Chapter 10. Multivariable integral Riemann integral over rectangles Rectangles and partitions

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1 Chapter 10 Multivariable integral 10.1 iemann integral over rectangles ote:??? lectures As in chapter chapter 5, we define the iemann integral using the Darboux upper and lower integrals. The ideas in this section are very similar to integration in one dimension. The complication is mostly notational. The differences between one and several dimensions will grow more pronounced in the sections following ectangles and partitions Definition Let (a 1,a 2,...,a n ) and (b 1,b 2,...,b n ) be such that a k b k for all k. A set of the form [a 1,b 1 ] [a 2,b 2 ] [a n,b n ] is called a closed rectangle. In this setting it is sometimes useful to allow a k = b k, in which case we think of [a k,b k ] = {a k } as usual. If a k < b k for all k, then a set of the form (a 1,b 1 ) (a 2,b 2 ) (a n,b n ) is called an open rectangle. For an open or closed rectangle := [a 1,b 1 ] [a 2,b 2 ] [a n,b n ] n or := (a 1,b 1 ) (a 2,b 2 ) (a n,b n ) n, we define the n-dimensional volume by V () := (b 1 a 1 )(b 2 a 2 ) (b n a n ). A partition P of the closed rectangle = [a 1,b 1 ] [a 2,b 2 ] [a n,b n ] is a finite set of partitions P 1,P 2,...,P n of the intervals [a 1,b 1 ],[a 2,b 2 ],...,[a n,b n ]. We will write P = (P 1,P 2,...,P n ). That is, for every k there is an integer l k and the finite set of numbers P k = {x k,0,x k,1,x k,2,...,x k,lk } such that a k = x k,0 < x k,1 < x k,2 < < x k,lk 1 < x k,lk = b k. Picking a set of n integers j 1, j 2,..., j n where j k {1,2,...,l k } we get the subrectangle [x 1, j1 1, x 1, j1 ] [x 2, j2 1, x 2, j2 ] [x n, jn 1, x n, jn ]. 75

2 76 CHAPTE 10. MULTIVAIABLE ITEGAL For simplicity, we order the subrectangles somehow and we say { 1, 2,..., } are the subrectangles corresponding to the partition P of. Or more simply, we say they are the subrectangles of P. In other words we subdivide the original rectangle into many smaller subrectangles. It is not difficult to see that these subrectangles cover our original, and their volume sums to that of. That is then When = j, and V () = V ( j ). k = [x 1, j1 1, x 1, j1 ] [x 2, j2 1, x 2, j2 ] [x n, jn 1, x n, jn ] V ( k ) = Δx 1, j1 Δx 2, j2 Δx n, jn = (x 1, j1 x 1, j1 1)(x 2, j2 x 2, j2 1) (x n, jn x n, jn 1). Let n be a closed rectangle and let f : be a bounded function. Let P be a partition of [a,b] and suppose that there are subrectangles. Let i be a subrectangle of P. Define m i := inf{ f (x) : x i }, M i := sup{ f (x) : x i }, L(P, f ) := U(P, f ) := i=1 i=1 m i V ( i ), M i V ( i ). We call L(P, f ) the lower Darboux sum and U(P, f ) the upper Darboux sum. The indexing in the definition may be complicated, fortunately we generally do not need to go back directly to the definition often. We start proving facts about the Darboux sums analogous to the one-variable results. Proposition Suppose n is a closed rectangle and f : is a bounded function. Let m,m be such that for all x we have m f (x) M. For any partition P of we have mv () L(P, f ) U(P, f ) MV (). Proof. Let P be a partition. Then note that m m i for all i and M i M for all i. Also m i M i for all i. Finally i=1 V ( i) = V (). Therefore, ) mv () = m ( V ( i ) i=1 = i=1 mv ( i ) i=1 i=1 M i V ( i ) m i V ( i ) i=1 MV ( i ) = M ( V ( i ) i=1 ) = MV ().

3 10.1. IEMA ITEGAL OVE ECTAGLES Upper and lower integrals By Proposition the set of upper and lower Darboux sums are bounded sets and we can take their infima and suprema. As before, we now make the following definition. Definition If f : is a bounded function on a closed rectangle n. Define f := sup{l(p, f ) : P a partition of }, We call the lower Darboux integral and the upper Darboux integral. As in one dimension we have refinements of partitions. f := inf{u(p, f ) : P a partition of }. Definition Let n be a closed rectangle and let P = (P 1,P 2,...,P n ) and P = ( P 1, P 2,..., P n ) be partitions of. We say P a refinement of P if as sets P k P k for all k = 1,2,...,n. It is not difficult to see that if P is a refinement of P, then subrectangles of P are unions of subrectangles of P. Simply put, in a refinement we took the subrectangles of P and we cut them into smaller subrectangles. Proposition Suppose n is a closed rectangle, P is a partition of and P is a refinement of P. If f : be a bounded function, then L(P, f ) L( P, f ) and U( P, f ) U(P, f ). Proof. Let 1, 2,..., be the subrectangles of P and 1, 2,..., M be the subrectangles of. Let I k be the set of indices j such that j k. We notice that k = j I k j, V ( k ) = j I k V ( j ). Let m j := inf{ f (x) : x j }, and m j := inf{ f (x) : j } as usual. otice also that if j I k, then m k m j. Then L(P, f ) = m k V ( k ) = m k V ( j ) j I k m j V ( j ) = j I k M m j V ( j ) = L( P, f ). The key point of this next proposition is that the lower Darboux integral is less than or equal to the upper Darboux integral. Proposition Let n be a closed rectangle and f : a bounded function. Let m,m be such that for all x we have m f (x) M. Then mv () f f MV (). (10.1)

