(E\) if there exists \(r>0\) such that \(B_r(x)\subset E\). The set of all interior points of E is called the interior of \(E\). Also, the closure \(\overline{E}\) of the \(E\) consists of the union of \(E\) and all its limit points. The boundary of a set \(E\), denoted by \(\partial E\), is the set of points which are in the closure of \(E\) but not in the interior of \(E\).
Note that the closure of a set is a closed set; every point in \(E\) is a limit point of \(E\); and a set is closed if and only if it contains all its limit points. Finally, a closed set \(E\) is perfect if \(E\) does not have any isolated points.
Lemma 1.1 If a rectangle is the almost disjoint union of finitely many other rectangles, say \(R=\bigcup_{k=1}^NR_k\), then \[ |R|=\sum_{k=1}^N |R_k| \]
Lemma 1.2 If \(R,R_1,\dots,R_N\) are rectangles, and \(R\subset\bigcup_{k=1}^NR_k\), then \[ |R|\leq\sum_{k=1}^N |R_k| \]
Proof.
Every open subset \(\cal{O}\) of \(\mathbb{R}\) can be written uniquely as a countable union of disjoint open intervals.
Every open subset \(\cal{O}\) of \(\mathbb{R}^d\), \(d\geq 1\), can be written uniquely as a countable union of almost disjoint closed cubes.
The Cantor set \(\cal{C}\) is by definition the intersection of all \(C_k\)’s: \[ \mathcal{C}=\bigcap_{k=0}^{\infty}C_k. \]
The set \(\cal{C}\) is not empty, since all end-points of the intervals in \(C_k\) (all \(k\)) belong to \(\cal{C}\). Despite its simple construction, the Cantor set enjoys many interesting topological and analytical properties. For instance, \(\cal{C}\) is closed and bounded, hence compact. Also, \(\cal{C}\) is totally disconnected: given any \(x,y\in \cal{C}\) there exists \(z\not\in \cal{C}\) that lies between \(x\) and \(y\). Finally, \(\cal{C}\) is perfect: it has no isolated points.
In terms of cardinality the Cantor set is rather large: it is not countable. Since it can be mapped to the interval \([0,1]\), the Cantor set has the cardinality of the continuum. However, from the point view of “length” the size f \(\cal C\) is small. Roughly speaking, it has length zero.
1.2 The exterior measure
Definition 1.1 If \(E\) is any subset of \(\mathbb{R}^d\), the exterior measure of \(E\) is \[ m_{\ast}(E) = \inf\sum_{j=1}^{\infty}|Q_j|, \] where the infimum is taken over all countable coverings \(E\subset \bigcup_{j=1}^{\infty}Q_j\) by closed cubes. In general we have \(0\leq m_{\ast}(E)\leq \infty\).
- It would not suffice to allow finite sums in the definition of \(m_{\ast}(E)\). The quantity that would be obtained if one considered only coverings of \(E\) by finite unions of cubes is in general larger than \(m_{\ast}(E)\).
- One can, however, replace the coverings by cubes, with coverings by rectangles; or with coverings by balls.
Example 1.1 The exterior measure of a closed cube is equal to its volume.
Proof.
We consider an arbitrary covering \(Q\subset \bigcup_{j=1}^{\infty}Q_j\) by cubes, where \(Q_j\) is closed for \(j=1,2,\dots\). Then it suffices to prove that \[ |Q|\leq \sum_{j=1}^{\infty}|Q_j|. \]Example 1.2 The exterior measure of a open cube is equal to its volume.
Example 1.3 The exterior measure of a rectangle is equal to its volume.
Proof.
Arguing as in the first example, we see that \(|R|\leq m_{\ast}(R)\). To obtain the reverse inequality, consider a grid in \(\mathbb{R}^d\) formed by cubes of side length \(1/k\), then let \(k\) tend to infinity yields \(m_{\ast}(R)\leq |R|\).Example 1.4 The Cantor set \(\cal C\) has exterior measure 0.
