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# AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUE INTEGRAL

AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUE INTEGRAL MARCO M. PELOSO Contents 1. First elements of measure theory Measure spaces Measures 4 2. Abstract integration theory Measurable
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AN INTRODUCTION TO MEASURE THEORY AND THE LEBESGUE INTEGRAL MARCO M. PELOSO Contents 1. First elements of measure theory Measure spaces Measures 4 2. Abstract integration theory Measurable functions Integration of non-negative functions Integration of complex-valued functions The space L 1 µ The Lebesgue measure on R Outer measures The Lebesgue measure on R Examples The Cantor set and its generalizations Integrals depending on a parameter More on L 1 m The Lebesgue measure in R n Product measure spaces and the Lebesgue integral in R n Product measure spaces Integration on product measure spaces More on the Lebesgue integral in R n Polar coordinates in R n Hausdorff measaures A quick review of submanifolds in R n Hausdorff measures Hausdorff measures in R n The fundamental theorem of calculus, integration by parts and the theorems of Gauss, Green and Stokes The fundamental theorem of calculus and Gauss theorem Extesion to domains with lower regularities Green s Theorem and identities Stokes s theorem 77 References 79 Notes for the course Analisi Matematica 4 per i corsi di Laurea presso Dipartimento di Matematica dell Università di Milano, a.a. 2018/189. The mark indicates the proofs and the parts that are not stricly required for the exam for the a.a. 2018/19. June 22, 2019. ii M. M. PELOSO MEASURE THEORY AND LEBESGUE INTEGRAL 1 In these notes we present a concise introduction to abstract measure theory and to the Lebesgue integral in euclidean spaces. These notes should be considered only as a support for the preparation for the exam, and not as detailed introduction to the subect. 1. First elements of measure theory We would like to indroduce a notion of measure as a function µ that assigns to every subset E of R n a value µe [0, + ] in such a way the following conditions are satisfied: i if E 1,..., E k... are disoint subsets, then µ k E k = k µe k; ii if F is congruent to E, that is, obtained from E by a translation, rotation or reflection, then µf = µe; iii µq = 1, where Q = [0, 1 n denotes the unit cube. Unfortunately, these three conditions are mutually incompatible, as the next example shows. Example 1.1. Assume that n = 1. Let us introduce an equivalent relation in [0, 1 by setting x y if x y is rational. Let N be a subset of [0, 1 containing exactly one element for each equivalent class. 1 Let R = Q [0, 1 and for r R define Then we have: N r = { x + r : x N [0, 1 r } { x + r 1 : x N [1 r, 1 }. a r R N r = [0, 1; b if r s are element of R, then N r N s =. In order to show a, we notice that clearly each N r [0, 1. Next, let x [0, 1 and we show that it belongs to at least one N r. Indeed, let y N be the element that is in the same equivalent class as x. If x y, then x N r, where x y = r R, while if x y, then x N r, where x y + 1 = r R. Thus, x N r for some r R. This proves a. Finally, if x N r N s, then we would have x r or x r + 1 and x s or x s + 1 as distinct elements in N but belonging to the same equivalent class, against the definition of N. Suppose then that µ : PR [0, + ] satisfy i-iii. By i and ii we have that µ {x + r : x N [0, 1 r } + µ {x + r : 1 x N [1 r, 1 } = µn r for every r R. Moreover, by a, i.e. since [0, 1 is the disoint union of the N r and these sets are countably many, by ii and iii we have 1 = µ [0, 1 = µ r R N r = µn. r R µn r = r R But this is impossible, since the right-hand-side equals + if µn 0, or it equals 0 if µn = 0. This example easily generilizes to the case n 1. 1 To obtain this, we need to assume the validity of the axiom of choice. 