x = (x 1,..., x d ), y = (y 1,..., y d ) x + y = (x 1 + y 1,..., x d + y d ), x = (x 1,..., x d ), t R tx = (tx 1,..., tx d ).


 Erik Bradley
 2 years ago
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1 1. Lebesgue Measure 2. Outer Measure 3. Radon Measures on R 4. Measurable Functions 5. Integration 6. CoinTossing Measure 7. Product Measures 8. Change of Coordinates 1. Lebesgue Measure Throughout R d is euclidean dspace, i.e. the real vector space of dtuples x = (x 1,..., x d ) with real entries x i, i = 1,..., d. Recall that x, y in R d can be added and scalar multiplied by t R, x = (x 1,..., x d ), y = (y 1,..., y d ) x + y = (x 1 + y 1,..., x d + y d ), x = (x 1,..., x d ), t R tx = (tx 1,..., tx d ). For A R d and a R d, the set A + a = {x + a : x A} is called the translate of A by a or more simply a translate of A. For t 0 the set ta = {tx : x A} is called the dilate of A by t or more simply a dilate of A. The set A = { x : x A} is the reflection of A about the origin. The operations of translation, reflection, and dilation, along with rotation, are key aspects of the geometry of R d. The unit cube is the set Q = Q(0, 1) = {x R d : 0 x i 1 for all 1 i d}. A cube in R d is any translate of any dilate of the unit cube. Thus a cube is a set of the form Q = Q(a, r) = {x R d : a i x i a i + r for all 1 i d} for some a R d and r 0. The number r 0 is the sidelength. Given Q = Q(a, r), we set Q = r d. In particular a single point is a cube. It is convenient to consider the empty set a cube, the empty cube. More generally, given a = (a 1, a 2,..., a d ), b = (b 1, b 2,..., b d ) in R d with a i b i, i = 1,..., d, a rectangle in R d is a set of the form R(a, b) = {x : a i x i b i for all 1 i d}. Of course every cube is a rectangle; in particular the empty set is a rectangle, the empty rectangle. We shall also allow rectangles to be unbounded, i.e. we allow a i = and b i = in the definition of R(a, b), in which case the inequalities a i x i b i appearing in the definition of R(a, b) are replaced by < x i b i, a i x i <, or < x i <, as the case may be, for each i = 1,..., d. Note the definition also allows degenerate rectangles; these are obtained by choosing a i = b i for at least one i = 1,..., d. Usually our cubes and rectangles are closed. However sometimes we will need open rectangles or cubes; these are obtained by using strict inequalities in the definitions. For 1
2 2 any open or closed rectangle, we define R, the volume of R, to be the product of b i a i over 1 i d. For example R d = and φ = 0. Unless specified explicitly, when we say rectangle or cube we always mean closed. In computing the volume R = d i=1 (b i a i ) of a degenerate unbounded rectangle R, we have to multiply 0 times, since degeneracy means b i a i = 0 for some i and unboundedness means b j a j = for some j, 1 i, j d. Because of this and many other occurrences, throughout we adopt the following convention: 0 = 0. Then the volume of a degenerate rectangle, i.e. a rectangle at least one of whose sides reduces to a point, is zero, even when the rectangle is unbounded. Problem 1.1. Show that the intersection of two rectangles is a rectangle. Show that if R 1 R 2 then R 1 R 2. Definition. Let A R d. The Lebesgue measure of A, or the ddimensional Lebesgue measure of A, or simply the measure of A, is the nonnegative number given by { A = inf R n : A } R n. Thus to obtain the measure of A, you cover A by a countable number of rectangles as efficiently as possible and you add up the volumes of the rectangles. Below (Theorem 1.8) we show that R = R for every rectangle, establishing that Lebesgue measure gives the correct answer for rectangles. To distinguish R and R for now, we call R the volume. Remark. Since {R, φ, φ,... } is a cover of R, it follows immediately from the definition that R R for all rectangles R. From this it follows that the Lebesgue measure of a degenerate rectangle is zero. In general, a cover of a set A is a collection of sets {B α } such that A α B α, i.e. A is contained in their union. A cover is countable if there are countably many sets {B α } = {B 1, B 2,... }. Problem 1.2. Let A R d. Show that A + a = A, A = A, and ta = t d A for t 0. Problem 1.3. Let A be the diagonal segment x 2 = x 1, 0 x 1 1, in R 2. Show that A = 0. Let denote the triangle in R 2 given by = {x R 2 : 0 x 1 1, 0 cx 1 x 2 c}. Here c > 0. Show that c/2. Let P denote the parallelogram in R 2 given by P = {x R 2 : 0 x 2 1, cx 2 x 1 cx 2 + 1}. Show that P 1. At the end of this section we show that T (A) = A for every rotation T : R d R d. We say then that Lebesgue measure is invariant under rotations, reflections, and translations i.e. euclidean motions. A hyperplane is a subset of R d of the form {x : n x = t}. Here and throughout n x = n 1 x 1 + +n d x d is the euclidean inner product of n and x in R d. Thus a hyperplane
3 depends on its unit normal n = (n 1,..., n d ) and its distance t 0 to the origin. A coordinate hyperplane is the hyperplane obtained by choosing n = e i = (0,..., 0, 1, 0,..., 0) (1 in the ith slot), i = 1,..., d, and t = 0. There are d coordinate hyperplanes in R d. We note that each coordinate hyperplane is a rectangle, since the ith coordinate hyperplane is the rectangle R(a, b) where a = (,...,, 0,,..., ), b = (,...,, 0,,..., ), with the zero appearing in the ith slot. Thus the volume of every coordinate hyperplane in R d is zero and hence so is its Lebesgue measure by the above remark. It follows from Problem 1.2 that if R = {x : x i = e i x = t} is a translate of a coordinate hyperplane then R = 0. Proposition 1.4. Lebesgue measure satisfies: (1) φ = 0; (2) monotonicity: A B A B ; (3) subadditivity: A A n A A n. Proof. Since φ Q(0, r) for all r we have φ r d by definition; letting r 0, (1) follows. If A B, then any cover of B is a cover of A and so A R n for any countable cover {R n } of B. Since B is the infimum of these summations over all possible covers of B, (2) follows. If the series in (3) is infinite, then (3) is true. So assume the series is finite. Then A n < for all n. By definition for all ɛ > 0 we can find a cover {R nm } of A n satisfying m=1 R nm < A n + ɛ2 n. Since {R nm, n, m = 1,... } is a cover of A we obtain A R nm < A n + ɛ. m=1 3 Since ɛ > 0 is arbitrary, (3) follows. Remark. Let R be an open rectangle and let R = R R c be its boundary. Then by monotonicity R R. Moreover there are finitely many translates of coordinate hyperplanes R 1, R 2,..., R N such that R R 1 R 2... R N. Thus by subadditivity R R R N = 0. Also R R R so by subadditivity R R + R = R + 0 = R. Hence R = 0 and R = R for all open rectangles R. Throughout x = x 2 = x x x2 d = x x is the euclidean norm of a point x = (x 1, x 2,..., x d ) R d, and d(x, y) = x y is the distance between x and y. For A R d, let diam A = sup{ x y : x, y A}. By definition we take diam φ = 0. Problem 1.5. Show that diam A = diam A for A R d. Problem 1.6. Let A R. Show that A diam A. A collection of rectangles {R α } is almost disjoint if R α R β lies in a hyperplane for all α β. For example a point a R d is integral if the coordinates of a are integers; the set of integral points is denoted Z d ; then the collection {Q(a, 1) : a Z d } is almost disjoint.