1.2.1 Properties of the exterior measure
From the definition we know that: - For every \(\epsilon>0\), there exists a covering \(E\subset \bigcup_{j=1}^{\infty}Q_j\) with \[ \sum_{j=1}^{\infty}m_{\ast}(Q_j)\leq m_{\ast}(E)+\epsilon. \]
Proposition 1.1
- (Monotonicity) If \(E_1\subset E_2\), then \(m_{\ast}(E_1)\leq m_{\ast}(E_2)\).
- (Countatble sub-additivity) If \(E= \bigcup_{j=1}^{\infty}E_j\), then \(m_{\ast}(E)\leq\sum_{j=1}^{\infty}m_{\ast}(E_j)\).
- If \(E\subset\mathbb{R}^d\), then \(m_{\ast}(E)=\inf m_{\ast}(\cal O)\), where the infimum is taken over all open sets \(\cal O\) containing \(E\).
Proof.
It suffices to prove that \(m_{\ast}(E)\geq \inf m_{\ast}(\cal O)\). Notice that for any closed cube \(Q_j\), there exists an open set \(S_j\) which contains \(Q_j\) and such that \(|S_j|\leq (1+\epsilon)|Q_j|\) for a fixed \(\epsilon>0\). - If \(E=E_1\cup E_2\),and \(d(E_1,E_2)>0\), then \[ m_{\ast}(E)= m_{\ast}(E_1)+ m_{\ast}(E_2). \]
- If a set \(E\) is the countable union of almost disjoint cubes \(E=\bigcup_{j=1}^{\infty}Q_j\), then
\[
m_{\ast}(E) = \sum_{j=1}^{\infty}|Q_j|.
\]
Proof.
Consider constructing a family of cubes that are at a finite distance from one another, so we can use proposition 4.
1.3 Measurable sets and the Lebesgue measure
The notion of measurability isolates a collection of subsets in \(\mathbb{R}^d\) for which the exterior measure satisfies all our desired properties, including additivity (and in fact countable additivity) for disjoint unions of sets.
There are a number of different ways of defining measurability, but these all turn out to be equivalent. Probably the simplest and most intuitive is the following: A subset \(E\) of \(\mathbb{R}^d\) is Lebesgue measurable, or simply measurable, if for any \(\epsilon > 0\) there exists an open set \(\cal O\) with \(E\subset\cal O\) and \[ m_{\ast}(\mathcal{O}-E) ≤ \epsilon. \] This should be compared to Proposition 1.1 3, which holds for all sets \(E\). If E is measurable, we define its Lebesgue measure (or measure) \(m(E)\) by \[ m(E)=m_{\ast}(E). \]
Immediately from the definition, we find:
Proposition 1.2
- Every open set in \(\mathbb{R}^d\) is measurable.
- If \(m_{\ast}(E)=0\), then \(E\) is measurable. In particular, if \(F\) is a subset of a set of exterior measure \(0\), then \(F\) is measurable.
- A countable union of measurable sets is measurable.
- Closed sets are measurable.
Proof.
Since any closed set \(F\) can be written as the union of compact sets, say \(F=\bigcup_{k=1}^{\infty}F\cap B_k\), where \(B_k\) denotes the closed ball of radius \(k\) centered at the origin, it suffices to prove that compact sets are measurable. - The complement of a measurable set is measurable.
Proof.
For every positive integer \(n\) we choose an open set \(\cal O_n\) with \(E\subset \cal O_n\) and \(m_{\ast}(\mathcal{O_n}-E)\leq 1/n\). Notice that \[ (E^c-\bigcup_{n=1}^{\infty}\mathcal{O_n^c}) \subset(\mathcal{O_n}-E) \] then \(E^c\) is measurable since \(E^c=(E^c-\bigcup_{n=1}^{\infty}\mathcal{O_n^c})\cup \bigcup_{n=1}^{\infty}\mathcal{O_n^c}\). - A countable intersection of measurable sets is measurable.
To prove the fourth proposition, we need the following lemma.
Lemma 1.3 If \(F\) is closed, \(K\) is compact, and these sets are disjoint, then \(d(F,K)>0\).
Theorem 1.1 If \(E_1,E_2,\dots\), are disjoint measurable sets, and \(E=\bigcup_{j=1}^{\infty}E_j\), then \[ m(E)=\sum_{j=1}^{\infty}m(E_j). \]
Proof.