4 78 CHAPTE 10. MULTIVAIABLE ITEGAL Proof. For any partition P, via Proposition mv () L(P, f ) U(P, f ) MV (). By taking suprema of L(P, f ) and infima of U(P, f ) over all P we obtain the first and the last inequality. The key of course is the middle inequality in (10.1). Let P = (P 1,P 2,...,P n ) and Q = (Q 1,Q 2,...,Q n ) be partitions of. Define P = ( P 1, P 2,..., P n ) by letting P k = P k Q k. Then P is a partition of as can easily be checked, and P is a refinement of P and a refinement of Q. By Proposition , L(P, f ) L( P, f ) and U( P, f ) U(Q, f ). Therefore, L(P, f ) L( P, f ) U( P, f ) U(Q, f ). In other words, for two arbitrary partitions P and Q we have L(P, f ) U(Q, f ). Via Proposition?? we obtain sup{l(p, f ) : P a partition of } inf{u(p, f ) : P a partition of }. In other words f f The iemann integral We now have all we need to define the iemann integral in n-dimensions over rectangles. Again, the iemann integral is only defined on a certain class of functions, called the iemann integrable functions. Definition Let n be a closed rectangle. Let f : be a bounded function such that f (x) dx = f (x) dx. Then f is said to be iemann integrable. The set of iemann integrable functions on is denoted by (). When f () we define the iemann integral f := f = When the variable x n needs to be emphasized we write f (x) dx, f (x 1,...,x n ) dx 1 dx n, or If 2, then often instead of volume we say area, and hence write f (x) da. Proposition implies immediately the following proposition. f. f (x) dv.

5 10.1. IEMA ITEGAL OVE ECTAGLES 79 Proposition Let f : be a iemann integrable function on a closed rectangle n. Let m,m be such that m f (x) M for all x. Then mv () f MV (). Example : A constant function is iemann integrable. Suppose f (x) = c for all x on. Then cv () f So f is integrable, and furthermore f = cv (). f cv (). The proofs of linearity and monotonicity are almost completely identical as the proofs from one variable. We therefore leave it as an exercise to prove the next two propositions. Proposition (Linearity). Let n be a closed rectangle and let f and g be in () and α. (i) α f is in () and (ii) f + g is in () and α f = α f ( f + g) = f + g. Proposition (Monotonicity). Let n be a closed rectangle and let f and g be in () and let f (x) g(x) for all x. Then f Again for simplicity if f : S is a function and S is a closed rectangle, then if the restriction f is integrable we say f is integrable on, or f () and we write f := Proposition For a closed rectangle S n, if f : S is integrable and S is a closed rectangle, then f is integrable over. Proof. Given ε > 0, we find a partition P such that U(P, f ) L(P, f ) < ε. By making a refinement of P we can assume that the endpoints of are in P, or in other words, is a union of subrectangles of P. Then the subrectangles of P divide into two collections, ones that are subsets of and ones whose intersection with the interior of is empty. Suppose that 1, 2..., K be the subrectangles g. f.

6 80 CHAPTE 10. MULTIVAIABLE ITEGAL that are subsets of and K+1,..., be the rest. Let P be the partition of composed of those subrectangles of P contained in. Then using the same notation as before. ε > U(P, f ) L(P, f ) = K Therefore f is integrable. K (M k m k )V ( k ) + (M k m k )V ( k ) = U( P, f ) L( P, f ) Integrals of continuous functions k=k+1 (M k m k )V ( k ) Later we will prove a much more general result, but it is useful to start with continuous functions only and prove that continuous functions are integrable. Before we get to continuous functions, let us state the following proposition, which has a very easy proof, but it is useful to emphasize as a technique. Proposition Let n be a closed rectangle and f : a bounded function. If for every ε > 0, there exists a partition P of such that then f (). U(P, f ) L(P, f ) < ε, Proof. Given an ε > 0 find P as in the hypothesis. Then f f U(P, f ) L(P, f ) < ε. As f f and the above holds for every ε > 0, we conclude f = f and f (). We say a rectangle = [a 1,b 1 ] [a 2,b 2 ] [a n,b n ] has longest side at most α if b k a k α for all k = 1,2,...,n. Proposition If a rectangle n has longest side at most α. Then for any x,y, x y nα. Proof. x y = (x 1 y 1 ) 2 + (x 2 y 2 ) (x n y n ) 2 (b 1 a 1 ) 2 + (b 2 a 2 ) (b n a n ) 2 α 2 + α α 2 = nα.