4 4 Proposition 1.7. Let R, R 1,..., R N, be bounded rectangles. If R R 1... R N then R R R N. If the rectangles R 1,..., R N are almost disjoint and their union equals R, then R = R R N. Proof. 1. If any R n is contained in some R m we can discard it, reducing the sum R R N but still covering R, so we can assume no R n is contained wholly in any other R m. 2. We prove this by induction on the dimension d. So assume we are in one dimension, i.e. R = [a, b] and R n = [a n, b n ] are intervals in R. Arrange things so that a 1 a Then a 1 a b b N. We now produce a new set of intervals J n = [c n, d n ], n = 1,..., N, such that (1) N J n R, (2) N R n N J n, (3) R n = J n, n = 1,..., N, if the R n s are almost disjoint, (4) the J n s are almost disjoint. To this end, note that we must have a 2 b 1 b 2 ; otherwise the R n s don t cover R or R 2 R 1. Define R 1 = [a 1, a 2 ] and R n = R n for n > 1. Then R n satisfy (1), (2), (3), and R 1 is almost disjoint from all the other R n s. Now repeat this procedure with R 2 replacing it by a smaller R 2 yielding R n satisfying (1), (2), (3) and R 1, R 2 are almost disjoint from all other R n and each other. Continuing in this manner at most N times we obtain the J n s. Hence R = b a d N c 1 = (d 1 c 1 ) + (d 2 c 2 ) + + (d N c N ) = J J N R R N. Note also all the inequalities in the last calculation are equalities if the R n s are almost disjoint and their union equals R. This establishes the result in dimension one. 3. Now assume the result is true in dimension d 1 and let s prove it for d. Let {R n } be a finite cover of R and suppose R n = I n R n and R = I R where R and R n are rectangles in R d 1 and I = [a 0, b 0 ], I n = [a n, b n ] are intervals in R. Then (1) {I n } must cover I; (2) { R n } must cover R; (3) if {R n } is an almost disjoint cover of R then {I n } is an almost disjoint cover of I and { R n } is an almost disjoint cover of R. If for some n = 0, 1,..., N the point a n lies in the interior of I m, m = 1,..., N, then perform the following operation: Replace R m by the two closed rectangles R m = [a m, a n ] R m and R m = [a n, b m ] R m. By doing this we obtain a new cover of R such that the sum of volumes of the old cover equals the sum of the volumes of the new cover, since R m = (b m a m ) R m = (bm a n ) R m + (an a m ) R m = R m + R m. Repeating this process over all n, m we obtain a sequence c 0 < c 1 < c 2 < < c M such that I [c 0, c M ] and if R n = I n R n is one of the rectangles in the cover, we have
5 I n = [c i 1, c i ] for some i. Note that the new cover we obtain is the same as the old cover if the old cover were almost disjoint and the union equalled R. Now by the induction hypothesis for each i (here the sums are over all n satisfying I n = [c i 1, c i ]) (c i c i 1 ) R (c i c i 1 ) R n = I n =[c i 1,c i ] [c i 1, c i ] R n = R n. I n =[c i 1,c i ] I n =[c i 1,c i ] Summing this inequality over i = 1,..., M, we obtain 5 R = (b 0 a 0 ) R (c M c 0 ) R = M N R n = R n. i=1 I n =[c i 1,c i ] M (c i c i 1 ) R i=1 Note also all inequalities in the last two calculations are equalities if the cover {R n } is almost disjoint and R 1... R N = R. Theorem 1.8. For any open or closed rectangle R, R = R. Proof. 1. Assume first R is closed and R <. We already know R R, so it enough to show R R. Given ɛ > 0 there are rectangles R n, n 1, such that R R n and R n < R + ɛ. Given θ > 1 let Rn θ be an open rectangle containing R n such that Rn θ < θ R n, n 1. Since R is compact and {Rn} θ is an open cover, there is a finite subcover Rn, θ n = 1,..., N. By Proposition 1.7 applied to R and the closures of the rectangles Rn, θ n = 1,..., N, we obtain R < θ N R θ n N R n θ < θ( R + ɛ). R n Since the numbers R and R do not depend on ɛ and θ letting ɛ 0 and θ 1 yields R R and hence the result for R closed. 2. If R is closed and R =, choose bounded rectangles R n, n 1, such that R n = R and R n. Then by monotonicity R n = R n R for all n 1 and so R =.
6 6 3. If R is open then R = R. By the remark after Proposition 1.4, we have R = R. The result follows. Problem 1.9. Show that any rectangle R is the union of an almost disjoint countable family of cubes Q n, n 1, satisfying R = Q n. Problem Show that { A = inf Q n : A Here Q n denotes a cube. Problem Let A R. Show that { A = inf diam B n : A } Q n. } B n. Here the sets B n need not be rectangles but may be arbitrary. Problem Let denote the triangle in R 2 given by = {x R 2 : 0 x 1 1, 0 cx 1 x 2 c}. Here c > 0. Show that = c/2 (what it should be use reflection, translation and subadditivity). Let P denote the parallelogram in R 2 given by P = {x R 2 : 0 x 2 1, cx 2 x 1 cx 2 + 1}. Show that P = 1 (what it should be). We turn to the rotationinvariance of Lebesgue measure. To this end, we recall a basic fact from linear algebra: Every invertible linear transformation T : R d R d can be expressed as a product T = T 1 T 2... T N of finitely many linear transformations T i, each T i one of the following three types: for some 1 j d and t 0, for some c R, T (x 1,..., x j,..., x d ) = (x 1,..., tx j,..., x d ) T (x 1, x 2,..., x j,..., x d ) = (x 1 + cx 2, x 2,..., x j,..., x d ) T (x 1,..., x j,..., x k,..., x d ) = (x 1,..., x k,..., x j,..., x d ) for some 1 j < k d. Let us call any transformation of one of the above types basic. Basic transformations of the second and third types are defined only if d 2. If d = 1 then every invertible linear transformation is basic of the first type. The above linear algebra fact is simply a reflection of the fact that any invertible matrix T can be reduced by elementary row operations to the identity matrix.
