Page 1 1. Review Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E has some upper bound, then E has exactly one least upper bound. Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is continuous on [a;b]. Then f is also uniformly continuous on [a;b]. Proposition 1.3. Let X be a subset ofR, let x 0 be a limit point of X, let f : X!R be a function, and let L be a real number. Then the following two statements are equivalent. f is dierentiable at x 0 on X with derivative L. For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx 0 j<, then jf(x) [f(x 0 ) +L(xx 0 )]j"jxx 0 j: Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such that f 0 (x) = f(b)f(a) ba : Proposition 1.5. Let X be a subset ofR, let x 0 be a limit point of X, and let f : X!R be a function. If f is dierentiable at x 0 , then f is also continuous at x 0 . Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx 0 be a limit point of X, and let f : X!R and g : X!R be functions. (iv) If f;g are dierentiable at x 0 , then fg is dierentiable at x 0 , and (fg) 0 (x 0 ) = f 0 (x 0 )g(x 0 ) +g 0 (x 0 )f(x 0 ). (Product Rule) Page 2 1. Review Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E has some upper bound, then E has exactly one least upper bound. Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is continuous on [a;b]. Then f is also uniformly continuous on [a;b]. Proposition 1.3. Let X be a subset ofR, let x 0 be a limit point of X, let f : X!R be a function, and let L be a real number. Then the following two statements are equivalent. f is dierentiable at x 0 on X with derivative L. For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx 0 j<, then jf(x) [f(x 0 ) +L(xx 0 )]j"jxx 0 j: Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such that f 0 (x) = f(b)f(a) ba : Proposition 1.5. Let X be a subset ofR, let x 0 be a limit point of X, and let f : X!R be a function. If f is dierentiable at x 0 , then f is also continuous at x 0 . Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx 0 be a limit point of X, and let f : X!R and g : X!R be functions. (iv) If f;g are dierentiable at x 0 , then fg is dierentiable at x 0 , and (fg) 0 (x 0 ) = f 0 (x 0 )g(x 0 ) +g 0 (x 0 )f(x 0 ). (Product Rule) 2. Riemann Sums Within calculus, the two most fundamental concepts are dierentiation and integration. We have covered dierentiation already, and we now move on to integration. Dening an integral is fairly delicate. In the case of the derivative, we created one limit, and the existence of this limit dictated whether or not the function in question was dierentiable. In the case of the Riemann integral, there is also a limit to discuss, but it is much more complicated than in the case of dierentiation. We should mention that there is more than one way to construct an integral, and the Riemann integral is only one such example. Within this course, we will only be discussing the Riemann integral. The Riemann integral has some deciencies which are improved upon by other integration theories. However, those other integration theories are more involved, so we focus for now only on the Riemann integral. Our starting point will be partitions of intervals into smaller intervals, which will form the backbone of the Riemann sum. The Riemann sum will then be used to create the Riemann integral through a limiting constructing. Denition 2.1 (Partition). Let a < b be real numbers. A partition P of the interval [a;b] is a nite subset of real numbers x 0 ;:::;x n such that a =x 0 <x 1 <<x n1 <x n =b: We write P =fx 0 ;x 1 ;:::;x n g. Remark 2.2. Let P;P 0 be partitions of [a;b]. Then the union P[P 0 of P and P 0 is also a partition of [a;b]. Denition 2.3 (Upper and Lower Riemann Sums). Let