7 10.1. IEMA ITEGAL OVE ECTAGLES 81 Theorem Let n be a closed rectangle and f : a continuous function, then f (). Proof. The proof is analogous to the one variable proof with some complications. The set is closed and bounded and hence compact. So f is not just continuous, but in fact uniformly continuous by Proposition??. Let ε > 0 be given. Find a δ > 0 such that x y < δ implies f (x) f (y) < ε V (). Let P be a partition of such that longest side of any subrectangle is strictly less than δ n. Then for all x,y k for a subrectangle k of P we have, by the proposition above, x y < n δ n = δ. Therefore f (x) f (y) f (x) f (y) < ε V (). As f is continuous on k, it attains a maximum and a minimum on this interval. Let x be a point where f attains the maximum and y be a point where f attains the minimum. Then f (x) = M k and f (y) = m k in the notation from the definition of the integral. Therefore, And so U(P, f ) L(P, f ) = M i m i = f (x) f (y) < ( M k V ( k ) ) ε V (). ( m k V ( k ) ) As ε > 0 was arbitrary, and f is iemann integrable on. = < ε V () b a (M k m k )V ( k ) f = V ( k ) = ε. b a f, Integration of functions with compact support Let U n be an open set and f : U be a function. We say the support of f is the set supp( f ) := {x U : f (x) 0}. That is, the support is the closure of the set of points where the function is nonzero. The closure is in U, that is in particular supp( f ) U. So for a point x U not in the support we have that f is constantly zero in a whole neighbourhood of x.

8 82 CHAPTE 10. MULTIVAIABLE ITEGAL A function f is said to have compact support if supp( f ) is a compact set. We will mostly consider the case when U = n. In light of the following exercise, this is not an oversimplification. Exercise : Suppose U n is open and f : U is continuous and of compact support. Show that the function f : n { f (x) if x U, f (x) := 0 otherwise, is continuous. Proposition Suppose f : n be a function with compact support. If is a closed rectangle such that supp( f ) o where o is the interior of, and f is integrable over, then for any other closed rectangle S with supp( f ) S o, the function f is integrable over S and S f = Proof. The intersection of closed rectangles is again a closed rectangle (or empty). Therefore we can take = S be the intersection of all rectangles containing supp( f ). If is the empty set, then supp( f ) is the empty set and f is identically zero and the proposition is trivial. So suppose that is nonempty. As, we know that f is integrable over. Furthermore S. Given ε > 0, take P to be a partition of such that f. U( P, f ) L( P, f ) < ε. ow add the endpoints of S to P to create a new partition P. ote that the subrectangles of P are subrectangles of P as well. Let 1, 2,..., K be the subrectangles of P and K+1,..., the new subrectangles. ote that since supp( f ), then for k = K + 1,..., we have supp( f ) k = /0. In other words f is identically zero on k. Therefore in the notation used previously we have U(P, f S ) L(P, f S ) = = K K (M k m k )V ( k ) + (M k m k )V ( k ) + k=k+1 k=k+1 = U( P, f ) L( P, f ) < ε. Similarly we have that L(P, f S ) = L( P, f ) and therefore f = f. Since we also get f = f, or in other words f = S f. S (M k m k )V ( k ) (0)V ( k )

9 10.1. IEMA ITEGAL OVE ECTAGLES 83 Because of this proposition, when f : n has compact support and is integrable over a rectangle containing the support we write f := f or f := n For example if f is continuous and of compact support then n f exists. f Exercises Exercise : Prove Proposition Exercise : Suppose that is a rectangle with the length of one of the sides equal to 0. Show that for any function f show that f () and f = 0. Exercise : Suppose and are two closed rectangles with. Suppose that f : is in (). Show that f ( ). Exercise : Suppose and are two closed rectangles with. Suppose that f : is in ( ) and f (x) = 0 for x /. Show that f () and Hint: see the previous exercise. f = Exercise : Suppose that n and n are two rectangles such that = is a rectangle, and is rectangle with one of the sides having length 0 (that is V ( ) = 0). Let f : be a function such that f ( ) and f ( ). Show that f () and Hint: see previous exercise. f. f = f + f. Exercise : Prove a stronger version of Proposition Suppose f : n be a function with compact support. Prove that if is a closed rectangle such that supp( f ) and f is integrable over, then for any other closed rectangle S with supp( f ) S, the function f is integrable over S and S f = f. Hint: notice that now the new rectangles that you add as in the proof can intersect supp( f ) on their boundary. Exercise : Suppose that and S are closed rectangles. Let f (x) := 1 if x and f (x) = 0 otherwise. Show that f is integrable over S and compute S f.