7 Problem Show that the inverse of a basic transformation is basic. Problem Let d 2 and let T be a basic transformation of the second type. Show that T (R) = R for every rectangle R. Hint  Start in dimension 2 and use Problem Theorem Let T : R d R d be any invertible linear transformation. Then T (A) = det(t ) A for all A R d. Proof. 1. We begin by noting T (R) = det(t ) R for each rectangle R and basic transformation T. For a basic transformation of the first and third type T (R) is a rectangle so we can use Theorem 1.8 to check this. For a basic transformation T of the second type this is Problem 1.14, since det(t ) = Suppose A is arbitrary and {R n } is a countable cover of A. Then T (A) T (R n) and so T (A) T (R n ) det(t ) R n. Taking the infimum over all covers of A by rectangles yields T (A) det(t ) A for T basic and A arbitrary. 3. Now suppose T is basic; then since T 1 is also basic, by step 2 we have T (A) det(t ) A, T 1 (A) det(t 1 ) A ; replacing A by T (A) in the second inequality yields A det(t ) 1 T (A) or T (A) det(t ) A ; combining this with the first inequality yields T (A) = det(t ) A for T basic and A arbitrary. 4. If the result is true for T and for S, then T S(A) = T (S(A)) = det(t ) S(A) = det(t ) det(s) A = det(t S) A and so the result is true for T S. Since any invertible T is a product T 1 T 2... T N with T i basic and by step 3 the result is true for each T i, we obtain the result for any invertible T. A rotation is a linear T : R d R d satisfying T T t = I. In particular such a T satisfies det(t ) = 1 and hence is invertible. Corollary Lebesgue measure is rotationinvariant: T (A) = A for all A and rotations T. Problem Show that the measure of any hyperplane in R d is zero. Problem Let T : R d R d be a linear transformation that is not invertible. Show that T (A) = det(t ) A for all A R d. Because of Theorem 1.15, a linear T : R d R d satisfying det(t ) = 1 is called volumepreserving. We close the section with some problems which will be useful later on. To this end recall that an open ball in R d is a set of the form {y : y a < r}; r > 0 is the radius and a R d is the center. The closure of the open ball {y : y a < r} is the closed ball B(a, r) = {y : y x r}. The unit ball is the set B(0, 1). Since B(0, 1) contains a nondegenerate rectangle and is contained in a bounded rectangle, we know 0 < B(0, 1) <. We denote B(0, 1) by α(d) and we compute α(d) in 7. 7
8 8 Problem Show that every set A R d is contained in a ball of radius diam A. Conclude that A α(d)(diam A) d. Problem Suppose f : R d R d satisfies f(x) f(y) C x y and let Q be a cube. Show that f(q) α(d)c d d d/2 Q. Show that A R d and A = 0 implies f(a) = 0. Remark. The first part of Problem 1.20 does not hold for a rectangle since the volume of a rectangle R can be related to its diameter only when R is a cube. Problem Suppose A R d has ddimensional Lebesgue measure zero. Then A R R d+1 has (d + 1)dimensional Lebesgue measure zero. Using the euclidean norm = 2 one can easily describe balls; to describe cubes easily, however, one has to resort to the sup norm x = max( x 1,..., x d ). Indeed one can easily check that the set {x : x c r} is the cube Q(a, 2r) with a i = c i r, i = 1,..., d. Thus {x : x c r} is the cube with center c and sidelength 2r. Problem Let f : R d R d be Lipschitz relative to the sup norm, f(x) f(y) C x y. Then f(a) C d A for A R d (Start with A a cube). When f : R d R d is Lipschitz relative to the euclidean norm, f(x) f(y) C x y, it is also the case that f(a) C d A ; however, to see this we will need the identification of Lebesgue measure in R d with ddimensional Hausdorff measure (?). 2. Outer Measure Let be a set and 2 the collection of all subsets of. Definition. An outer measure on is a function µ : 2 [0, ], defined on all subsets of, and satisfying (1) µ(φ) = 0; (2) A A n µ(a) µ(a n). For example, in the previous section, we saw that Lebesgue measure is an outer measure on R d. Remark. It follows that if µ is an outer measure on and A B, then µ(a) µ(b). Definition. A set A is measurable or µmeasurable if for every set B, µ(b) = µ(a B) + µ(a c B). Remark. If µ(a) = 0 then A is µmeasurable: To see this, note by monotonicity µ(a B) µ(a) = 0 and µ(a c B) µ(b). Thus µ(b) µ(a B) + µ(a c B). The reverse inequality follows by subadditivity. Remark. A set A is µmeasurable iff A c is µmeasurable.
9 Remark. Although the relevance of this definition (due to Caratheodory) is not clear at this point, it turns out to be exactly what we need, as we shall soon see. Below we use the notation A B = A B c. Definition. An algebra on a set is a collection A of subsets of such that (1) φ, A; (2) A A implies A c A; (3) A A, B A implies A B A and A B A. By induction (3) yields: A n A, n = 1,..., N, implies N A n A and N A n A. Problem 2.1. Let A be the collection of all sets which are finite disjoint unions of the form A = (C 1 G 1 )... (C N G N ) where G n are open and C n are closed in R d, n = 1,..., N. Show that A is an algebra. Definition. Let A be an algebra. A map µ : A [0, ] is finitely additive if µ(a B) = µ(a) + µ(b) for A, B disjoint in A; by induction, this is the same as µ(a 1... A N ) = µ(a 1 ) + + µ(a N ) for A 1,..., A N disjoint in A, for all N 1. Theorem 2.2. Properties of Measurable Sets: Let A 1, A 2,... be a sequence of µmeasurable sets. (1) The sets A n and A n are µmeasurable. (2) If A 1, A 2,... are disjoint, then (3) If A 1 A 2..., then ( ) µ A n = µ(a n ). ( ) lim µ(a n) = µ A n. n (4) If A 1 A 2... and µ(a 1 ) <, then ( ) lim µ(a n) = µ A n. n 9 Proof. 1. Since µ(b) µ(b A) + µ(b A c ) for all A, B, it suffices to show the opposite inequality in order to prove the set A is µmeasurable.
10 10 2. For each B, µ(b) = µ(b A 1 ) + µ(b A c 1) = µ(b A 1 ) + µ(b A c 1 A 2 ) + µ(b A c 1 A c 2) µ(b (A 1 A 2 )) + µ(b (A 1 A 2 ) c ), and so A 1 A 2 is measurable by step 1. By induction the union of finitely many µ measurable sets is µmeasurable. 3. Since (A 1 A 2 ) c = A c 1 A c 2, the intersection of two and hence finitely many measurable sets is measurable. Thus the measurable sets form an algebra. 4. If A 1, A 2,... are disjoint and measurable let B N denote the union of A n, n = 1,..., N, let A = A n, and let B. Since A N is measurable, µ(b B N ) = µ(b B N A N ) + µ(b B N A c N) = µ(b A N ) + µ(b B N 1 ). By induction it follows that µ(b B N ) = N µ(b A n) and so, since the sets B N are measurable, µ(b) = µ(b B N ) + µ(b B c N ) N µ(b A n ) + µ(b A c ) since B N A. Letting N, subadditivity yields µ(b) µ(b A n ) + µ(b A c ) µ(b A) + µ(b A c ) µ(b). All the inequalities in the last calculation are thus equalities. This shows A = A n is measurable; choosing B = shows (2) holds. 5. If A 1, A 2,... are measurable but not necessarily disjoint, then B 1 = A 1, B 2 = A 2 B 1, B 3 = A 3 (B 1 B 2 ),..., are disjoint and measurable since the measurable sets form an algebra. Since B n is measurable by step 4 and A n = B n, we obtain A n measurable. Since ( c A n = An) c, A n is also measurable and thus (1) holds. 6. (3) follows from (2): If A 1 A 2 A 3... are measurable and A 0 = φ, then A n A c n 1, n 1, are disjoint and measurable; using (2) we obtain lim µ(a N ) = lim N N N µ(a n A c n 1) = µ(a n A c n 1) ( ) ( ) = µ (A n A c n 1) = µ A n.
