Lecture 8. Q = [a 1, b 1 ] [a n, b n ], (3.10) width(q) = sup(b i a i ). (3.12) i. from which we define the lower and upper Riemann sums,

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1 Lecture iemann Integral of Several Variables Last time we defined the iemann integral for one variable, and today we generalize to many variables. Definition 3.3. A rectangle is a subset of n of the form where a i, b i. = [a 1, b 1 ] [a n, b n ], (3.10) Note that x = (x 1,..., x n ) a i x i b i for all i. The volume of the rectangle is v() = (b 1 a 1 ) (bn a n ), (3.11) and the width of the rectangle is width() = sup(b i a i ). (3.12) i ecall (stated informally) that given [a, b], a finite subset of [a, b] is a partition of [a, b] if a, b and you can write = {t i : i = 1,..., N}, where t 1 = a < t 2 <... < t N = b. An interval I belongs to if and only if I is one of the intervals [t i, t i+1 ]. Definition 3.4. A partition of is an n tuple ( 1,..., n ), where each i is a partition of [a i, b i ]. Definition 3.5. A rectangle = I 1 I n belongs to if for each i, the interval I i belongs to i. Let f : be a bounded function, let be a partition of, and let be a rectangle belonging to. We define m f = inf f = g.l.b. {f(x) : x } M f = sup f = l.u.b. {f(x) : x }, from which we define the lower and upper iemann sums, L(f, ) = m (f)v() (3.13) (3.14) U(f, ) = M (f)v(). 1

2 It is evident that L(f, ) U(f, ). (3.15) Now, we will take a sequence of partitions that get finer and finer, and we will define the integral to be the limit. Let = ( 1,..., n ) and = ( 1,..., n ) be partitions of. We say that refines if i i for each i. Claim. If refines, then L(f, ) L(f. ). (3.16) roof. We let j = j for j i, and we let i = i { a}, where a [a i, b i ]. We can create any refinement by multiple applications of this basic refinement. is If a rectangle belonging to, then either 1. belongs to, or 2. =, where, belong to. In the first case, the contribution of to L(f, ) equals the contribution of to L(f, ), so the claim holds. In the second case, m v() = m (v( ) + v( )) (3.17) and So, Taken altogether, this shows that Thus, and belong to. Claim. If refines, then m r = inf f inf f, inf f. (3.18) m m, m (3.19) m v() m v( ) + m v( ) (3.20) U(f, ) U(f, ) (3.21) The proof is very similar to the previous proof. Combining the above two claims, we obtain the corollary Corollary 2. If and are partitions, then U(f, ) L(f, ) (3.22) 2

3 roof. Define = ( 1 1,..., n n ). So, refines and. We have shown that U(f, ) U(f, ) L(f, ) L(f, ) (3.23) L(f, ) U(f, ). Together, these show that U(f, ) L(f, ). (3.24) With this result in mind, we define the lower and upper iemann integrals: f = sup L(f, ) (3.25) f = inf U(f, ). Clearly, we have f Finally, we define iemann integrable. f, (3.26) Definition 3.6. A function f is iemann integrable over if the lower and upper iemann integrals coincide (are equal). 3.3 Conditions for Integrability Our next problem is to determine under what conditions a function is (iemann) integrable. Let s look at a trivial case: Claim. Let F : be the constant function f(x) = c. Then f is. integrable over, and c = cv(). (3.27) roof. Let be a partition, and let be a rectangle belonging to. We see that m (f) = M (f) = c, so U(f, ) = M (f)v() = c v() (3.28) = cv(). 3

4 Similarly, L(f, ) = cv(). (3.29) Corollary 3. Let be a rectangle, and let { i : i = 1,..., N} be a collection of rectangles covering. Then v() v( i ). (3.30) Theorem 3.7. If f : is continuous, then f is. integrable over. roof. We begin with a definition Definition 3.8. Given a partition of, we define emember that mesh width( ) = sup width(). (3.31) compact = f : is uniformly continuous. (3.32) That is, given ɛ > 0, there exists δ > 0 such that if x, y and x y < δ, then f(x) f(y) < ɛ. Choose a partition of with mesh width less than δ. Then, for every rectangle belonging to and for every x, y, we have x y < δ. By uniform continuity we have, M (f) m (f) ɛ, which is used to show that U(f, ) L(f, ) = (M (f) m (f))v() We can take ɛ 0, so which shows that f is integrable. ɛ v() ɛv(). (3.33) sup L(f, ) = inf U(f, ), (3.34) We have shown that continuity is sufficient for integrability. However, continuity is clearly not necessary. What is the general condition for integrability? To state the answer, we need the notion measure of zero. Definition 3.9. Suppose A n. The set A is of measure zero if for every ɛ > 0, there exists a countable covering A of by rectangles 1, 2, 3,... such that v( i ) < ɛ. i 4

5 Theorem Let f : be a bounded function, and let A be the set of points where f is not continuous. Then f is. integrable if and only A if is of measure zero. Before we prove this, we make some observations about sets of measure zero: 1. Let A, B n and suppose B A. If A is of measure zero, then B is also of measure zero. 2. Let A i n for i = 1, 2, 3,..., and suppose the A i s are of measure zero. Then A i is also of measure zero. 3. ectangles are not of measure zero. We prove the second observation: For any ɛ > 0, choose coverings i,1, i,2,... of A i such that each covering has total volume less than ɛ/2 i. Then { i,j } is a countable covering of A i of total volume ɛ 2 i i=1 = ɛ. (3.35) 5