If \(F_1,F_2,\dots, F_N\), are compact and disjoint, then obviously for any \(j,k,j\ne k\), \(d(F_j,F_k)>0\), so \(m\left(\bigcup_{j=1}^NF_j\right)=\sum_{j=1}^Nm(F_j)\). If each \(E_j\) is bounded, we can choose a closed subset \(F_j\) for \(E_j\) with \(m_{\ast}(E_j-F_j)\leq \epsilon/2^j\) for each \(j\). Then \[ m(E) \geq \sum_{j=1}^Nm(F_j) \ge \sum_{j=1}^Nm(E_j)-\epsilon \] Letting \(N\) tend to infinity, since \(\epsilon\) is arbitrary we find that \[ m(E)\ge\sum_{j=1}^{\infty}m(E_j). \] In the general case, we select any sequence of cubes \(\{Q_k\}_{k=1}^{\infty}\) that increases to \(\mathbb{R}^d\), in the sense that \(Q_k\subset Q_{k+1}\) for all \(k\ge 1\) and \(\bigcup_{k=1}^{\infty}Q_k=\mathbb{R}^d\). We then let \(S_1=Q_1\) and \(S_k=Q_k-Q_{k-1}\) for \(k\ge 2\). If we define measurable sets by \(E_{j,k}=E_j\cap S_k\), then \[ m(E) =\sum_{j,k}m(E_{j,k})=\sum_j\sum_km(E_{j,k})=\sum_jm(E_j). \]Corollary 1.1 Suppose \(E_1,E_2,...\) are measurable subsets of \(\mathbb{R}^d\).
- If \(E_k \nearrow E\), then \(m(E) =\lim_{N\to \infty}m(E_N)\).
- If \(E_k \searrow E\) and \(m(E_k)<\infty\) for some \(k\), then \(m(E) =\lim_{N\to \infty}m(E_N)\).
Theorem 1.2 Suppose \(E\) is measurable subset of \(\mathbb{R}^d\). Then for every \(\epsilon>0\):
- There exists an open set \(\cal O\) with \(E\subset \cal O\) and \(m(\mathcal{O}-E)\leq\epsilon\).
- There exists a closed set \(F\) with \(F\subset E\) and \(m(E-F)\leq\epsilon\).
- If \(m(E)\) is finite, there exists a compact set \(K\) with \(K\subset E\) and \(m(E-K)\leq \epsilon\).
- If \(m(E)\) is finite, there exists a finite union \(F=\bigcup_{j=1}^NQ_j\) of closed cubes such that
\[
m(E\triangle F)\leq \epsilon.
\]
Proof.
Choose a family of closed cubes \(\{Q_j\}_{j=1}^{\infty}\) so that \[ E\subset \bigcup_{j=1}^{\infty}Q_j \text{ and } \sum_{j=1}^{\infty}|Q_j|\le m(E) +\epsilon/2. \] Since \(m(E) <\infty\), then series converges and there exists \(N>0\) such that \(\sum_{j=n+1}^{\infty}|Q_j|\le \epsilon/2\). If \(F=\bigcup_{j=1}^NQ_j\), then \[ \begin{aligned} m(E\triangle F) &= m(E-F)+m(F-E)\\ &\le m\left( \bigcup_{j=1}^{\infty}Q_j - F\right)+m\left( \bigcup_{j=1}^{\infty}Q_j-E\right)\\ &\le m\left( \bigcup_{j=n+1}^{\infty}Q_j\right)+ m\left( \bigcup_{j=n+1}^{\infty}Q_j\right)-m(E)\\ &\le \sum_{j=n+1}^{\infty}|Q_j|+\sum_{j=1}^{\infty}|Q_j|-m(E)\\ &\le \epsilon. \end{aligned} \]
1.3.2 \(\sigma\)-algebra and Borel sets
Borel \(\sigma\)-algebra in \(\mathbb{R}^d\), denoted by \(\cal B_{\mathbb{R}^d}\), is the smallest \(\sigma\)-algebra in \(\mathbb{R}^d\) that contains all open sets. Elements of this \(\sigma\)-algebra are called Borel sets. Since we observe that any intersection (not necessarily countable) of \(\sigma\)-algebra is again a \(\sigma\)-algebra, we may define \(\cal B_{\mathbb{R}^d}\) as the intersection of all \(\sigma\)-algebras that contain the open sets. This shows the existence and uniqueness of the Borel \(\sigma\)-algebra.