11 7. (4) follows from (3): If A 1 A 2 A 3... are measurable then A 1 A c 1 A 1 A c 2 A 1 A c 3... are measurable; using (3), finite additivity on measurable sets, and subadditivity, we obtain µ(a 1 ) lim µ(a n) = lim µ(a 1 A c n) n n ( ) ( ( )) = µ (A 1 A c n) = µ A 1 A n ( ) µ(a 1 ) µ A n 11 which yields µ ( A n ) lim n µ(a n). The reverse inequality follows easily from monotonicity. This shows (4). Remark. (4) in Theorem 2.2 is not necessarily true if µ(a 1 ) =. For example take A n = (n, ) R and µ equal Lebesgue measure: Then µ(a n ) = for all n 1, but A n = φ! Problem 2.3. Let A, B be µmeasurable and let A B (A B) (B A) be their symmetric difference. Show that µ(a) µ(b) µ(a B). Definition. A collection of subsets B 2 is a σalgebra provided (1) φ, B; (2) B B implies B c B; (3) B n B, n 1, implies B n B and B n B. Of course every σalgebra is an algebra. We have shown above that the µmeasurable sets form a σalgebra B. In the case of Lebesgue measure, we call the measurable sets simply the Lebesgue measurable sets. Problem 2.4. Let A R d be Lebesgue measurable with A <. Then for all ɛ > 0 there is a finite union B of rectangles satisfying A B < ɛ. The intersection of a collection of σalgebras on is again a σalgebra on. In particular 2 is a σalgebra; given any collection of sets C 2 we set σ(c) to be the smallest σalgebra containing C. Equivalently we can set σ(c) = {B : C B} to be the intersection of all σalgebras containing C. The σalgebra σ(c) is also said to be generated by C. Definition. Let B be a σalgebra on a set. A map µ : B [0, ] is a measure if (1) µ(φ) = 0,
12 12 (2) {B n } B disjoint implies ( ) µ B n = µ(b n ). (2) is called countable additivity. Of course if µ is countably additive then µ is finitely additive. We can now summarize what we have shown above: If µ is an outer measure on a set, the collection B of µmeasurable sets is a σalgebra, and µ restricted to B is a measure. In particular, Lebesgue measure restricted to the Lebesgue measurable sets M is a measure. Now the next question is: Are reasonable sets such as rectangles, open sets, etc., in R d Lebesgue measurable? Definition. The Borel σalgebra B(R d ) of R d is the smallest σalgebra on R d containing the open sets. An element of B(R d ) is called a Borel set. Problem 2.5. B(R d ) is generated by rectangles. Problem 2.6. Every open set is an almost disjoint countable union of cubes. Below we will see that every Borel set is Lebesgue measurable i.e. B(R d ) M. Given x R d, B R d, let d(x, B) = inf{ x y : y B}. Given A R d, let d(a, B) = inf{d(x, B) : x A} = inf{ x y : x A, y B}. Problem 2.7. Show that d(a, B) = d(a, B) where A denotes the closure of A. Theorem 2.8 (Caratheodory s Criterion). Let µ be an outer measure on R d. If µ(a B) = µ(a) + µ(b) for all sets A, B R d satisfying d(a, B) > 0 then every Borel set is µmeasurable: If M is the σalgebra of µmeasurable sets then B(R d ) M. Proof. Since the complement of an open set is closed, it is enough to show every closed C is µmeasurable. By subadditivity it is enough to show (2.1) µ(b) µ(b C) + µ(b C) for all B R d. If µ(b) = this is obvious, so assume µ(b) <. Define { C n = x R d : d(x, C) 1 }. n Then d(b C, B C n ) 1/n > 0 and so (2.2) µ(b) µ((b C n ) (B C)) = µ(b C n ) + µ(b C).
13 We claim lim n µ(b C n ) = µ(b C). If this claim were true, then we could take the limit in (2.2) and obtain (2.1), concluding the proof. To prove the claim, let R n = { x B : 1 n + 1 < d(x, C) 1 }. n Then B C = (B C n ) ( k=n R k) (since C is closed) so (2.3) µ(b C n ) µ(b C) µ(b C n ) + µ(r k ). If we can show µ(r n) < then lim n k=n µ(r k) = 0 and so letting n in (2.3) yields the claim. Now d(r i, R j ) > 0 if j i + 2. Hence by induction we find ( m m ) µ(r 2k ) = µ R 2k µ(b), k=1 k=1 k=n ( m m ) µ(r 2k+1 ) = µ R 2k+1 µ(b). k=0 k=0 Letting m and adding these inequalities yields 13 µ(r n ) 2µ(B) <. Definition. A Borel measure on R d is an outer measure µ on R d satisfying (1) every Borel set is µmeasurable, (2) for every A R d there is a Borel B A such that µ(a) = µ(b). Problem 2.9. Let µ be a Borel measure. Then A R d is µmeasurable iff there is a Borel set B R d with µ(a B) = 0 (Problem 2.3). Theorem Lebesgue measure is a Borel measure on R d. Proof. Suppose A, B R d with d(a, B) > 0. By Theorem 2.8 we have to show A B A + B. We can assume A B <. Let d(a, B) = δ > 0 and let R n, n 1, be a cover of A B by rectangles such that (*) R n < A B + ɛ.