10 84 CHAPTE 10. MULTIVAIABLE ITEGAL Exercise : Let = [0,1] [0,1] 2. a) Suppose f : is defined by Show that f () and compute f. b) Suppose f : is defined by Show that f / (). f (x,y) := f (x,y) := { 1 if x = y, 0 else. { 1 if x Q or y Q, 0 else.

11 10.2. ITEATED ITEGALS AD FUBII THEOEM Iterated integrals and Fubini theorem ote:??? lectures The iemann integral in several variables is hard to compute from the definition. For onedimensional iemann integral we have the fundamental theorem of calculus and we can compute many integrals without having to appeal to the definition of the integral. We will rewrite a a iemann integral in several variables into several one-dimensional iemann integrals by iterating. However, if f : [0,1] 2 is a iemann integrable function, it is not immediately clear if the three expressions [0,1] 2 f, f (x,y)dxdy, are equal, or if the last two are even well-defined. and f (x,y)dydx Example : Define f (x,y) := { 1 if x = 1/2 and y Q, 0 otherwise. Then f is iemann integrable on := [0,1] 2 and f = 0. Furthermore, f (x,y)dxdy = 0. However 1 f (1/2,y)dy 0 does not exist, so we cannot even write f (x,y)dydx. Proof: Let us start with integrability of f. We simply take the partition of [0,1] 2 where the partition in the x direction is {0, 1/2 ε, 1/2 + ε,1} and in the y direction {0,1}. The subrectangles of the partition are 1 := [0, 1/2 ε] [0,1], 2 := [1/2 ε, 1/2 + ε] [0,1], 3 := [1/2 + ε,1] [0,1]. We have m 1 = M 1 = 0, m 2 = 0, M 2 = 1, and m 3 = M 3 = 0. Therefore, and L(P, f ) = m 1 (1/2 ε) 1 + m 2 (2ε) 1 + m 3 (1/2 ε) 1 = 0, U(P, f ) = M 1 (1/2 ε) 1 + M 2 (2ε) 1 + M 3 (1/2 ε) 1 = 2ε. The upper and lower sum are arbitrarily close and the lower sum is always zero, so the function is integrable and f = 0. For any y, the function that takes x to f (x,y) is zero except perhaps at a single point x = 1/2. We know that such a function is integrable and 1 0 f (x,y)dx = 0. Therefore, f (x,y)dxdy = 0. However if x = 1/2, the function that takes y to f (1/2,y) is the nonintegrable function that is 1 on the rationals and 0 on the irrationals. See Example??.

12 86 CHAPTE 10. MULTIVAIABLE ITEGAL We will solve this problem of undefined inside integrals by using the upper and lower integrals, which are always defined. We split n+m into two parts. That is, we write the coordinates on n+m = n m as (x,y) where x n and y m. For a function f (x,y) we write f x (y) := f (x,y) when x is fixed and we wish to speak of the function in terms of y. We write f y (x) := f (x,y) when y is fixed and we wish to speak of the function in terms of x. Theorem (Fubini version A). Let S n m be a closed rectangle and f : S be integrable. The functions g: and h: defined by are integrable over and In other words g(x) := S f x and h(x) := g = h = S f. f x S ( ) ( ) f = f (x,y)dy dx = f (x,y)dy dx. S S S If it turns out that f x is integrable for all x, for example when f is continuous, then we obtain the more familiar f = f (x,y)dydx. S S Proof. Let P be a partition of and P be a partition of S. Let 1, 2,..., be the subrectangles of P and 1, 2,..., K be the subrectangles of P. Then P P is the partition whose subrectangles are j k for all 1 j and all 1 k K. Let m j,k := inf (x,y) j k f (x,y). We notice that V ( j k ) = V ( j)v ( k ) and hence L(P P, f ) = K m j,k V ( j k ) = ( K m j,k V ( k ) )V ( j ).

13 10.2. ITEATED ITEGALS AD FUBII THEOEM 87 If we let m k (x) := inf f (x,y) = inf f x (y), y k y k then of course if x j then m j,k m k (x). Therefore K j,k V ( m K k ) m k (x)v ( k ) = L(P, f x ) f x = g(x). S As we have the inequality for all x j we have K m j,k V ( k ) inf g(x). x j We thus obtain L(P P, f ) ( ) inf g(x) V ( j ) = L(P,g). x j Similarly U(P P, f ) U(P,h), and the proof of this inequality is left as an exercise. Putting this together we have L(P P, f ) L(P,g) U(P,g) U(P,h) U(P P, f ). And since f is integrable, it must be that g is integrable as U(P,g) L(P,g) U(P P, f ) L(P P, f ), and we can make the right hand side arbitrarily small. Furthermore as L(P P, f ) L(P,g) U(P P, f ) we must have that g = S f. Similarly we have L(P P, f ) L(P,g) L(P,h) U(P,h) U(P P, f ), and hence U(P,h) L(P,h) U(P P, f ) L(P P, f ). So if f is integrable so is h, and as L(P P, f ) L(P,h) U(P P, f ) we must have that h = S f. We can also do the iterated integration in opposite order. The proof of this version is almost identical to version A, and we leave it as an exercise to the reader.