a < b be real numbers, let f : [a;b]!R be a bounded function, and letP =fx 0 ;:::;x n g be a partition of [a;b]. For ev- ery integer 1in, the functionfj [x i1 ;x i ] is also a bounded function. So, sup x2[x i1 ;x i ] f(x) and inf x2[x i1 ;x i ] f(x) exist by the Least Upper Bound property (Theorem 1.1). We therefore dene the upper Riemann sum U(f;P ) by U(f;P ) := n X i=1 sup x2[x i1 ;x i ] f(x) (x i x i1 ): We also dene the lower Riemann sum L(f;P ) by L(f;P ) := n X i=1 inf x2[x i1 ;x i ] f(x) (x i x i1 ): Remark 2.4. For each integer 1 i n, we dene a function g : [a;b]! R such that g(x) := sup y2[x i1 ;x i ] f(y) for all x i1 x < x i , with g(b) := f(b). Then g is constant on [x i1 ;x i ) for all 1 i n, and f(x) g(x) for all x2 [a;b]. The upper Riemann sum U(f;P ) then represents the area under the function g, which is meant to upper bound the area under the function f. Similarly, for each integer 1 i n, we dene a function h: [a;b]!R such that h(x) := inf y2[x i1 ;x i ] f(y) for all x i1 x < x i , with h(b) := f(b). Then h is constant on [x i1 ;x i ) for all 1 i n, and h(x) g(x) for all x2 [a;b]. The lower Riemann sum L(f;P ) then represents the area under the function g, which is meant to lower bound the area under the function f. Page 3 1. Review Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E has some upper bound, then E has exactly one least upper bound. Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is continuous on [a;b]. Then f is also uniformly continuous on [a;b]. Proposition 1.3. Let X be a subset ofR, let x 0 be a limit point of X, let f : X!R be a function, and let L be a real number. Then the following two statements are equivalent. f is dierentiable at x 0 on X with derivative L. For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx 0 j<, then jf(x) [f(x 0 ) +L(xx 0 )]j"jxx 0 j: Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such that f 0 (x) = f(b)f(a) ba : Proposition 1.5. Let X be a subset ofR, let x 0 be a limit point of X, and let f : X!R be a function. If f is dierentiable at x 0 , then f is also continuous at x 0 . Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx 0 be a limit point of X, and let f : X!R and g : X!R be functions. (iv) If f;g are dierentiable at x 0 , then fg is dierentiable at x 0 , and (fg) 0 (x 0 ) = f 0 (x 0 )g(x 0 ) +g 0 (x 0 )f(x 0 ). (Product Rule) 2. Riemann Sums Within calculus, the two most fundamental concepts are dierentiation and integration. We have covered dierentiation already, and we now move on to integration. Dening an integral is fairly delicate. In the case of the derivative, we created one limit, and the existence of this limit dictated whether or not the function in question was dierentiable. In the case of the Riemann integral, there is also a limit to discuss, but it is much more complicated than in the case of dierentiation. We should mention that there is more than one way to construct an integral, and the Riemann integral is only one such example. Within this course, we will only be discussing the Riemann integral. The Riemann integral has some deciencies which are improved upon by other integration theories. However, those other integration theories are more involved, so we focus for now only on the Riemann integral. Our starting point will be partitions of intervals into smaller intervals, which will form the backbone of the Riemann sum. The Riemann sum will then be used to create the Riemann integral through a limiting constructing. Denition 2.1 (Partition). Let a < b be real numbers. A partition P of the interval [a;b] is a nite subset of real numbers x 0 ;:::;x n such that a =x 0 <x 1 <<x n1 <x n =b: We write P =fx 0 ;x 1 ;:::;x n g. Remark 