Remark. There exists Lebesgue measurable sets that are not Borel sets. (See Exercise ???ref(exr: #35)
1.4 Measurable functions
1.4.1 Definition and basic properties
The starting point is the notion of a characteristic function of a set E, which is defined by \[ \chi_E(x)= \begin{cases} 1& \text{if }x\in E,\\ 0& \text{if }x\not\in E. \end{cases} \] For the Riemann integral it is in effect theclass of step functions that build the blocks of integration theory , with each give as a finite sum
\[ f = \sum_{k=1}^Na_k\chi_{R_k} \] where each \(R_k\) is a rectangle, and the \(a_k\) are constants.
For the lebesgue integral we need a more general notion. A simple function is a finite sum \[ f = \sum_{k=1}^Na_k\chi_{E_k} \] where each \(E_k\) is a measurable set of finite measure, and the \(a_k\) are constants.
A function \(f\) defined on a measurable subset \(E\) of \(\mathbb{R}^d\) is measurable, if for all \(a\in\mathbb{R}\), the set \[ f^{-1}([-\infty,a))=\{x\in E: f(x)<a\} \] is measurable. Note that this definition applies to extended-valued functions, so we use \(f^{-1}([-\infty,a))\) instead of \(f^{-1}((-\infty,a))\).
Proposition 1.3
- The finite-valued function \(f\) is measurable iff \(f^{-1}(\cal O)\) is measurable for every open set \(\cal O\), and iff \(f^{-1}(F)\) is measurable for every closed set \(F\). (Note that this property also applies to extended-valued functions, if we make the additional hypothesis that both \(f^{-1}(\infty)\) and \(f^{-1}(-\infty)\) are measurable sets since \([-\infty,a)=\{-\infty\}\cup\left(\bigcup_{n=1}^{\infty}(-n,a)\right)\).)
- If \(f\) is continuous on \(\mathbb{R}^d\), then \(f\) is measurable. If \(f\) is measurable and finite-valued, and \(\Phi\) is continuous, then \(\Phi\circ f\) is measurable. (Note that it is not true that \(f\circ\Phi\) is measurable. See exercise 35.)
Proof.
\(\Phi\) is continuous, so \(\Phi^{-1}((-\infty,a))\) is open set \(\cal O\). - Suppose \(\{f_n\}_{n=1}^{\infty}\) is a sequence of measurable functions. Then \[ \sup_nf_n(x),\quad \inf_nf_n(x),\quad\limsup_{n\to\infty}f_n(x)\quad \text{and} \quad\liminf_{n\to\infty}f_n(x) \] are measurable.
- If \(\{f_n\}_{n=1}^{\infty}\) is a collection of measurable functions, and \[ \lim_{n\to\infty}f_n(x)=f(x), \] then \(f\) is measurable.
- If \(f\) and \(g\) are measurable, then
- The integer powers \(f^k\), \(k\ge 1\) are measurable.
- \(f+g\) and \(fg\) are measurable if both \(f\) and \(g\) are finite-valued.
- Suppose \(f\) is measurable, and \(f(x)=g(x)\) for a.e. \(x\). Then \(g\) is measurable.
1.4.2 Approximation by simple functions or step functions
Theorem 1.3
Suppose \(f\) is a non-negative measurable function on \(\mathbb{R}^d\). Then there exists an increasing sequence of non-negative simple functions \(\{\varphi_n\}_{n=1}^{\infty}\) that converges pointwise to \(f\), namely, \[ \varphi_k(x)\le\varphi_{k+1}(x) \text{ and } \lim_{k\to \infty}\varphi_k(x)=f(x),\ \text{for all }x. \]Proof.