14 88 CHAPTE 10. MULTIVAIABLE ITEGAL Theorem (Fubini version B). Let S n m be a closed rectangle and f : S be integrable. The functions g: S and h: S defined by are integrable over S and That is we also have g(y) := ( f = S S f y and h(y) := g = h = S S f (x,y)dx S f. ) ( dy = S f y f (x,y)dx ) dy. ext suppose that f x and f y are integrable for simplicity. For example, suppose that f is continuous. Then by putting the two versions together we obtain the familiar f = f (x,y)dydx = f (x,y)dxdy. S S Often the Fubini theorem is stated in two dimensions for a continuous function f : on a rectangle = [a, b] [c, d]. Then the Fubini theorem states that f = b d a c f (x,y)dydx = S d b c a f (x,y)dxdy. And the Fubini theorem is commonly thought of as the theorem that allows us to swap the order of iterated integrals. epeatedly applying Fubini theorem gets us the following corollary: Let := [a 1,b 1 ] [a 2,b 2 ] [a n,b n ] n be a closed rectangle and let f : be continuous. Then f = b1 b2 a 1 a 2 bn a n f (x 1,x 2,...,x n )dx n dx n 1 dx 1. Clearly we can also switch the order of integration to any order we please. We can also relax the continuity requirement by making sure that all the intermediate functions are integrable, or by using upper or lower integrals Exercises Exercise : Compute xexy dxdy in a simple way. Exercise : Prove the assertion U(P P, f ) U(P,h) from the proof of Theorem Exercise : Prove Theorem

15 10.2. ITEATED ITEGALS AD FUBII THEOEM 89 Exercise : Let = [a,b] [c,d] and f (x,y) := g(x)h(y) for two continuous functions g: [a,b] and h: [a,b]. Prove ( b )( d ) f = g h. a c Exercise : Compute x 2 y 2 dxdy and (x 2 + y 2 2 ) x 2 y 2 (x 2 + y 2 ) 2 dydx. You will need to interpret the integrals as improper, that is, the limit of 1 ε as ε 0. Exercise : Suppose f (x,y) := g(x) where g: [a,b] is iemann integrable. Show that f is iemann integrable for any = [a, b] [c, d] and Exercise : Define f : [ 1,1] [0,1] by f (x,y) := b f = (d c) g. a { x if y Q, 0 else. Show a) f (x,y)dxdy exists, but f (x,y)dydx does not. b) Compute f (x,y)dydx and f (x,y)dydx. c) Show f is not iemann integrable on [ 1,1] [0,1] (use Fubini).

16 90 CHAPTE 10. MULTIVAIABLE ITEGAL 10.3 Outer measure and null sets ote:??? lectures Outer measure and null sets Before we characterize all iemann integrable functions, we need to make a slight detour. We introduce a way of measuring the size of sets in n. Definition Let S n be a subset. Define the outer measure of S as m (S) := inf V ( j ), where the infimum is taken over all sequences { j } of open rectangles such that S j. In particular S is of measure zero or a null set if m (S) = 0. We will only need measure zero sets and so we focus on these. ote that S is of measure zero if for every ε > 0 there exist a sequence of open rectangles { j } such that S j and V ( j ) < ε. (10.2) Furthermore, if S is measure zero and S S, then S is of measure zero. We can in fact use the same exact rectangles. We can also use balls and it is sometimes more convenient. In fact we can choose balls no bigger than a fixed radius. Proposition Let δ > 0 be given. A set S n is measure zero if and only if for every ε > 0, there exists a sequence of open balls {B j }, where the radius of B j is r j < δ such that S B j and ote that the volume of B j is proportional to r n j. r n j < ε. Proof. First note that if is a (closed or open) cube (rectangle with all sides equal) of side s, then is contained in a closed ball of radius ns by Proposition , and therefore in an open ball of size 2 ns. Let s be a number that is less than the smallest side of and also so that 2 ns < δ. We claim is contained in a union of closed cubes C 1,C 2,...,C k of sides s such that k V (C j ) 2 n V ().