2 M. M. PELOSO 1.1. Measure spaces. Led by the previous example, we try to define such function µ on a domain which is strictly contained in PR n but still satisfying i-iii. To this end we introduce the following definitions. Definition 1.2. Given a set X we call algebra a non-empty collection A of subsets of X that is closed under finite unions and complements, that is, if the following conditions are satisfied: i if E 1,..., E m A, then m k=1 E k A; ii if E A, then c E A. Notice that, if E A, X = E c E A, so that also = c X A. A non-empty collection A of subsets of X is called a σ-algebra if it is an algebra and it is closed under countable unions. In other words, if the following conditions are satisfied: i if E 1, E 2,..., A, then + k=1 E k A; ii if E A, then c E A. = c + Notice that a σ-algebra A is closed under countable intersections, since + E c E A. We observe that an algebra A that is closed under countable unions of disoint subsets of X, is a σ-algebra. Indeed, given a sequence {E k } of elements in A, define F 1 = E 1, F k = E k \ k 1 E k 1 = E c k E. 1 Then the F k are disoint and + k=1 E k = + k=1 F k, which belongs to A by assumption. Example 1.3. Let X be any set. 1 PX is a σ-algebra; 2 if {, X } is a σ-algebra; 3 if X is uncountable and we set A = { E : E is countable, or c E is countable 2}, then A is a σ-algebra. It is easy to see that the intersection of σ-algebras is again a σ-algebra. following definition makes sense. Therefore, the Definition 1.4. Given any subset E of PX, we call ME the σ-algebra generated by E as the smallest σ-algebra containing E, that is, the intersection of all σ-algebras containing E. Observe that PX is a σ-algebra containing E, so the above intersection is not empty. The following lemma is elementary, but it deserves its own statement. Lemma 1.5. If A is a σ-algebra, and E MF A, then ME MF. Proof. Since MF is a σ-algebra containing E, it contains the smallest σ-algebra containing E, i.e. ME. The above definition leads us to the following fundamental notion. Definition 1.6. If X is a topological space, we call the Borel σ-algebra in X, and we denote it by B X, the σ-algebra generated by the collections of open sets in X, and a set E B X a Borel set in X. 2 In this case we say that E is co-countable. MEASURE THEORY AND LEBESGUE INTEGRAL 3 Observe that if F is a closed set, then F B X, so are countable unions of closed sets, their complements, countable unions of such sets, etc. We call a countable intersection of open sets a G δ -set, and countable union of closed sets an F σ -set. If X = R is endowed with its natural topology, we denote by B R the σ-algebra of Borel sets in R with respect to this topology. It is going to play a fundamental role in remeinder of these notes. The following are elementary consequences of the definition. Proposition 1.7. The σ-algebra B R is generated by each of the following sets: 1 the collection of open intervals E 1 = { a, b : a b } ; 2 the collection of closed intervals E 2 = { [a, b] : a b } ; 3 the collection of half-open intervals E 3 = { a, b] : a b }, or E 3 = { [a, b : a b } ; 4 the collection of open rays E 4 = {, b : b R }, or E 4 = { a, + : a R } ; 5 the collection of closed rays E 5 = {, b] : b R }, or E 5 = { [a, + : a R }. Proof. The elements of the E and E are all open, or closed, or intersection of open and closed sets, so that each of the σ-algebras in 1-5 is contained in B R. Conversely, it is clear that ME 1 B R since each open set is countable union of elements of E 1. In all remaining cases, it suffices to show that open sets are contained in ME or ME. By symmetry, we consider only the cases of ME, = 1,..., 5. Clearly, a, b = + [ k=1 a 1 k, b 1 ] k, so 2 follows. Next, a, b = + k=n a, b 1 k ], where N is chosen so that a b 1 N. This shows 3. To prove 4 we consider [b, + = c, b and, a [b, + = [b, a, if b a. Thus, 4 follows from the case 3 for E 3. Case 5 is analogous and left to the reader. A similar description can be given for the σ-algebra of Borel sets in R n. If X 1,..., X n is a collection of non-empty sets, X = n X, we denote by π : X X the coordinate maps. If M is a σ-algebra on X, = 1,..., n, we set n {π 1 M = M E : E M, = 1,..., n } that is, the σ-algebra