14 14 Now take each rectangle and subdivide it as an almost disjoint union of smaller rectangles, each of diameter less than δ. We obtain a new cover of A B such that the sum in (*) is unchanged (this is crucial) by Proposition 1.7. Moreover each of the rectangles in the new cover intersects either A or B or neither but not both. Thus if {R n} denote the rectangles which intersect A, and {R n} denote the rectangles which intersect B then {R n} is a cover of A, {R n} is a cover of B and hence A + B R n + R n R n A B + ɛ. Letting ɛ 0 yields the result. To prove the second assertion, given A such that A =, choose B = R d. Otherwise assume A < and choose, for each m 1, a cover Rn m, n 1, of A such that Rm n < A + 1/m. Set B m = Rm n. Then B m is Borel, B m A, and B m Rm n = Rm n < A + 1/m. Let B denote the intersection of the B m s. Then B is Borel, B A, and A B < A + 1/m for all m. Remark. In particular Problem 2.9 applies to Lebesgue measure. To recap, outer measures are defined on all subsets of while measures are defined only on certain collections B of subsets of called σalgebras. For example if you think of Lebesgue measure as defined on all subsets of R d, it is an outer measure. If you restrict Lebesgue measure to the Lebesgue measurable sets M then it is a measure on M. We have just seen above that B(R d ) M. So we can restrict Lebesgue measure further to B(R d ) and obtain a measure on B(R d ). Definition. Let µ be an outer measure on. We say a µmeasurable set A is finite with respect to µ if µ(a) <. We say A is σfinite with respect to µ if A = A n with A n µmeasurable and finite. If is finite or σfinite with respect to µ then every B is so; in this case, we say µ is finite or σfinite respectively. If µ() = 1 we say µ is a probability (outer) measure. For example Lebesgue measure is σfinite. Our next item is to establish uniqueness of Lebesgue measure. To this end, we will need to introduce a concept that will be useful in later sections as well. Definition. Let be a set. A collection C 2 is a monotone class if (1) A n C, n 1, increasing, A 1 A 2..., implies A n C, (2) A n C, n 1, decreasing, A 1 A 2..., implies A n C. The intersection of a family of monotone classes is a monotone class. Thus given any collection A 2, we can speak of the smallest monotone class containing A. Remark. Let µ, ν be two finite measures on a σalgebra B on a set. Because of (3) and (4) of Theorem 2.2 it follows that C = {B B : µ(b) = ν(b)} is a monotone class. Here we need to know the finiteness of the measures to apply (4). Problem Let C be a monotone class that is an algebra. Show that C is a σalgebra.
15 Lemma Let A be an algebra and let C be a monotone class containing A. Then C contains σ(a). Proof. Let C 0 denote the smallest monotone class containing A. Then C C 0 A. If we show that C 0 σ(a), we are done. For A C 0, let C A = {B C 0 : A B, B A, and A B are in C 0 }. Check that C A is a monotone class. Now let A A. Since A is an algebra, we know C A A and so C A C 0. Thus for all A A and B C 0, we have A B, B A, and A B in C 0. Now let M be the intersection of C B over all B C 0. Then M is a monotone class and A M by the previous statement. Thus C 0 M, or for all A C 0 and B C 0, we have A B, B A, and A B in C 0. Thus C 0 is an algebra and hence C 0 is a σalgebra (Problem 2.11) containing A and hence containing σ(a). Corollary Let µ, ν be two measures on a σalgebra B on a set. Let A B be an algebra and suppose µ = ν on A. Suppose in addition there are sets A n A, n 1, satisfying A 1 A 2..., = A n, and µ(a n ) = ν(a n ) <. Then µ = ν on σ(a) B. Proof. When µ (and hence ν) is finite this follows from Lemma 2.12 and the Remark preceding it. In general let µ n (B) = µ(b A n ), ν n (B) = ν(b A n ). Then µ n = ν n on A and both are finite measures on B. Hence they agree on σ(a) which yields µ(b A n ) = ν(b A n ) for B B and n 1. Now let n and use Theorem 2.2(3). We now turn to uniqueness of Lebesgue measure. Theorem Let µ be a Borel measure on R d. If µ(r) = R for each rectangle R, then µ is Lebesgue measure. Proof. 1. If A R n, then µ(a) µ(r n) = R n. Taking the infimum over all covers yields µ(a) A. In particular this shows that the µmeasure of the boundary of a rectangle equals zero (recall the remark after Proposition 1.4). Thus µ(a) = A for any A obtained as an almost disjoint union of rectangles. Hence (Problem 2.6) µ(g) = G for every open set G. 2. Let R n, n 1, be an increasing sequence of bounded open rectangles with union R d. Given C closed and G open, let G n = {x R d : d(x, C) < 1/n}. Then G n is open and C = G n and hence C G = (G n G). By (4) of Theorem 2.2, we obtain µ(c G) = C G if G is bounded. Hence µ(c G R n ) = C G R n for every n 1; letting n, we obtain µ(a) = A for every A A where A is the algebra in Problem Since σ(a) = B(R d ), we obtain by Corollary 2.13 µ(a) = A for every Borel set A. 4. If A is any subset of R d, let B 1, B 2 be Borel sets each containing A and satisfying A = B 1, µ(a) = µ(b 2 ). Let B = B 1 B 2 ; then B is Borel, B A, and µ(b) = µ(a), B = A. Hence µ(a) = µ(b) = B = A. 15
16 16 The outer measure µ in Caratheodory s criterion need not be of the form { } (2.4) µ(a) = inf ρ(b n ) : A B n as in the definition of Lebesgue measure in 1. However, Caratheodory s criterion is not completely general: It applies only to sets, such as = R d, which are metric spaces. For outer measures given by (2.4) on arbitrary sets, there is an alternative criterion to that of Caratheodory s, which we now explain. Problem Let B be a collection of subsets of containing φ, and let ρ : B [0, ] be any function such that ρ(φ) = 0. Let µ be given by (2.4) where the cover {B n } is always taken from B. Show that µ is an outer measure on. Definition. Let B be an algebra (not a σalgebra) and let ρ : B [0, ] be finitely additive on B. We say ρ is a premeasure if ρ(b) = ρ(b 1 ) + ρ(b 2 ) +... for all B B and countable disjoint sequences {B n } B with union equal to B, i.e. ρ is countably additive on B. Problem Let ρ be a premeasure on an algebra B and let µ denote the outer measure induced by ρ according to (2.4). Show that (1) µ(b) = ρ(b) for B B; (2) every B B is µmeasurable. Problem Let be a set, B a σalgebra on, and µ a measure on B. Show that there is exactly one outer measure µ on satisfying (1) µ (B) = µ(b) for B B; (2) every B B is µ measurable; (3) for each A, there is a B B containing A, B A, and satisfying µ (B) = µ (A). Remark. Problem 2.17 shows that the concepts of measure and outer measure are more or less equivalent: If µ is a Borel measure on R d then its restriction to B(R d ) is a measure. Conversely, if µ is a measure on B(R d ), then by Problem 2.17 there is exactly one Borel measure on R d whose restriction to B(R d ) is µ. Because of this a Borel measure is frequently defined to be a measure on B(R d ) as an alternative to the equivalent definition given above. Problem Let be any set (e.g. = R) and fix a point p (e.g. p = 0). For A set µ(a) = 1 if p A and µ(a) = 0 if p / A. (1) Show that µ is an outer measure. (2) Which subsets A are µmeasurable? Definition. Let µ be an outer measure on R d. µ is outer regular if µ(a) = inf{µ(g) : A G}