2.2. Let P;P 0 be partitions of [a;b]. Then the union P[P 0 of P and P 0 is also a partition of [a;b]. Denition 2.3 (Upper and Lower Riemann Sums). Let a < b be real numbers, let f : [a;b]!R be a bounded function, and letP =fx 0 ;:::;x n g be a partition of [a;b]. For ev- ery integer 1in, the functionfj [x i1 ;x i ] is also a bounded function. So, sup x2[x i1 ;x i ] f(x) and inf x2[x i1 ;x i ] f(x) exist by the Least Upper Bound property (Theorem 1.1). We therefore dene the upper Riemann sum U(f;P ) by U(f;P ) := n X i=1 sup x2[x i1 ;x i ] f(x) (x i x i1 ): We also dene the lower Riemann sum L(f;P ) by L(f;P ) := n X i=1 inf x2[x i1 ;x i ] f(x) (x i x i1 ): Remark 2.4. For each integer 1 i n, we dene a function g : [a;b]! R such that g(x) := sup y2[x i1 ;x i ] f(y) for all x i1 x < x i , with g(b) := f(b). Then g is constant on [x i1 ;x i ) for all 1 i n, and f(x) g(x) for all x2 [a;b]. The upper Riemann sum U(f;P ) then represents the area under the function g, which is meant to upper bound the area under the function f. Similarly, for each integer 1 i n, we dene a function h: [a;b]!R such that h(x) := inf y2[x i1 ;x i ] f(y) for all x i1 x < x i , with h(b) := f(b). Then h is constant on [x i1 ;x i ) for all 1 i n, and h(x) g(x) for all x2 [a;b]. The lower Riemann sum L(f;P ) then represents the area under the function g, which is meant to lower bound the area under the function f. Denition 2.5 (Upper and Lower Integrals). Leta<b be real numbers, letf : [a;b]! R be a bounded function. We dene the upper Riemann integral R b a f of f on [a;b] by Z b a f := inffU(f;P ): P is a partition of [a;b]g: We also dene the lower Riemann integral R b a f of f on [a;b] by Z b a f := supfL(f;P ): P is a partition of [a;b]g: Lemma 2.6. Let f : [a;b]!R be a bounded function, so that there exists a real number M such thatMf(x)M for all x2 [a;b]. Then M(ba) Z b a f Z b a fM(ba) In particular, R b a and R b a exist as real numbers. Proof. If we choose P to be the partition P =fa;bg, then U(f;P ) = (ba) sup x2[a;b] f(x) and L(f;P ) = (ba) inf x2[a;b] f(x). So, U(f;P ) (ba)M and L(f;P ) (ba)(M). So,M(ba) R b a f and R b a f M(ba) by the denition of supremum and inmum, respectively. We now show that R b a f R b a f. Let P be any partition of [a;b]. By the denition of L(f;P ) and U(f;P ), we have1 < L(f;P ) U(f;P ) < +1. So, we know that the setfU(f;P ): P is a partition of [a;b]g is nonempty and bounded from below. Similarly, the setfL(f;P ): P is a partition of [a;b]g is nonempty and bounded from above. Then, by the least upper bound property (Theorem 1.1), R b a f and R b a f exist as real numbers. So, given any "> 0, choose a partition P such that L(f;P ) R b a f". (Such a partition P exists by the denition of the supremum.) We then have R b a f L(f;P ) +" U(f;P ) +". Taking the inmum over partitions P of [a;b] of both sides of this inequality, we get R b a f R b a f +". Since "> 0 is arbitrary, we conclude that R b a f R b a f, as desired. 3. Riemann Integral Denition 3.1 (Riemann Integral). Let a < b be real numbers, let f : [a;b]!R be a bounded function. If R b a f = R b a f we say that f is Riemann integrable on [a;b], and we dene Z b a f := Z b a f = Z b a f: Remark 3.2. Dening the Riemann integral of an unbounded function takes more care, and we defer this issue to later courses. Page 4 1. Review Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E has some upper bound, then E has exactly one least upper bound. Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is continuous on [a;b]. Then f is also uniformly continuous on [a;b]. Proposition 1.3. Let X be a subset ofR, let x 0 be a limit point of X, let f : X!R be a function, and let L be a real number. Then the following two statements are equivalent. f is dierentiable at x 0 on X with derivative L. For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx 0 j<, then jf(x) [f(x 0 ) +L(xx 0 )]j"jxx 0 j: Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such that f 0 (x) = f(b)f(a) ba : Proposition 1.5. Let X be a subset ofR, let x 0 be a limit point of X, and let f : X!R be a function. If f is dierentiable at x 0 , then f is also continuous at x 0 . Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx 0 be a limit point of X, and let f : X!R and g : X!R be functions. (iv) If f;g are dierentiable at x 0 , then fg is dierentiable at x 0 , and (fg) 0 (x 0 ) = f 0 (x 0 )g(x 0 ) +g 0 (x 0 )f(x 0 ). (Product Rule) 2. Riemann Sums Within calculus, the two most fundamental concepts are dierentiation and integration. We have covered dierentiation already, and we now move on to integration. Dening an integral is fairly delicate. In the case of the derivative, we created one limit, and the existence of this limit dictated whether or not the function in question was dierentiable. In the case of the Riemann integral, there is also a limit to discuss, but it is much more complicated than in the case of dierentiation. We should mention that there is more than one way to construct an integral, and the Riemann integral is only one such example. Within this course, we will only be discussing the Riemann integral. The Riemann integral has some deciencies which are improved upon by other integration theories. However, those other integration theories are more involved, so we focus for now only on the Riemann integral. Our starting point will be partitions of intervals into smaller intervals, which will form the backbone of the Riemann sum. The Riemann sum will then be used to create the Riemann integral through a limiting constructing. Denition 2.1 (Partition). Let a < b be real numbers. A partition P of the interval [a;b] is a nite subset of real numbers x 0 ;:::;x n such that a =x 0 <x 1 <<x n1 <x n =b: We write P =fx 0 ;x 1 ;:::;x n g. Remark 2.2. Let P;P 0 be partitions of [a;b]. Then the union P[P 0 of P and P 0 is also a partition of [a;b]. Denition 2.3 (Upper and Lower Riemann Sums). Let a < b be real numbers, let f : [a;b]!R be a bounded function, and letP =fx 0 ;:::;x n g be a partition of [a;b]. For ev- ery integer 1in, the functionfj [x i1 ;x i ] is also a bounded function. So, sup x2[x i1 ;x i ] f(x) and inf x2[x i1 ;x i ] f(x) exist by the Least Upper Bound property (Theorem 1.1). We therefore dene the upper Riemann sum U(f;P ) by U(f;P ) := n X i=1 sup x2[x i1 ;x i ] f(x) (x i x i1 ): We also dene the lower Riemann sum L(f;P ) by L(f;P ) := n X i=1 inf x2[x i1 ;x i ] f(x) (x i x i1 ): Remark 2.4. For each integer 1 i n, we dene a function g : [a;b]! R such that g(x) := sup y2[x i1 ;x i ] f(y) for all x i1 x < x i , with g(b) := f(b). Then g is constant on [x i1 ;x i ) for all 1 i n, and f(x) g(x) for all x2 [a;b]. The upper Riemann sum U(f;P ) then represents the area under the function g, which is meant to upper bound the area under the function f. Similarly, for each integer 1 i n, we dene a function h: [a;b]!R such that h(x) := inf y2[x i1 ;x i ] f(y) for all x i1 x < x i , with h(b) := f(b). Then h is constant on [x i1 ;x i ) for all 1 i n, and h(x) g(x) for all x2 [a;b]. The lower Riemann sum L(f;P ) then represents the area under the function g, which is meant to lower bound the area under the function f. Denition 2.5 (Upper and Lower Integrals). Leta<b be real numbers, letf : [a;b]! R be a bounded function. We dene the upper Riemann integral R b a f of f on [a;b] by Z b a f := inffU(f;P ): P is a partition of [a;b]g: We also dene