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1 The subset R of R n is a closed rectangle if there are n non-empty closed intervals {[a 1, b 1 ], [a 2, b 2 ],..., [a n, b n ]} so that R = [a 1, b 1 ] [a 2, b 2 ] [a n, b n ]. The subset R of R n is an open rectangle if there are n non-empty open intervals {(a 1, b 1 ), (a 2, b 2 ),..., (a n, b n )} so that R = (a 1, b 1 ) (a 2, b 2 ) (a n, b n ). The volume µ(r) of the rectangle R, in both cases, is µ(r) := (b 1 a 1 ) (b 2 a 2 ) (b n a n ). Definition 1. The subset S of R n has (Lebesgue) measure zero if for every > 0 there exist open rectangles R 1, R 2, R 3,... so that S R i and µ(r i ) <. Note that if T S and S has measure zero then T has measure zero. Lemma 1. If S 1, S 2, S 3, S 4,... are countably many sets of measure zero then the set S = i N S i has measure zero. Proof. Let > 0 be given. There exists, for every i N, a set R i of open rectangles with S i R i and Then S i N Ri and i N R R i µ(r) < 2 i. µ(r) < 2 =. i R R i i N Lemma 2. Let P be a partition of the rectangle R of R n. Then the set of points on the boundaries of the rectangles in the partition P has measure zero. 1

2 Proof. Let R be an (n 1)-dimensional rectangle in R n. Then for every > 0 there exists an n-dimensional rectangle R of R n with µ(r ) < and with R R. Hence R has measure zero. The set of points on the boundaries of the rectangles in the partition P can be covered by finitely many (n 1)-dimensional rectangles. The Lemma follows now from Lemma 1. Definition 2. The function f : R n R has bounded support if there exists a rectangle R so that f(x 1, x 2,..., x n ) = 0 whenever (x 1, x 2,..., x n ) R. Definition 3. Let f : R n R be a function. The oscillation of the function f at a R n, denoted by O(f, a), is O(f,a) := inf U { } sup{ f(x) f(y) : x, y U} : where U is a neighborhood of a. Note that O(f, a) 0. Lemma 3. Let O(f, a) <. There exists an open rectangle R containing a so that sup{f(x) : x R} inf{f(x) : x R} <. (1) Proof. If O(f, a) < then there exists an open rectangle R so that sup{ f(x) f(y) : x, y R} = δ for some δ > 0 and hence f(x) f(y) δ for all x, y R which in turn implies (1). Lemma 4. Let > 0. If for every open rectangle R containing a there are points x, y R with f(x) f(y) > then O(f, a). If O(f, a) > then for every open rectangle R containing a there are points x, y R with f(x) f(y). Proof. For every open rectangle R containing a there are points x, y R with f(x) f(y) > implies that for every open rectangle R containing a there are points x, y R with f(x) f(y) > implies that sup{ f(x) f(y) : x, y R} > for every open rectangle R containing a implies that O(f, a). O(f, a) > implies that sup{ f(x) f(y) : x, y R} > for every open rectangle R containing a implies that for every open rectangle R containing a there are points x, y R with f(x) f(y) > implies that for every open rectangle R containing a there are points x, y R with f(x) f(y) >. 2

3 Lemma 5. Let > 0 and O(f, a). containing a Then for every open rectangle sup{f(x) : x R} inf{f(x) : x R}. Proof. If the Lemma does not hold then there exists an open rectangle R containing a with sup{f(x) : x R} inf{f(x) : x R} = δ for some δ > 0. Hence f(x) f(y) δ for all points x, y R which in turn implies sup{ f(x) f(y) < in contradiction to O(f, a). Lemma 6. Let f : R n R be a function. Then O(f, a) = 0 if and only if f is continuous at a. Proof. Let f be continuous at a and > 0. There exists an open rectangle R containing a so that f(x) f(a) < /2 for all x R. Hence for all x, y R f(x) f(y) = f(x) f(a) + f(a) f(x) f(x) f(a) + f(a) f(y) /2 + /2 =. Hence sup{ f(x) f(y) : x, y R} which in turn implies that O(f, a). Because > 0 was chosen arbitrary it follows that O(f, a) = 0. Let O(f, a) = 0 and > 0 and hence O(f, a) <. There exists, according to Lemma 3, an open rectangle R containing a so that sup{f(x) : x R} inf{f(x) : x R} <. Hence f(x) f(a) < for all x R. Lemma 7. Let f : R n R be a function and > 0 and S the set of points a for which O(f, a). The set S is closed. Proof. Let a be an accumulation point of S. Assume for a contradiction that a S, that is O(f, a) <. Then there exists an open rectangle R so that sup{ f(x) f(y) : x, y R} = δ for some δ > 0 and hence f(x) f(y) δ for all x, y R. (2) 3