17 for all µmeasurable A R d ; here G R d is open. µ is inner regular if µ(a) = sup{µ(k) : K A} for all µmeasurable A R d ; here K R d is compact. An outer measure that is outer regular and inner regular is regular. Theorem Lebesgue measure is regular. Proof. Let µ(a) = inf{ G : A G}. By mimicking the proof of Proposition 1.4, check that µ is an outer measure. Check that µ(g) = G for G open. Check that for all A R d, there is a Borel B A satisfying µ(b) = µ(a). We now use Caratheodory s criterion to check that Borel sets are µmeasurable. To this end, let A, B be such that δ = d(a, B) > 0. We have to show µ(a B) µ(a)+µ(b). If µ(a B) = this is obvious; if µ(a B) <, for each ɛ > 0 choose G A B open such that G < µ(a B) + ɛ. Let G be the set of x G satisfying d(x, A) < δ/2. Let G be the set of x G satisfying d(x, B) < δ/2. Then G A, G B are open and disjoint and G G G. Thus µ(a) + µ(b) G + G G < µ(a B) + ɛ. Since ɛ is arbitrary, let ɛ 0. To summarize we have shown µ is a Borel measure agreeing with Lebesgue measure on open rectangles. Use this to check that µ and agree on degenerate bounded rectangles i.e. bounded hyperplane segments; this implies µ( R) = 0 for all open rectangles R (see Remark after Proposition 1.4) and hence µ(r) = R for all closed rectangles. Since µ is a Borel measure, by Theorem 2.14, µ =. This shows outer regularity. To show inner regularity, let R be a (closed) rectangle and let A R be Lebesgue measurable. Since µ is outer regular and R A <, given ɛ > 0 we can choose G R A satisfying G < R A +ɛ. Let K = R G. Then K is closed and bounded hence compact, K A, and A = R R A < R G + ɛ R R G + ɛ K + ɛ. Here we used the Lebesgue measurability of A and subadditivity. This shows inner regularity if A is contained in a rectangle. In general choose R 1 R 2 R 3... rectangles with union R d and let A be Lebesgue measurable. Then by what we just learned for each n 1 there is a compact K n A R n A satisfying K n > A R n + 1/n. Thus This shows A = sup{ K : K A}. A lim n K n lim n A R n = A. Remark. In the above proof we showed A = inf{ G : A G} for all sets A R d, not just the measurable sets. In Theorem 2.19 we used uniqueness to establish outer regularity for Lebesgue measure. This method is convenient because it extends to show outer regularity for any Radon measure (see below). However for Lebesgue measure there is a simple direct proof of outer regularity. Problem Show directly from the definition that Lebesgue measure is outer regular. Definition. A Radon measure is a Borel measure µ on R d such that µ(k) < for all compact K R d. For example Lebesgue measure is a Radon measure. 17
18 18 Problem Let µ, ν be Radon measures on R d agreeing on all open sets. Show that µ, ν then agree on all sets A in the algebra A of Problem 2.1. Conclude that µ = ν on all sets. Problem By mimicking the proof of Theorem 2.19 and using Problem 2.21, show that every Radon measure on R d is regular. Problem Let A R d have positive Lebesgue measure, A > 0, and let 0 < α < 1. Show that there is a rectangle R such that R A > α R. In other words, no matter how complicated the set A is, there always exists some rectangle R so that A covers 99% of R! Problem Let A R d be Lebesgue measurable with A > 0. Show that there is an ɛ > 0 such that x ɛ implies (x + A) A > 0. (Hint  This is true if A is a rectangle: Use Problem 2.23 and start in dimension one first). Problem Let A R d be Lebesgue measurable with A > 0. Let Ã = {y z : y, z A} be the set of differences. Note 0 Ã. Show that Ã contains a neighborhood of the origin: There is an ɛ > 0 such that {x : x ɛ} Ã. In other words, no matter how complicated the set A is, you can always find some small rectangle R with 0 R and R Ã! Problem Let A 1 A 2 A 3... be any subsets of R d. Show that A n = lim n A n. The sets A n need not be measurable! Problem Show that the converse of Theorem 2.8 holds: Let µ be an outer measure on R d such that every Borel set is µmeasurable; show that µ(a B) = µ(a) + µ(b) for all A, B R d satisfying d(a, B) > 0. Problem Let be a set, B a σalgebra on, and µ n, n 1, measures on B. Let a n, n 1, be nonnegative. Show that is a measure on B. µ = a n µ n Problem Counting measure on a set is defined by setting µ(a) to be the number of points in A, when A is a finite set, and µ(a) = otherwise. Show that µ is a Borel measure on = R d. Exactly which sets are measurable? Is µ regular? Is µ Radon? Problem Let f : R d R d be Lipschitz, f(x) f(y) C x y. Use inner regularity and Problem 1.20 to show A Lebesgue measurable implies f(a) Lebesgue measurable. Problem Let µ be an outer measure on a set and let A be µmeasurable. Set ν(b) = µ(a B). Then B is νmeasurable iff B A is µmeasurable.
19 3. Radon Measures on R We now apply the results of the previous section to construct Borel measures on R which are finite on bounded intervals, i.e. Radon measures. They turn out to be in onetoone correspondence, more or less, with increasing functions F : R R. Recall F : R R is increasing if x y implies F (x) F (y). Such an F has left and righthand limits at each point: F (a+) = lim x a F (x) = inf x>a F (x), F (a ) = lim F (x) = sup F (x). x a Moreover F ( ) = lim x F (x) = sup x R F (x) and F ( ) = lim x F (x) = inf x R F (x) exist; these limits could possibly equal and respectively. The function F is rightcontinuous if F (a+) = F (a) for all a R. Theorem 3.1. If F : R R is increasing and rightcontinuous there is a unique Radon measure µ F on R such that x<a µ F ((a, b]) = F (b) F (a), a < b. If G is another such function, then µ F = µ G iff F G is constant. Conversely, if µ is a Radon measure on R and we define µ((0, x]) if x > 0 F (x) = 0 if x = 0, µ((x, 0]) if x < 0 then F is increasing and rightcontinuous and µ = µ F. Proof. Given the function F, set ρ((a, b]) = F (b) F (a), a < b, and ρ(φ) = 0. For A R set { } (3.1) µ F (A) = inf ρ((a n, b n ]) : A (a n, b n ]. By Problem 2.15, µ F is an outer measure. Just as we showed in the proof of Theorem 2.10 that Caratheodory s criterion holds for Lebesgue measure, we can show, by subdividing intervals, that Caratheodory s criterion holds for µ F. Thus every Borel set is µ F measurable. We now show µ F ((a, b]) = ρ((a, b]). It follows from (3.1) that µ((a, b]) ρ((a, b]). To show the reverse inequality µ((a, b]) ρ((a, b]), we proceed as in Theorem 1.8. Suppose (a n, b n ], n 1, are intervals whose union contains (a, b] and let ɛ > 0. Since F is rightcontinuous, there exists δ > 0 and δ n > 0, such that F (a + δ) F (a) < ɛ and F (b n + δ n ) F (b n ) < ɛ2 n = ɛ n, n 1. Note ɛ n = ɛ. Then the open intervals (a n, b n +δ n ), n 1, cover the compact interval [a+δ, b] and so there is a finite subcover. By 19
20 20 discarding any (a n, b n + δ n ) which is contained in a larger one and relabeling the index n, we may assume that (i) (a n, b n + δ n ), n = 1,..., N cover [a + δ, b], (ii) a 1 < a 2 < < a N, (iii) a n+1 < b n + δ n < b n+1 + δ n+1 for n = 1,..., N 1. Thus ρ((a, b]) = F (b) F (a) F (b) F (a + δ) + ɛ F (b N + δ N ) F (a 1 ) + ɛ N 1 = F (b N + δ N ) F (a N ) + [F (a n+1 ) F (a n )] + ɛ N 1 F (b N + δ N ) F (a N ) + [F (b n + δ n ) F (a n )] + ɛ = N [F (b n + δ n ) F (a n )] + ɛ N [F (b n ) F (a n )] + ρ((a n, b n ]) + 2ɛ. N ɛ n + ɛ If we now take the infimum of the right side over all covers of (a, b] as in (3.1), we obtain ρ((a, b]) µ F ((a, b])+2ɛ. Letting ɛ 0, this shows F (b) F (a) = ρ((a, b]) = µ F ((a, b]). To conclude that µ F is a Borel measure we need to verify that for each A R there is a Borel B A with µ F (A) = µ F (B). So let A R; if µ F (A) = we are done. If µ F (A) < for each ɛ there are intervals (a n, b n ], n 1, such that A (a n, b n ] B ɛ and µ F (B ɛ ) µ F ((a n, b n ]) = ρ((a n, b n ]) < µ F (A) + ɛ. Let B = B 1/n; since B ɛ is Borel so is B; moreover B A and µ F (A) µ F (B) < µ F (A) + 1/n for all n 1. It follows that µ F is a Borel measure that is finite on bounded intervals, i.e. µ F is a Radon measure. The uniqueness of µ F follows from Corollary 2.13 as in the proof of Theorem 2.14, where now A is the algebra in Problem 3.3 below. If G is another function such that F G is a constant then µ F = µ G follows from (3.1). If µ F = µ G, then F (x) F (0) = µ F ((0, x]) = µ G ((0, x]) = G(x) G(0) and hence F G is a constant. Finally, given µ define F as above; it follows that F (b) F (a) = µ((a, b]) for all a < b which implies F is increasing. If x n x then (0, x n] = (0, x] and so F is rightcontinuous by (4) of Theorem 2.2. Hence µ = µ F by uniqueness. The simplest Borel measure on R is the Dirac mass (or unit mass) µ at the origin: { 1 if 0 A, µ(a) = 0 if 0 / A.