17 10.3. OUTE MEASUE AD ULL SETS 91 It is clearly true (without the 2 n ) if has sides that are integer multiples of s. So if a side is of length (l + α)s, for l and 0 α < 1, then (l + α)s 2ls. Increasing the side to 2ls we obtain a new larger rectangle of volume at most 2 n times larger, but whose sides are multiples of s. So suppose that there exist { j } as in the definition such that (10.2) is true. As we have seen above, we can choose closed cubes {C k } with C k of side s k as above that cover all the rectangles { j } and so s n k = V (C k ) 2 n V ( k ) < 2 n ε. Covering C k with balls B k of radius r k = 2 ns k we obtain rk n < 22n nε. And as S j j kc k k B k, we are finished. Suppose that we have the ball condition above for some ε > 0. Without loss of generality assume that all r j < 1. Each B j is contained a in a cube (rectangle with all sides equal) j of side 2r j. So V ( j ) = (2r j ) n < 2 n r j. Therefore S j and V ( j ) < 2 n r j < 2 n ε. In fact the definition of outer measure could have been done with open balls as well, not just null sets. Although we leave this generalization to the reader Examples and basic properties Example : The set Q n n of points with rational coordinates is a set of measure zero. Proof: The set Q n is countable and therefore let us write it as a sequence q 1,q 2,... For each q j find an open rectangle j with q j j and V ( j ) < ε2 j. Then Q n j and V ( j ) < ε2 j = ε. In fact, the example points to a more general result. Proposition A countable union of measure zero sets is of measure zero. Proof. Suppose S = S j

18 92 CHAPTE 10. MULTIVAIABLE ITEGAL where S j are all measure zero sets. Let ε > 0 be given. For each j there exists a sequence of open rectangles { j,k } such that S j and Then j,k V ( j,k ) < 2 j ε. S j,k. As V ( j,k ) is always positive, the sum over all j and k can be done in any order. In particular, it can be done as V ( j,k ) < 2 j ε = ε. The next example is not just interesting, it will be useful later. Example : Let P := {x n : x k = c} for a fixed k = 1,2,...,n and a fixed constant c. Then P is of measure zero. Proof: First fix s and let us prove that P s := {x n : x k = c, x j s for all j k} is of measure zero. Given any ε > 0 define the open rectangle It is clear that P s. Furthermore := {x n : c ε < x k < c + ε, x j < s + 1 for all j k} V () = 2ε ( 2(s + 1) ) n 1. As s is fixed, we can make V () arbitrarily small by picking ε small enough. ext we note that P = P j and a countable union of measure zero sets is measure zero. Example : If a < b, then m ([a,b]) = b a. Proof: In the case of, open rectangles are open intervals. Since [a,b] (a ε,b + ε) for all ε > 0. Hence, m ([a,b]) b a.

19 10.3. OUTE MEASUE AD ULL SETS 93 Let us prove the other inequality. Suppose that {(a j,b j )} are open intervals such that [a,b] (a j,b j ). We wish to bound (b j a j ) from below. Since [a,b] is compact, then there are only finitely many open intervals that still cover [a,b]. As throwing out some of the intervals only makes the sum smaller, we only need to take the finite number of intervals still covering [a,b]. If (a i,b i ) (a j,b j ), then we can throw out (a i,b i ) as well. Therefore we have [a,b] k (a j,b j ) for some k, and we assume that the intervals are sorted such that a 1 < a 2 < < a k. ote that since (a 2,b 2 ) is not contained in (a 1,b 1 ) we have that a 1 < a 2 < b 1 < b 2. Similarly a j < a j+1 < b j < b j+1. Furthermore, a 1 < a and b k > b. Thus, m ([a,b]) k (b j a j ) k 1 (a j+1 a j ) + (b k a k ) = b k a 1 > b a. Proposition Suppose E n is a compact set of measure zero. Then for every ε > 0, there exist finitely many open rectangles 1, 2,..., k such that E 1 2 k and k V ( j ) < ε. Also for any δ > 0, there exist finitely many open balls B 1,B 2,...,B k of radii r 1,r 2,...,r k < δ such that E B 1 B 2 B k and k r n j < ε. Proof. Find a sequence of open rectangles { j } such that E j and V ( j ) < ε. By compactness, finitely many of these rectangles still contain E. That is, there is some k such that E 1 2 k. Hence k V ( j ) V ( j ) < ε. The proof that we can choose balls instead of rectangles is left as an exercise.

20 94 CHAPTE 10. MULTIVAIABLE ITEGAL Images of null sets Before we look at images of measure zero sets, let us see what a continuously differentiable function does to a ball. Lemma Suppose U n is an open set, B U is an open or closed ball of radius at most r, f : B n is continuously differentiable and suppose f (x) M for all x B. Then f (B) B, where B is a ball of radius at most Mr. Proof. Without loss of generality assume B is a closed ball. The ball B is convex, and hence via Proposition 8.4.2, that f (x) f (y) M x y for all x,y in B. In particular, suppose B = C(y,r), then f (B) C ( f (y),mr ). The image of a measure zero set using a continuous map is not necessarily a measure zero set. However if we assume the mapping is continuously differentiable, then the mapping cannot stretch the set too much. The proposition does not require compactness, and this is left as an exercise. Proposition Suppose U n is an open set and f : U n is a continuously differentiable mapping. If E U is a compact measure zero set, then f (E) is measure zero. Proof. We must first handle a couple of techicalities. First let us replace U by a smaller open set to make f (x) bounded. At each point x E pick an open ball B(x,r x ) such that the closed ball C(x,r x ) U. By compactness we only need to take finitely many points x 1,x 2,...,x q to still conver E. Define q q U := B(x j,r x j ), K := C(x j,r x j ). (10.3) We have E U K U. The set K is compact. The function that takes x to f (x) is continuous, and therefore there exists an M > 0 such that f (x) M for all x K. So without loss of generality we may replace U by U and from now on suppose that f (x) M for all x U. At each point x E pick a ball B(x,δ x ) of maximum radius so that B(x,δ x ) U. Let δ = inf x E δ x. Take a sequence {x j } E so that δ x j δ. As E is compact, we can pick the sequence to be convergent to some y E. Once x j y < δ y 2, then δ x j > δ y 2 by the triangle inequality. Therefore δ > 0. Given ε > 0, there exist balls B 1,B 2,...,B k of radii r 1,r 2,...,r k < δ such that E B 1 B 2 B k and k r n j < ε. Suppose B 1,B 2,...,B k are the balls of radius Mr 1,Mr 2,...,Mr k from Lemma f (E) f (B 1 ) f (B 2 ) f (B k ) B 1 B 2 B k and k Mr n j < Mε.