generated by π 1 E where E M, = 1,..., n. Proposition 1.8. We have that B R n = n B R. Proof. Let E B R, = 1,..., n. Then π1 1 E 1 = E 1 R R, and similarly for π 1 E. Then n E = n π 1 E. Hence, n B R contains the σ-algebra generated by { n E : E open in R } =: F, i.e. n B R MF. On the other hand, for each = 1,..., n, the set { E R : π 1 E McF } is a σ-algebra, that contains B R. Hence, n { n B R = M E : E open in R }. This implies that n B R B R n. 4 M. M. PELOSO To see the reverse inclusion, let E be any open set in R n and consider its points with rational coordinates, which is a dense subset in E. Take a ngbh, contained in E, of any such point that is a cartesian product of open sets in R in each coordinate. Then E is union of open sets that are of the form n E with E open in R. Thus, { n B R n M E : E open in R } n = B R. This proves the proposition Measures. Let X be a set and M a σ-algebra in PX. The pair X, M is called a measurable space. Definition 1.9. Given a measurable space X, M, a function µ : M [0, + ] is called a measure on X if i µ = 0; + ii if E 1, E 2, M, are disoint, then µ E = µe. The triple X, M, µ is called a measure space. Property ii is called countable additivity, in contrast with an analogous property m ii if E 1, E 2,..., E m M, are disoint, then µ E = m µe, which will be called finite additivity. Notice that a measure µ satisfies also ii since we may take E k = for all k 1. Given a measure space X, M, µ, µ is said to be finite if µx + and is said to be σ-finite if there exists a collection of sets {E }, with E M, + E = X and such that µe +. In other words, µ is said to be σ-finite if X is countable union of sets of finite measure. Our treatment will essentially concern only with σ-finite measures. Here are some simple examples of measure spaces. Example Let X be any set, and let M = PX. Define the counting measure on X by setting + if E is infinite µe = m if E contains exactly m elements 0 if E =. Then µ is a measure. Notice that µ{x} = 1 for all x X, and that it is σ-finite if X is countable, and it is finite if X is finite. 2 A particular case of 1 is when X is countable, or more in particular, equals N or Z. 3 Instead, a generalization of 1 is the following. Let f : X [0, + ] be given and define µe = x E fx. Since fx is non-negative, there is no ambiguity in the above definition even if E is uncountable. For, if there exist uncountably many x E such that fx 0, then there exists n N for which E n = { x E : fx 1/n } is infinite. Then, x E fx MEASURE THEORY AND LEBESGUE INTEGRAL 5 x E n fx = +. If there exist at most countably many x E such that fx 0, then the sum becomes a series, and the notion of convergence is the standard one. It is easy to chech again that such µ is a measure. A particular case is when f is the function that equals 1 at a given x 0 X and it is 0 anywhere else. Such measure, is called the Dirac delta at x 0. 4 if X is uncountable, let A be the σ-algebra of countable or co-countable sets. Define { 0 if E is countable µe = 1 if E is co-countable. Again, it is easy to check that such µ is a measure on A. The first elementary properties of measures are given in the next result. Proposition Let X, M, µ be a measure space. Then, the following properties hold true. i Monotonicity: If E, F M with E F, then µe µf. ii Subadditivity: If {E } M, then + µ E µe. iii Continuity from below: If {E } M and E 1 E 2, then + lim µe = µ E. + iv Continuity from above: If {E } M and E 1 E 2 and µe 1 + then Proof. i Since F E, we have that + lim µe = µ E. + µf = µ F \ E F E = µ F \ E E = µf \ E + µe µe. ii Given {E }, we define {F } as in 1, that is, we set F 1 = E 1, F k = E k \ k 1 E for k 2. Since the {F } are disoint and + F = + E, by i we have that + µ + E = µ F = µf µe. iii Let now {E } M and E 1 E 2. Setting E 0 :=, we have that the sets E \ E 1 are disoint and + E \ E 1 = + E. Then, + n µ E = µ E \ E 1 = lim µ E \ E 1 = lim µe n. =0 =0 6 M. M. PELOSO iv Set F = E 1 \ E. Then F 1 F 2, and µe 1 = µf + µe, since E and F are disoint, and + F = E 1 \ + E. Finally, using iii