21 In this case µ = µ H where H(x) = 0 if x < 0, H(x) = 1 if x 0. Note that the increasing function H is continuous except at the origin. Similarly one has the Dirac mass µ a at any given point a R; the corresponding function in this case is H(x a) where H is as above. Of course the Dirac mass makes sense on an arbitrary set (Problem 2.18). Problem 3.2. Establish the uniqueness of µ F. Below if a =, (a, b] = (, b], if b =, (a, b] = (a, ), and if a = and b =, (a, b] = R. Problem 3.3. Let A be the collection of sets which are finite disjoint unions of intervals of the form (a, b], where a < b. For A A, A = (a 1, b 1 ]... (a N, b N ], set ρ(a) = N [F (b n) F (a n )], and set ρ(φ) = 0. Show that (1) A is an algebra (Problem 2.1), (2) ρ is welldefined on A: If A = N (a n, b n ] = M m=1 (c m, d m ], then ρ(a) = M m=1 [F (d m) F (c m )]. (3) ρ is finitely additive on A, (4) ρ is a premeasure on A. (The proof of (4) is very similar to that of µ F ((a, b]) = ρ((a, b]) appearing above). Now repeat the proof of Theorem 3.1 using premeasures instead of Caratheodory s criterion. Remark. Of course, by Theorem 2.14 with d = 1, µ F for F (x) = x is Lebesgue measure. In general, the measure µ F is the LebesgueStieltjes measure associated to F. There is a completely parallel development to Theorem 3.1 involving leftcontinuous increasing functions; here the connection between µ and F reads µ([a, b)) = F (b) F (a). Remark. If µ is a finite measure on R, then F can be chosen such that F (x) = µ((, x]). To see this, suppose µ = µ G and note G( ) = lim x G(x) = lim µ((x, 0]) = µ((, 0]) x is finite and hence µ = µ F where F (x) = G(x) G( ) = µ((, x]). This F is called the cumulative distribution function of µ. Remark. If µ is a probability measure and µ((, x]) = F (x), then 0 F (x) 1, F ( ) = 0, and F ( ) = Remark. By Problem 2.22 µ F particular for any Borel A R) is always regular: For any µ F measurable A R (in µ F (A) = inf{µ F (G) : A G}, µ F (A) = sup{µ F (K) : K A}.
22 22 Here G is open and K is compact. Problem 3.4. If µ F (A) < is measurable, then for every ɛ > 0 there is a finite union B = N (a n, b n ] of halfopen intervals such that µ F (A B) < ɛ (see Problems 2.3, 2.4). Problem 3.5. Show that µ({a}) = F (a) F (a ). a R iff µ({a}) = 0. Hence F is continuous at a point Problem 3.6. Construct an increasing rightcontinuous F : R R such that (1) 0 F (x) 1, (2) F is discontinuous at the rationals, (3) F is continuous at the irrationals; (Hint  Problems 2.18 and 2.28). 4. Measurable Functions Since we deal with complexvalued functions, we recall standard notation for complex numbers. If z = x + iy C, i = 1, x = Rz R, y = Iz R, then z = x iy, z = zz = x 2 + y 2, and sgn(z) = z/ z if z 0, sgn(0) = 0. We begin with a simple but fundamental observation. Let f : Y be a map and let B Y. If we set f 1 (B) = {x : f(x) B}, we obtain a map f 1 : 2 Y 2 which commutes with unions, intersections, and complements: f 1 ( α B α ) = α f 1 (B α ), f 1 ( α B α ) = α f 1 (B α ), f 1 (B c ) = ( f 1 (B) ) c. If B is a collection of subsets of Y, we let f 1 (B) denote the collection of subsets of given by f 1 (B) = {f 1 (B) : B B}. It follows that if B is a σalgebra on Y then f 1 (B) is a σalgebra on. Definition. A measurable space is a pair (, B) where is a set and B a σalgebra on. A function f : R is measurable if f 1 (G) B for every open set G in R. Similarly a function f : C is measurable if f 1 (G) B for every open set G C. More generally let Y be a metric space. A map f : Y is measurable if f 1 (G) B for every open G Y. Remark. Since (a, ) = [a+1/n, ), f 1 ([a, )) B for all a implies f 1 ((a, )) B for all a. Since [a, ) = (a 1/n, ), f 1 ((a, )) B for all a implies f 1 ([a, )) B for all a. By taking complements, we see that (1) f 1 ((a, )) B for all a iff (2) f 1 ([a, )) B for all a iff (3) f 1 ((, b)) B for all b iff (4) f 1 ((, b]) B for all b.