4 Because a is an accumulation point of S there exists an element, say b S, with b R. It follows from Lemma 4 and from O(f, b) > δ/2 that there are points x, y R with f(x) f(y) δ/2 in contradiction to Inequality (2). Theorem 1 (Lebesgue s Theorem for Riemann Integrability). Let f : R n R be a bounded function with bounded support. Then f is Riemann integrable if and only if the set of points in R n at which f is discontinuous has (Lebesgue) measure zero. Proof. Let S be the set of discontinuities of f and R a closed rectangle so that f(x 1, x 2,..., x n ) = 0 whenever (x 1, x 2,..., x n ) R. Let f < M R. Assume first that S has measure zero. We have to show that f is Riemann integrable. Let > 0 be given. We have to find a partition P = (P 1, P 2,..., P n ) for which the difference of the upper sum and the lower sum is smaller than. Let S be the set of points a in R n for which the oscillation O(f, a) 2µ(R). Let S be the set of points a in R n for which the oscillation O(f, a). It follows that S is bounded because S R and because it is closed according to Lemma 7 it is compact. Also S S according to Lemma 6. Hence S has measure zero. Let R 1, R 2, R 3,... be open rectangles so that S R i and µ(r i ) < 4M. Because S is compact there exist finitely many of those rectangles, say {T 1, T 2,..., T m }, with S T 1 T 2 T m := T. The set T is open and hence the complement T := R \ T of T in R is closed and bounded and hence compact. Let b T and hence not in S. According to Lemma 3 there exists an open rectangle R b containing b so that sup{f(x) : x R b } inf{f(x) : x R b } < 4 2µ(R).

5 The set of rectangles {R b : b T } is an open cover of the compact set T and hence there is a finite subcover, say {T 1, T 2,..., T r}, of open rectangles of T. Let A := {T 1, T 2,..., T m } {T 1, T 2,..., T r}. Then A is an open cover of R. Let P be a partition of the rectangle R so that every rectangle of the partition P has the property that if it has a point in common with some rectangle A A then it is a subset of the closure of A. Let A 1 be the set of rectangles in A which are a subset of the closure of a rectangle in the set {T 1, T 2,..., T m }. Let A 2 be the rectangles in A not in A 1. Note that every rectangle in A 2 is a subset of the closure a rectangle in the set {T 1, T 2,..., T r}. Then U(P, f) L(P, f) 2Mµ(A) + 2µ(R) µ(a) = A A 1 A A 2 2M µ(a) + µ(a) 2M 2µ(R) 4M + µ(r) =. 2µ(R) A A 1 A A 2 For the other direction we assume that f is Riemann integrable. We have to show that S has measure zero. For r > 0 let S r be the set of elements a S with O(f, a) r. It follows from Lemma 6 that S = S 1. n n N Hence it follows from Lemma 1 that it suffices to prove that S r has measure zero for r > 0. Let > 0 be given and assume for a contradiction that whenever {R i : i N} is a countable set of rectangles which cover S r then µ(r i ). i N Let P be a partition of R. Let A be the set of rectangles of P which contain an element of S r in the interior. It follows from Lemma 2 that µ(a) 2. A A 5

6 It follows from Lemma 5 that for all A A. Hence sup{f(x) : x A} inf{f(x) : x A} r U(P, f) L(P, f) = A P( sup{f(x) : x A} inf{f(x) : x A} ) µ(a) ( ) sup{f(x) : x A} inf{f(x) : x A} µ(a) A A A A rµ(a) = r A A µ(a) r 2. In contradiction to the assumption that f is Riemann integrable we obtained that U(P, f) L(P, f) r/2 for every partition P. 6

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