we have + µe 1 = µ + E + lim µf = µ E + lim µe 1 µe. + + Since µe 1 +, we can substract it from both side of the equation to obtain iv. We remark that the assumption µe 1 + could be replaced by µe n + for some n, but the finiteness of some µe n is necessary. For, consider the case of N, PN, µ, where µ is the counting measure, and the sets E = {n : n }. Then, µe = + for all but µ + E = µ = 0. Let X, M, µ be a measure space. A very important class of sets, is the class of sets of measure zero, also called null sets, that is, the sets E such that µe = 0. By countable subadditivity, a countable union of null sets is again a null set. By monotonicity, if E M, F E, F M and µe = 0, then also µf = 0. But, in general, it is not true that F M. Definition A measure space X, M, µ is said to be complete if M contains all subsets of null sets, that is, for every E M with µe = 0 and F E, we have F M. The next result shows that any measure space can be extended to a complete measure space. Theorem Let X, M, µ be a measure space and let M = M N = { Ẽ : Ẽ = E N with E M, N N }, where N = { N PX : there exists F M with N F and µf = 0 }. If E N M, with E M and N N, define µ E N = µe. Then X, M, µ is a complete measure space, called the completion of X, M, µ. Observe that the completion of X, M, µ is uniquely determined, in the sense that once M is constructed, there exists a unique measure µ on M that is complete and restricted to M coincides with µ. Proof. We begin by showing that M is a σ-algebra. If {E } M, then E = E N, where E M and N N, for all. Since N N, there exists F M such that N F and µf = 0. Therefore, µf = 0. Therefore, + N + F =: F, where µf + + N N. Hence, + E = + E N = + + E N M N = M. Next, let E M and N N. We need to show that c E N M. Then there exists F M such that N F and µf = 0. By possibly replacing N by N \ E, we may assume that E N =, and then, replacing F by F c E, also that E F =. In this case, E N = E F c F N and 3 MEASURE THEORY AND LEBESGUE INTEGRAL 7 c E N = c E F F \ N. Notice that c E F M, while F \ N F, with µf = 0, so that F \ N N. This shows that M is a σ-algebra. We now show that µ is well defined and is a complete measure. If E 1 N 1 = E 2 N 2 with E M and N N, with N F and µf = 0, = 1, 2, then E 1 E 2 F 2 and µe 1 µe 2 + µf 2 = µe 2. Thus, µe 1 µe 2. Arguing in the same way we obtain the reverse inequality so that µe 1 = µe 2. Therefore, µ is well defined. Next, let {E N } be a sequence of disoint sets in M. If F M, N F and µf = 0, then µ E N = µ E + N = µ E = It follows that µ is a measure, and it is clear that it is complete. µe = µe N. 2. Abstract integration theory In order to begin our approach to the theory of integration, we need to discuss the notion of measurable functions Measurable functions. In analogy with the definition of continuous functions, as morphisms between topological spaces, we have the following definition. Definition 2.1. Let X, M, Y, N be measurable spaces, and f : X Y be given. We say that f is M, N -measurable if f 1 E M for all E N. It is clear that if f : X Y is M, N -measurable and g : Y Z is N, O-measurable, then g f : X Z is M, O-measurable. Also, observe that if N is a σ-algebra in PY, then { f 1 E : E N } is a σ-algebra in PX. In fact, f 1 E = f 1 E and c f 1 E = f 1 c E, and the conclusion follows. Proposition 2.2. Let f : X Y be M, N -measurable and suppose that N is generated by E. Then f is M, N -measurable if and only if f 1 E M for all E E. Hence, if X, Y are topological spaces, and f : X Y continuous, then f is B X, B Y - measurable. Proof. First of all, we observe that the implication only if is trivial. Conversely, suppose that f 1 E M for all E E. It is easy to check that { E Y : f 1 E M } is a σ-algebra in PY that contains E. Hence, it contains the σ-algebra generated by E, that is, it contains N and f is M, N -measurable. The second conclusion now is obvious. 