23 Since every open set in R is a countable union of open intervals and an interval (a, b) = (a, ) (, b), we see a function f : R is measurable iff one of (1), (2), (3), (4) holds. Remark. Let (, B) be a measurable space, let B, and set f(x) = 1 if x B, f(x) = 0 if x / B. Then f : R is measurable iff B B. This function f is the characteristic or support or indicator function of B and is denoted f = 1 B. Problem 4.1. For A, B in, 1 A 1 B = 1 A B where A B is the symmetric difference (Problem 2.3). Let be a metric space with distance d. The open ball with center x and radius r > 0 is the set B(x, r) = {y : d(x, y) < r}. A set G is open if for every x G there is an r > 0 such that B(x, r) G. The Borel σalgebra of is the smallest σalgebra B() containing the open sets in. A set D is dense if for each x there is a sequence x n D, n 1, satisfying x n x. A metric space is separable if it has a countable dense subset. For example R d is separable. Remark. Note that B() depends only on the open sets (i.e. the topology) in the metric space ; B() does not depend on the choice of metric as long as the open sets remain unchanged. Problem 4.2. Let be a separable metric space. Show that B() is the smallest σ algebra containing the open balls. Remark. The notion of Borel measure, defined in 2 for R d, makes sense for any metric space. Similarly Caratheodory s Criterion (Theorem 2.8) is valid for any metric space. In 7 we apply these remarks to a special metric space Ω. Problem 4.3. Let (, B) be a measurable space and let Y be a metric space. Show that f : Y is measurable iff f 1 (B(Y )) B. Definition. If both and Y are metric spaces, we say f : Y is Borel or Borel measurable if f is measurable relative to the measurable space (, B()). In particular let µ denote an outer measure on R d. Then f : R d C is µmeasurable if f is measurable relative to (R d, M) where M is the σalgebra of µmeasurable sets. If µ is Lebesgue measure we call f Lebesgue measurable. Remark. It follows immediately that every continuous f : Y is Borel. Moreover for any Borel measure µ on R d and continuous f : R d C, f is µmeasurable. Remark. Let f : (, B) Y be measurable and let g : Y Z be continuous where Y, Z are metric spaces. Then g f : (, B) Z is measurable since for G Z open (g f) 1 (G) = f 1 (g 1 (G)), f is measurable, and g 1 (G) is open. Definition. Let (, A), (Y, B) be measurable spaces. The σalgebra A B on Y is the smallest σalgebra on Y containing all products of the form A B with A A, B B, where A B = {(x, y) : x A, y B}. 23
24 24 Remark. If, Y are metric spaces, then there are many choices of metrics on Y which yield the product topology. A choice convenient for the proof of Lemma 4.4 below is d((x, y), (x, y )) = max(d (x, x ), d Y (y, y )); with this choice we have B((a, b), r) = B (a, r) B Y (b, r); here (a, b) Y. Remark. If, Y are metric spaces, then B() B(Y ) is the smallest σalgebra on Y relative to which the projections π : Y, π (x, y) = x, π Y : Y Y, π Y (x, y) = y, are measurable. Lemma 4.4. Let, Y be metric spaces. Then B() B(Y ) B( Y ). If, Y are separable, then B() B(Y ) = B( Y ). Proof. If A, B Y are open, then so is A B Y ; hence in this case A B B( Y ). For each A B() let C A = {B Y : A B B( Y )}. Check that C A is a σalgebra. By the first remark, when A is open C A contains the open sets B in Y. Since C A is a σalgebra, C A must contain B(Y ); hence A B B( Y ) for all A open and B Borel. Now let C = {A : A B B( Y ) for all B B(Y )}. Check that C is a σalgebra; since we just showed open sets A are in C, we obtain B() C. Thus A B B( Y ) for all A, B Borel. Hence B() B(Y ) B( Y ). If, Y are separable, then so is Y. By Problem 4.2 to obtain the reverse inequality it is enough to show B B() B(Y ) for all open balls B in Y. But by the above remark a ball in Y is of the form B B where B, B are balls in, Y respectively. Thus balls in Y are in B() B(Y ). Problem 4.5. Show that f = u + iv : (, B) C is measurable iff u = Rf and v = If are measurable from (, B) to R. More generally let, Y be separable metric spaces and let u : (Z, C), v : (Z, C) Y. Show that u and v are measurable iff (u, v) : (Z, C) Y is measurable. Use 4.3 and 4.4. Proposition 4.6. If f, g : (, B) C are measurable then so are f ± g and fg. Proof. Define A : C 2 C and M : C 2 C by A(x, y) = x + y, M(x, y) = xy. Then A and M are continuous and F = (f, g) : C 2 is measurable (Problem 4.5). By the remark above, A F = f + g and P F = fg are measurable. In particular cg is measurable for any measurable g : C and c C; hence g and also f g = f + ( g) are also measurable. Remark. It is convenient to allow functions to take on the values ± ; let f : [, ] be such a function and set A = f 1 ( ), B = f 1 ( ), C = (A B). We say f is measurable if the sets A, B are measurable and f1 C : R is measurable. Check that f : [, ] is measurable iff f 1 ([, a)) B for all a iff f 1 ((a, ]) B for all a. We do not allow complexvalued functions to take on the value.
25 Proposition 4.7. If f n : [, ] is a sequence of measurable functions on (, B), then the functions 25 g 1 (x) = sup f n (x) n 1 g 3 (x) = lim sup f n (x) n g 2 (x) = inf n 1 f n(x) g 4 (x) = lim inf n f n(x) are all measurable. If lim n f n (x) exists for all x, then f is measurable. Proof. Since g1 1 ((a, ]) = f n 1 ((a, ]) for all a R we see g 1 is measurable. Similarly g2 1 ([, a)) = f n 1 ([, a)) for a R so g 2 is measurable. Thus fn = sup k n f k is measurable for every n 1 and hence g 3 = inf n 1 fn is measurable. Similarly f n = inf k n f k is measurable and hence so is g 4 = sup n 1 f n. Problem 4.8. Let f n : R d R, n 1, be a sequence of continuous functions and let B = {x : lim n f n (x) exists }. Show that B R d is Borel. Let f : R be realvalued. We set f + = max(f, 0), f = max( f, 0); then f = f + + f and f = f + f. f + is the positive part of f and f is the negative part of f. Corollary 4.9. If f, g are measurable and realvalued so are max(f, g), min(f, g), f +, f, and f. Definition. A function f : C is simple if f is measurable and the range of f is a finite set {c 1,..., c N }. In this case (4.1) f = N c n 1 Bn where B n = f 1 ({c n }), n = 1,..., N. When the numbers c n are distinct, the sets B n are disjoint, and (4.1) is called the standard representation of the simple function f. Proposition Let (, B) be a measurable space. (1) If f : [0, ] is measurable, there is a sequence of simple functions f n : [0, ), n 1, such that 0 f 1 f 2 f, f n f pointwise, and f n f uniformly on any set on which f is bounded. (2) If f : C is measurable, there is a sequence of simple functions f n, n 1, such that 0 f 1 f 2 f, f n f pointwise, and f n f uniformly on any set on which f is bounded. Proof. 1. Suppose f 0. For each n 1 and 1 k n2 n, let B n,k = {x : (k 1)2 n f(x) < k2 n }. Then for each n 1 the sets B n,k, k = 1,..., n2 n, are disjoint and their union is B n = {x : 0 f(x) < n}. Define f n by setting f n equal to (k 1)2 n on B n,k, k = 1,..., n2 n, and setting f n = n on B c n. Then f n 0 is simple, f 2 n < f n f
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