the lower Riemann integral R b a f of f on [a;b] by Z b a f := supfL(f;P ): P is a partition of [a;b]g: Lemma 2.6. Let f : [a;b]!R be a bounded function, so that there exists a real number M such thatMf(x)M for all x2 [a;b]. Then M(ba) Z b a f Z b a fM(ba) In particular, R b a and R b a exist as real numbers. Proof. If we choose P to be the partition P =fa;bg, then U(f;P ) = (ba) sup x2[a;b] f(x) and L(f;P ) = (ba) inf x2[a;b] f(x). So, U(f;P ) (ba)M and L(f;P ) (ba)(M). So,M(ba) R b a f and R b a f M(ba) by the denition of supremum and inmum, respectively. We now show that R b a f R b a f. Let P be any partition of [a;b]. By the denition of L(f;P ) and U(f;P ), we have1 < L(f;P ) U(f;P ) < +1. So, we know that the setfU(f;P ): P is a partition of [a;b]g is nonempty and bounded from below. Similarly, the setfL(f;P ): P is a partition of [a;b]g is nonempty and bounded from above. Then, by the least upper bound property (Theorem 1.1), R b a f and R b a f exist as real numbers. So, given any "> 0, choose a partition P such that L(f;P ) R b a f". (Such a partition P exists by the denition of the supremum.) We then have R b a f L(f;P ) +" U(f;P ) +". Taking the inmum over partitions P of [a;b] of both sides of this inequality, we get R b a f R b a f +". Since "> 0 is arbitrary, we conclude that R b a f R b a f, as desired. 3. Riemann Integral Denition 3.1 (Riemann Integral). Let a < b be real numbers, let f : [a;b]!R be a bounded function. If R b a f = R b a f we say that f is Riemann integrable on [a;b], and we dene Z b a f := Z b a f = Z b a f: Remark 3.2. Dening the Riemann integral of an unbounded function takes more care, and we defer this issue to later courses. Theorem 3.3 (Laws of integration). Let a<b be real numbers, and let f;g : [a;b]!R be Riemann integrable functions on [a;b]. Then (i) The function f +g is Riemann integrable on [a;b], and R b a (f +g) = ( R b a f) + ( R b a g). (ii) For any real number c, cf is Riemann integrable on [a;b], and R b a (cf) =c( R b a f). (iii) The function fg is Riemann integrable on [a;b], and R b a (fg) = ( R b a f) ( R b a g). (iv) If f(x) 0 for all x2 [a;b], then R b a f 0. (v) If f(x)g(x) for all x2 [a;b], then R b a f R b a g. (vi) If there exists a real number c such that f(x) =c for x2 [a;b], then R b a f =c(ba). (vii) Let c;d be real numbers such that c a < b d. Then [c;d] contains [a;b]. Dene F (x) := f(x) for all x 2 [a;b] and F (x) := 0 otherwise. Then F is Riemann integrable on [c;d], and R d c F = R b a f. (viii) Let c be a real number such that a < c < b. Then fj [a;c] and fj [c;b] are Riemann integrable on [a;c] and [c;b] respectively, and Z b a f = Z c a fj [a;c] + Z b c fj [c;b] : Exercise 3.4. Prove Theorem 3.3. Remark 3.5. Concerning Theorem 3.3(viii), we often write R c a f instead of R c a fj [a;c] . 3.1. Riemann integrability of continuous functions. So far we have discussed some properties of Riemann integrable functions, but we have not shown many functions that are actually Riemann integrable. In this section, we show that a continuous function on a closed interval is Riemann integrable. Theorem 3.6. Let a<b be real numbers, and let f : [a;b]!R be a continuous function on [a;b]. Then f is Riemann integrable. Proof. We will produce a family of partitions of the interval [a;b] such that the upper and lower Riemann integrals of f are arbitrarily close to each other. From Theorem 1.2, f is uniformly continuous on [a;b]. Let " > 0. Then, by uniform continuity of f, there exists = (") > 0 such that, if x;y2 [a;b] satisfyjxyj < , then jf(x)f(y)j<". By the Archimedean property, there exists a positive integer N such that (ba)=N <. Consider the partition P of the interval [a;b] of