3 Indeed, recall that A \ B = A c B, then, since E F =, E N = E F \ F \ N = E F c F c N = E F c F N. 8 M. M. PELOSO Definition 2.3. If X, M is a measurable space, then f : X R is said to be measurable if it is M, B R -measurable. It is however convenient to consider functions that take value in the extended reals, that is, in { R = R {± } = } [, + ]. In this case, the σ-algebra of Borel sets B R is defined as E R : E R BR. Then we say that f : X R is measurable if it is M, BR -measurable. Proposition 2.4. Let X, M be a measurable space, and let f : X R be given. Then the following conditions are equivalent. i f is measurable, that is, it is M, B R -measurable. ii f 1 [, b M for every b R. iii f 1 [, b] M for every b R. iv f 1 a, + ] M for every a R. v f 1 [a, + ] M for every a R. Proof. If f : X R, the conclusions follow at once from Prop. s 1.7 and 2.2. If f takes values in [, + ], the conclusions are a simple consequence of Def The following simple result will be needed later on. Lemma 2.5. Let X, M be a measurable space, and let f : X C be given. measurable if and only if Re f and Im f are measurable. Then f is Proof. We identify f with f 1, f 2 : X R 2. Observe that the coordinate maps π : R 2 R, = 1, 2 are measurable. Since composition of measurable functions is measurable and f 1 = Re f = π 1 f, f 2 = Im f = π 2 f, f measurable implies Re f, Im f are measurable. Conversely, suppose Re f and Im f are measurable. By Prop. 1.8 we know that B C = B R 2 = B R B R. Then, if E B C, E = E 1 E 2, with E 1, E 2 B R and f 1 E = f1 1 E 1 f2 1 E 2 that is in M by assumption. We remark that when we require that f takes values in R we include the case of f having finite values. Theorem 2.6. Let X, M be a measurable space. i If f, g : X R are measurable, then f + g, fg are measurable. ii If f : X R are measurable, = 1, 2,..., then are measurable. g 1 = sup f, g 2 = inf iii f, g : X R are measurable, then are measurable. f, g 3 = lim sup f, + maxf, g, minf, g g 4 = lim inf + f iv If f : X C are measurable, = 1, 2,..., and gx = lim + f x exists, then g is measurable. Proof. i Observe that f + g = S F, where F x = fx, gx and S : C C C is the sum-function, i.e. Sz + w = z + w. Arguing as in Lemma 2.5 we see that F is measurable. MEASURE THEORY AND LEBESGUE INTEGRAL 9 Since S is continuous is measurable, and then f + g is measurable. Replacing S by P, where P z, w = zw, we obtain the measurability of fg. ii Notice that while g 1 1 { a, + ] = x X : sup f x a } = + f 1 a, + ], g2 1 { [, b = x X : inf f x b } + = f 1 [, b. The measurability of g 1 and g 2 follows from Prop Next, we observe that 4 g 3 x = lim sup f x = inf + k sup f x k Then h k x = sup k f x are measurable, so is inf k h k x = g 3 x. The argument for g 4 is analogous, since g 4 x = lim inf f x = sup inf f x. + k k Finally, iii and iv are trivial consequences of ii Corollary 2.7. If f : X R is measurable, then f + = maxf, 0 and f = minf, 0 are measurable. If g : X C is measurable, then g and sgn g := g/ g are measurable. We point out that f = f + f, f = f + + f, and g = sgn g g. Proof. We only need to prove the statements for g and sgn g. But these are elementary and we leave the details to the reader. Definition 2.8. Let X, M be a measurable space, and let E M. We define the characteristic function of E as the function { 1 if x E χ E x = 0 otherwise. We call a simple function a finite linear combination with complex coefficients of characteristic funcitons of measurable sets E n fx = c χ E x. 2 Observe that simple functions can characterized as the measurable functions whose range is finite that is, they attain at most finitely many distinct values. We remark that the representation 2 of a simple function f is not unique, but it becomes unique if we write m fx = d χ F x. 4 We recall that, given a sequence {an}, lim sup n a n = s, where s is sup of the limit points of {a n}. Then, s = inf k sup n k a n = lim k sup n k a n. Analogously, if we set s = lim inf n a n, then s = sup k inf n k a n = lim k inf n k a n. 10 M. M. PELOSO where F = f 1 {d }, d 0, and { d 1,..., d m } = f X \ {0} is the range of f taken away the value 0. Notice also that finite sums and products of simple
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