21 10.3. OUTE MEASUE AD ULL SETS Exercises Exercise : Finish the proof of Proposition , that is, show that you can use balls instead of rectangles. Exercise : If A B then m (A) m (B). Exercise : Show that if n is a closed rectangle then m () = V (). Exercise : Prove a version of Proposition without using compactness: a) Mimic the proof to first prove that the proposition holds only if E is relatively compact; a set E U is relatively compact if the closure of E in the subspace topology on U is compact, or in other words if there exists a compact set K with K U and E K. Hint: The bound on the size of the derivative still holds, but you may need to use countably many balls. Be careful as the closure of E need no longer be measure zero. b) ow prove it for any null set E. Hint: First show that {x U : d(x,y) 1/m for all y / U and d(0,x) m} is a compact set for any m > 0. Exercise : Let U n be an open set and let f : U be a continuously differentiable function. Let G := {(x,y) U : y = f (x)} be the graph of f. Show that f is of measure zero. Exercise : Given a closed rectangle n, show that for any ε > 0 there exists a number s > 0 and finitely many open cubes C 1,C 2,...,C k of side s such that C 1 C 2 C k and k V (C j ) V () + ε. Exercise : Show that there exists a number k = k(n,r,δ) depending only on n, r and δ such the following holds. Given B(x,r) n and δ > 0, there exist k open balls B 1,B 2,...,B k of radius at most δ such that B(x,r) B 1 B 2 B k. ote that you can find k that really only depends on n and the ratio δ/r.

22 96 CHAPTE 10. MULTIVAIABLE ITEGAL 10.4 The set of iemann integrable functions ote:??? lectures Oscillation and continuity Let S n be a set and f : S a function. Instead of just saying that f is or is not continuous at a point x S, we we need to be able to quantify how discontinuous f is at a function is at x. For any δ > 0 define the oscillation of f on the δ -ball in subset topology that is B S (x,δ) = B n(x,δ) S as o( f,x,δ) := sup y B S (x,δ) f (y) inf f (y) = y B S (x,δ) sup y 1,y 2 B S (x,δ) ( f (y1 ) f (y 2 ) ). That is, o( f,x,δ) is the length of the smallest interval that contains the image f ( B S (x,δ) ). Clearly o( f,x,δ) 0 and notice o( f,x,δ) o( f,x,δ ) whenever δ < δ. Therefore, the limit as δ 0 from the right exists and we define the oscillation of a function f at x as o( f,x) := lim o( f,x,δ) = inf o( f,x,δ). δ 0 + δ>0 Proposition f : S is continuous at x S if and only if o( f,x) = 0. Proof. First suppose that f is continuous at x S. Then given any ε > 0, there exists a δ > 0 such that for y B S (x,δ) we have f (x) f (y) < ε. Therefore if y 1,y 2 B S (x,δ) then f (y 1 ) f (y 2 ) = f (y 1 ) f (x) ( f (y 2 ) f (x) ) < ε + ε = 2ε. We take the supremum over y 1 and y 2 o( f,x,δ) = sup ( f (y1 ) f (y 2 ) ) 2ε. y 1,y 2 B S (x,δ) Hence, o(x, f ) = 0. On the other hand suppose that o(x, f ) = 0. Given any ε > 0, find a δ > 0 such that o( f,x,δ) < ε. If y B S (x,δ) then f (x) f (y) sup y 1,y 2 B S (x,δ) ( f (y1 ) f (y 2 ) ) = o( f,x,δ) < ε. Proposition Let S n be closed, f : S, and ε > 0. The set {x S : o( f,x) ε} is closed. Proof. Equivalently we want to show that G = {x S : o( f,x) < ε} is open in the subset topology. As inf δ>0 o( f,x,δ) < ε, find a δ > 0 such that o( f,x,δ) < ε