the form P =fx 0 ;:::;x N g =fa;a+(ba)=N;a+2(ba)=N;a+3(ba)=N;:::;a+(N1)(ba)=N;bg: Note that x i x i1 = (ba)=N for all 1iN. Since f is continuous on [a;b], f is also continuous on [x i1 ;x i ] for each 1 i N. In particular, fj [x i1 ;x i ] achieves its maximum and minimum for all 1iN. So, for each 1iN, there exist m i ;M i 2 [x i1 ;x i ] such that inf x2[x i1 ;x i ] f(x) =f(m i ); and sup x2[x i1 ;x i ] f(x) =f(M i ): Since x i x i1 = (ba)=N < , we havejm i M i j < for each 1 i n. Since f is uniformly continuous, we conclude that inf x2[x i1 ;x i ] f(x) =f(m i )>f(M i )" = ( sup x2[x i1 ;x i ] f(x))"; 8 1in: () Page 5 1. Review Theorem 1.1 (Least Upper Bound Property). Let E be a nonempty subset ofR. If E has some upper bound, then E has exactly one least upper bound. Theorem 1.2. Let a < b be real numbers, and let f : [a;b]! R be a function which is continuous on [a;b]. Then f is also uniformly continuous on [a;b]. Proposition 1.3. Let X be a subset ofR, let x 0 be a limit point of X, let f : X!R be a function, and let L be a real number. Then the following two statements are equivalent. f is dierentiable at x 0 on X with derivative L. For every"> 0, there exists a =(")> 0 such that, ifx2X satisesjxx 0 j<, then jf(x) [f(x 0 ) +L(xx 0 )]j"jxx 0 j: Corollary 1.4 (Mean Value Theorem). Let a<b be real numbers, and let f : [a;b]!R be a continuous function which is dierentiable on (a;b). Then there exists x2 (a;b) such that f 0 (x) = f(b)f(a) ba : Proposition 1.5. Let X be a subset ofR, let x 0 be a limit point of X, and let f : X!R be a function. If f is dierentiable at x 0 , then f is also continuous at x 0 . Theorem 1.6 (Properties of Derivatives). LetX be a subset ofR, letx 0 be a limit point of X, and let f : X!R and g : X!R be functions. (iv) If f;g are dierentiable at x 0 , then fg is dierentiable at x 0 , and (fg) 0 (x 0 ) = f 0 (x 0 )g(x 0 ) +g 0 (x 0 )f(x 0 ). (Product Rule) 2. Riemann Sums Within calculus, the two most fundamental concepts are dierentiation and integration. We have covered dierentiation already, and we now move on to integration. Dening an integral is fairly delicate. In the case of the derivative, we created one limit, and the existence of this limit dictated whether or not the function in question was dierentiable. In the case of the Riemann integral, there is also a limit to discuss, but it is much more complicated than in the case of dierentiation. We should mention that there is more than one way to construct an integral, and the Riemann integral is only one such example. Within this course, we will only be discussing the Riemann integral. The Riemann integral has some deciencies which are improved upon by other integration theories. However, those other integration theories are more involved, so we focus for now only on the Riemann integral. Our starting point will be partitions of intervals into smaller intervals, which will form the backbone of the Riemann sum. The Riemann sum will then be used to create the Riemann integral through a limiting constructing. Denition 2.1 (Partition). Let a < b be real numbers. A partition P of the interval [a;b] is a nite subset of real numbers x 0 ;:::;x n such that a =x 0 <x 1 <<x n1 <x n =b: We write P =fx 0 ;x 1 ;:::;x n g. Remark 2.2. Let P;P 0 be partitions of [a;b]. Then the union P[P 0 of P and P 0 is also a partition of [a;b]. Denition 2.3 (Upper and Lower Riemann Sums). Let a < b be real numbers, let f : [a;b]!R be a bounded function, and letP =fx 0 ;:::;x n g be a partition of [a;b]. For ev- ery integer 1in, the functionfj [x i1 ;x i ] is also a bounded function. So, sup x2[x i1 ;x i ] f(x) and inf x2[x i1 ;x i ] f(x) exist by the Least Upper Bound property (Theorem 