\(\Phi\) is continuous, so \(\Phi^{-1}((-\infty,a))\) is open set \(\cal O\).Theorem 1.4
Suppose \(f\) is a measurable function on \(\mathbb{R}^d\). Then there exists a sequence of simple functions \(\{\varphi_n\}_{n=1}^{\infty}\) that satisfies \[ |\varphi_k(x)|\le|\varphi_{k+1}(x)| \text{ and } \lim_{k\to \infty}\varphi_k(x)=f(x),\ \text{for all }x. \]Proof.
We use the decomposition of \(f\): \(f(x)=f^{+}(x)-f^{-}(x)\).Theorem 1.5
Suppose \(f\) is measurable on \(\mathbb{R}^d\). Then there exists an sequence of step functions \(\{\psi_n\}_{n=1}^{\infty}\) that converges pointwise to \(f(x)\) for almost every \(x\), \[ \varphi_k(x)\le\varphi_{k+1}(x) \text{ and } \lim_{k\to \infty}\varphi_k(x)=f(x),\ \text{for all }x. \]1.4.3 Littlewood’s three principles
Theorem 1.6 (Egorov) Suppose \(\{f_k\}_{k=1}^{\infty}\) is a sequence of measurable functions defined on a measurable set \(E\) with \(m(E)<\infty\), and assume that \(f_k\to f\) a.e. on \(E\) and \(f_1,f_2,\dots,f_k,f\) are finite valued a.e. on \(E\). Given \(\epsilon>0\), we can find a closed set \(A_{\epsilon}\subset E\) such that \(m(E-A_{\epsilon})\le \epsilon\) and \(f_k\to f\) uniformly on \(A_{\epsilon}\).
Remark. Note that \(f_1,f_2,\dots,f_k,f\) are finite valued a.e. on \(E\). Indeed, if it is not satisfied, then we cannot construct \(\{E_k^n\}_{k=1}^\infty\) such that \(E_k^n\nearrow E\). A counterexample is that \[ f_k(x)=\begin{cases} k&|x|\le k,\\ \infty &|x|>k \end{cases} \] and \(f(x)=\infty\) on \(\mathbb{R}\).
Remark. Note that \(m(E)<\infty\) and it is easy to construct counterexamples when \(m(E)=\infty\). Indeed, if \(m(E)=\infty\), then we cannot find \(k_n\) such that \(m(E-E_{k_n}^n)<1/2^n\) since \(m(E-E_{k_n}^n)=m(E)-m(E_{k_n}^n)\).
Theorem 1.7 (Lusin) Suppose \(f\) is measurable and finite valued a.e. on \(E\) with \(E\) of finite measure. Then for every \(\epsilon>0\) there exists a closed set \(F_{\epsilon}\), with \[ F_{\epsilon}\subset E, \text{ and }m(E-F_{\epsilon})\le \epsilon \] and such that \(f|_{F_{\epsilon}}\) is continuous.
1.5 The Brunn-Minkowski inequality
Remark.
Let \(a,b\ge0\), then \[ \begin{aligned} (a+b)^\gamma&\ge a^\gamma+b^\gamma\text{ if }\gamma\ge 1,\\ (a+b)^\gamma&\le a^\gamma+b^\gamma\text{ if }0<\gamma< 1 \end{aligned} \]Proof.
Let \(f(\gamma)=(1+x)^\gamma-(1+x^\gamma)\), where \(x> 0\). Then \[ \begin{aligned} f^\prime (\gamma)&=(1+x)^\gamma\ln(1+x)-x^\gamma\ln x\\ &=[(1+x)^\gamma-x^\gamma]\ln(1+x)+x^\gamma\ln(1+1/x)>0, \end{aligned} \] notice that \(f(1)=0\), so when \(\gamma\ge 1\), \((1+x)^\gamma\ge(1+x^\gamma)\) and when \(0<\gamma< 1\), \((1+x)^\gamma<(1+x^\gamma)\). With this result, the original inequality is obvious.???ref(exr:19)
Theorem 1.8 Suppose \(A\) and \(B\) are measurable sets in \(\mathbb{R}^d\) and their sum \(A+B\) is also measurable. Then \[ m(A+B)^{1/d}\ge m(A)^{1/d}+m(B)^{1/d}. \]