23 10.4. THE SET OF IEMA ITEGABLE FUCTIOS 97 Take any ξ B S (x, δ/2). otice that B S (ξ, δ/2) B S (x,δ). Therefore, ( o( f,ξ, δ/2) = sup f (y1 ) f (y 2 ) ) ( sup f (y1 ) f (y 2 ) ) = o( f,x,δ) < ε. y 1,y 2 B S (ξ,δ/2) y 1,y 2 B S (x,δ) So o( f,ξ ) < ε as well. As this is true for all ξ B S (x, δ/2) we get that G is open in the subset topology and S \ G is closed as is claimed The set of iemann integrable functions We have seen that continuous functions are iemann integrable, but we also know that certain kinds of discontinuities are allowed. It turns out that as long as the discontinuities happen on a set of measure zero, the function is integrable and vice versa. Theorem (iemann-lebesgue). Let n be a closed rectangle and f : a bounded function. Then f is iemann integrable if and only if the set of discontinuities of f is of measure zero (a null set). Proof. Let S be the set of discontinuities of f. That is S = {x : o( f,x) > 0}. The trick to this proof is to isolate the bad set into a small set of subrectangles of a partition. There are only finitely many subrectangles of a partition, so we will wish to use compactness. If S is closed, then it would be compact and we could cover it by small rectangles as it is of measure zero. Unfortunately, in general S is not closed so we need to work a little harder. For every ε > 0, define S ε := {x : o( f,x) ε}. By Proposition S ε is closed and as it is a subset of which is bounded, S ε is compact. Furthermore, S ε S and S is of measure zero. Via Proposition there are finitely many open rectangles S 1,S 2,...,S k that cover S ε and V (S j ) < ε. The set T = \(S 1 S k ) is closed, bounded, and therefore compact. Furthermore for x T, we have o( f,x) < ε. Hence for each x T, there exists a small closed rectangle T x with x in the interior of T x, such that sup f (y) inf f (y) < 2ε. y T x y T x The interiors of the rectangles T x cover T. As T is compact there exist finitely many such rectangles T 1,T 2,...,T m that covers T. ow take all the rectangles T 1,T 2,...,T m and S 1,S 2,...,S k and construct a partition out of their endpoints. That is construct a partition P with subrectangles 1, 2,..., p such that every j is contained in T l for some l or the closure of S l for some l. Suppose we order the rectangles so that 1, 2,..., q are those that are contained in some T l, and q+1, q+2,..., p are the rest. In particular, we have q V ( j ) V () and p V ( j ) ε. j=q+1

24 98 CHAPTE 10. MULTIVAIABLE ITEGAL Let m j and M j be the inf and sup over j as before. If j T l for some l, then (M j m j ) < 2ε. Let B be such that f (x) B for all x, so (M j m j ) < 2B over all rectangles. Then U(P, f ) L(P, f ) = = p (M j m j )V ( j ) ( q (M j m j )V ( j ) ( q 2εV ( j ) ) ) + ( p j=q+1 + ( p j=q+1 2BV ( j ) 2εV () + 2Bε = ε(2v () + 2B). (M j m j )V ( j ) Clearly, we can make the right hand side as small as we want and hence f is integrable. For the other direction, suppose that f is iemann integrable over. Let S be the set of discontinuities again and now let S k := {x : o( f,x) 1/k}. Fix a k. Given an ε > 0, find a partition P with subrectangles 1, 2,..., p such that U(P, f ) L(P, f ) = p ) (M j m j )V ( j ) < ε Suppose that 1, 2,..., p are order so that the interiors of 1, 2,..., q intersect S k, while the interiors of q+1, q+2,..., p are disjoint from S k. If x j S k and x is in the interior of j so sufficiently small balls are completely inside j, then by definition of S k we have M j m j 1/k. Then ε > p (M j m j )V ( j ) q (M j m j )V ( j ) 1 k q V ( j ) In other words q V ( j) < kε. Let G be the set of all boundaries of all the subrectangles of P. The set G is of measure zero (see Example ). Let j denote the interior of j, then S k 1 2 q G. As G can be covered by open rectangles arbitrarily small volume, S k must be of measure zero. As S = S k and a countable union of measure zero sets is of measure zero, S is of measure zero. )

25 10.4. THE SET OF IEMA ITEGABLE FUCTIOS Exercises Exercise : Suppose that f : (a,b) (c,d) is a bounded continuous function. Show that the integral of f over = [a,b] [c,d] makes sense and is uniquely defined. That is, set f to be anything on the boundary of and compute the integral. Exercise : Suppose n is a closed rectangle. Show that (), the set of iemann integrable functions, is an algebra. That is, show that if f,g () and a, then a f (), f + g () and f g (). Exercise : Suppose n is a closed rectangle and f : is a bounded function which is zero except on a closed set E of measure zero. Show that f exists and compute it. Exercise : Suppose n is a closed rectangle and f : and g: are two iemann integrable functions. Suppose f = g except for a closed set E of measure zero. Show that f = g.