1.1). We therefore dene the upper Riemann sum U(f;P ) by U(f;P ) := n X i=1 sup x2[x i1 ;x i ] f(x) (x i x i1 ): We also dene the lower Riemann sum L(f;P ) by L(f;P ) := n X i=1 inf x2[x i1 ;x i ] f(x) (x i x i1 ): Remark 2.4. For each integer 1 i n, we dene a function g : [a;b]! R such that g(x) := sup y2[x i1 ;x i ] f(y) for all x i1 x < x i , with g(b) := f(b). Then g is constant on [x i1 ;x i ) for all 1 i n, and f(x) g(x) for all x2 [a;b]. The upper Riemann sum U(f;P ) then represents the area under the function g, which is meant to upper bound the area under the function f. Similarly, for each integer 1 i n, we dene a function h: [a;b]!R such that h(x) := inf y2[x i1 ;x i ] f(y) for all x i1 x < x i , with h(b) := f(b). Then h is constant on [x i1 ;x i ) for all 1 i n, and h(x) g(x) for all x2 [a;b]. The lower Riemann sum L(f;P ) then represents the area under the function g, which is meant to lower bound the area under the function f. Denition 2.5 (Upper and Lower Integrals). Leta<b be real numbers, letf : [a;b]! R be a bounded function. We dene the upper Riemann integral R b a f of f on [a;b] by Z b a f := inffU(f;P ): P is a partition of [a;b]g: We also dene the lower Riemann integral R b a f of f on [a;b] by Z b a f := supfL(f;P ): P is a partition of [a;b]g: Lemma 2.6. Let f : [a;b]!R be a bounded function, so that there exists a real number M such thatMf(x)M for all x2 [a;b]. Then M(ba) Z b a f Z b a fM(ba) In particular, R b a and R b a exist as real numbers. Proof. If we choose P to be the partition P =fa;bg, then U(f;P ) = (ba) sup x2[a;b] f(x) and L(f;P ) = (ba) inf x2[a;b] f(x). So, U(f;P ) (ba)M and L(f;P ) (ba)(M). So,M(ba) R b a f and R b a f M(ba) by the denition of supremum and inmum, respectively. We now show that R b a f R b a f. Let P be any partition of [a;b]. By the denition of L(f;P ) and U(f;P ), we have1 < L(f;P ) U(f;P ) < +1. So, we know that the setfU(f;P ): P is a partition of [a;b]g is nonempty and bounded from below. Similarly, the setfL(f;P ): P is a partition of [a;b]g is nonempty and bounded from above. Then, by the least upper bound property (Theorem 1.1), R b a f and R b a f exist as real numbers. So, given any "> 0, choose a partition P such that L(f;P ) R b a f". (Such a partition P exists by the denition of the supremum.) We then have R b a f L(f;P ) +" U(f;P ) +". Taking the inmum over partitions P of [a;b] of both sides of this inequality, we get R b a f R b a f +". Since "> 0 is arbitrary, we conclude that R b a f R b a f, as desired. 3. Riemann Integral Denition 3.1 (Riemann Integral). Let a < b be real numbers, let f : [a;b]!R be a bounded function. If R b a f = R b a f we say that f is Riemann integrable on [a;b], and we dene Z b a f := Z b a f = Z b a f: Remark 3.2. Dening the Riemann integral of an unbounded function takes more care, and we defer this issue to later courses. Theorem 3.3 (Laws of integration). Let a<b be real numbers, and let f;g : [a;b]!R be Riemann integrable functions on [a;b]. Then (i) The function f +g is Riemann integrable on [a;b], and R b a (f +g) = ( R b a f) + ( R b a g). (ii) For any real number c, cf is Riemann integrable on [a;b], and R b a (cf) =c( R b a f). (iii) The function fg is Riemann integrable on [a;b], and R b a (fg) = ( R b a f) ( R b a g). (iv) If f(x) 0 for all x2 [a;b], then R b a f 0. (v) If f(x)g(x) for all x2 [a;b], then R b a f R b a g. (vi) If there exists a real number c such that f(x) =c for x2 [a;b], then R b a f =c(ba). (vii) Let c;d be real numbers such that c a < b d. Then [c;d] contains [a;b]. Dene F (x) := f(x) for all x 2 [a;b] and F (x) := 0 otherwise. Then F is Riemann integrable on [c;d], and R d c F = R b a f. (viii) Let c be a real number such that a < c < b. Then fj [a;c] and fj [c;b] are Riemann integrable on [a;c] and [c;b] respectively, and Z b a f = Z c a fj [a;c] + Z b c fj [c;b] : Exercise 3.4. Prove Theorem 3.3. Remark 3.5. Concerning Theorem 3.3(viii), we often write R c a f instead of R c a fj [a;c] . 3.1. Riemann integrability of continuous functions. So far we have discussed some properties of Riemann integrable functions, but we have not shown many functions that are actually Riemann integrable. In this section, we show that a continuous function on a closed interval is Riemann integrable. Theorem 3.6. Let a<b be real numbers, and let f : [a;b]!R be a continuous function on [a;b]. Then f is Riemann integrable. Proof. We will produce a family of partitions of the interval [a;b] such that the upper and lower Riemann integrals of f are arbitrarily close to each other. From Theorem 1.2, f is uniformly continuous on [a;b]. Let " > 0. Then, by uniform continuity of f, there exists = (") > 0 such that, if x;y2 [a;b] satisfyjxyj < , then jf(x)f(y)j<". By the Archimedean property, there exists a positive integer N such that (ba)=N <. Consider the partition P of the interval [a;b] of the form P =fx 0 ;:::;x N g =fa;a+(ba)=N;a+2(ba)=N;a+3(ba)=N;:::;a+(N1)(ba)=N;bg: Note that x i x i1 = (ba)=N for all 1iN. Since f is continuous on [a;b], f is also continuous on [x i1 ;x i ] for each 1 i N. In particular, fj [x i1 ;x i ] achieves its maximum and minimum for all 1iN. So, for each 1iN, there exist m i ;M i 2 [x i1 ;x i ] such that inf x2[x i1 ;x i ] f(x) =f(m i ); and sup x2[x i1 ;x i ] f(x) =f(M i ): Since x i x i1 = (ba)=N < , we havejm i M i j < for each 1 i n. Since f is uniformly continuous, we conclude that inf x2[x i1 ;x i ] f(x) =f(m i )>f(M i )" = ( sup x2[x i1 ;x i ] f(x))"; 8 1in: () We now estimate U(f;P ) andL(f;P ). By the denition of U(f;P ) andL(f;P ), we have L(f;P )U(f;P ): () However, L(f;P ) is also close to U(f;P ) by (): L(f;P ) = ba N N X i=1 ( inf x2[x i1 ;x i ] f(x))> ba N N X i=1 [( sup x2[x i1 ;x i ] f(x))"] =(ba)" +U(f;P ): By the denition of R b a f, we conclude that Z b a f >(ba)" +U(f;P ): By the denition of R b a f, we conclude that Z b a f >(ba)" + Z b a f: Since "> 0 is arbitrary, we get Z b a f Z b a f: Combining this inequality with Lemma 2.6, we conclude that R b a f = R b a f. That is, f is Riemann integrable. Exercise 3.7. Let a < b be real numbers. Let f : [a;b]!R be a bounded function. Let c2 [a;b]. Assume that, for each > 0, we know that f is Riemann integrable on the set fx2 [a;b]: jxcjg. Then f is Riemann integrable on [a;b]. 3.2. Piecewise Continuous Functions. We can now expand a bit more the family of functions that are Riemann integrable. Proposition 3.8. Let a < b be real numbers. Assume that f : [a;b]!R is continuous at every point of [a;b], except for a nite number of points. Then f is Riemann integrable. Proof. By Theorem 3.3(viii) and an inductive argument, it suces to consider the case that f is discontinuous at a single point c2 [a;b]. Let > 0. Then f is continuous on the set E :=fx2 [a;b]: jxcj g. Note that E consists of either one or two closed intervals. Since fj E is continuous, we then conclude that fj E Riemann integrable by Theorem 3.6. Then Exercise 3.7 says that f is Riemann integrable on [a;b], as desired. 3.3. Monotone Functions. It turns out that monotone functions are Riemann integrable as well. There exist monotone functions that are not piecewise continuous, so the current section is not subsumed by the previous one. Proposition 3.9. Let a<b be real numbers, and let f : [a;b]!R be a monotone function. Then f is Riemann integrable.Read More

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