lecture11

lecture11 - Lecture 11 We review some basic properties of...

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Unformatted text preview: Lecture 11 We review some basic properties of the Riemann integral. Let Q ⊆ Rn be a rectangle, and let f, g : q → R be bounded functions. Assume that f, g are R. integrable. We have the following properties of R. integrals: • Linearity: a, b ∈ R =⇒ af + bg is R. integrable and � � � af + bg = a f + b g. Q Q Q (3.82) • Monotonicity: If f ≤ g , then � f≤ Q Q � g. (3.83) • Maximality Property: Let h : Q → R be a function defined by h(x) = max(f (x), g (x)). Theorem 3.14. The function h is R. integrable and �� � �� h ≥ max f, g . Q Q Q (3.84) Proof. We need the following lemma. Lemma 3.15. If f and g are continuous at some point x0 ∈ Q, then h is continuous at x0 . Proof. We consider the case f (x0 ) = g (x0 ) = h(x0 ) = r. The functions f and g are continuous at x0 if and only if for every � > 0, there exists a δ > 0 such that |f (x) − f (x0 )| < � and |g (x) − g (x0 )| < � whenever |x − x0 | < δ . Substitute in f (x0 ) = g (x0 ) = r. The value of h(x) is either f (x) or g (x), so |h(x) − r| < � for |x − x0 | < δ . That is |h(x) − h(x0 )| < � for |x − x0 | < δ , so h is continuous at x0 . The proofs of the other cases are left to the student. We defined h = max(f, g ). The lemma tells is that h is integrable. Define E, F , and G as follows: E = Set of points in Q where f is discontinuous, F = Set of points in Q where g is discontinuous, G = Set of points in Q where h is discontinuous. (3.85) (3.86) (3.87) 1 The functions f, g are integrable over Q if and only if E, F are of measure zero. The lemma shows that G ⊆ E ∪ F , so h is integrable over Q. To finish the proof, we notice that h = max(f, g ) ≥ f , g. (3.88) Then, by monotonicity, � h ≥ max Q Q �� f, � Q � g. (3.89) Remark. Let k = min(f, g ). Then k = − max(−f, −g ). So, the maximality property also implies that k is integrable and �� � � � k ≤ min f, g . (3.90) Q Q Q A useful trick for when dealing with functions is to change the sign. The preceding remark and the following are examples where such a trick is useful. Let f : Q → R be a R. integrable function. Define f+ = max(f, 0), f− = max(−f, 0). (3.91) Both of these functions are R. integrable and non­negative: f+ , f− ≥ 0. Also note that f = f+ − f− . This decomposition is a trick we will use over and over again. Also note that |f | = f+ + f− , so |f | is R. integrable. By monotonicity, � � � |f | = f+ + f− Q Q Q � � ≥ f+ − f− (3.92) Q Q � = f. Q By replacing f by −f , we obtain � |f | ≥ Q � −f � = − f. Q Q (3.93) Combining these results, we arrive at the following claim Claim. � Q �� � � � |f | ≥ � f � � � Q (3.94) Proof. The proof is above. 2 3.6 Integration Over More General Regions So far we’ve been defining integrals over rectangles. Let us now generalize to other sets. Let S be a bounded set in Rn , and let f : S → R be a bounded function. Let fS : Rn → R be the map defined by � f (x) if x ∈ S , fS (x) = (3.95) 0 if x ∈ S . / ¯ Let Q be a rectangle such that Int Q ⊃ S . Definition 3.16. The map f is Riemann integrable over S if fS is Riemann integrable over Q. Additionally, � � f= S Q fS . (3.96) One has to check that this definition does not depend on the choice of Q, but we do not check this here. Claim. Let S be a bounded set in Rn , and let f, g : S → R be bounded functions. Assume that f, g are R. integrable over S . Then the following properties hold: • Linearity: If a, b ∈ R, then af + bg is R. integrable over S , and � � � af + bg = a f + b g. S S S (3.97) • Monotonicity: If f ≤ g , then � f≤ S S � g. (3.98) • Maximality: If h = max(f, g ) (over the domain S ), then h is R. integrable over S , and �� � �� h ≥ max f, g . (3.99) S S S Proof. The proofs are easy, and we outline them here. • Linearity: Note that afS + bgS = (af + bg )S , so � � af + bg = (af + bg )S S Q � � = a fS + b gS �Q � Q = a f + b g. S S (3.100) 3 • Monotonicity: Use f ≤ g =⇒ fS ≤ gS . • Maximality: Use h = max(f, g ) =⇒ hS = max(fS , gS ). Let’s look at some nice set theoretic properties of the Riemann integral. Claim. Let S, T be bounded subsets of Rn with T ⊆ S . Let f : S → R be bounded and non­negative. Suppose that f is R. integrable over both S and T . Then � � f≤ f. (3.101) T S ¯ Proof. Clearly, fT ≤ fS . Let Q be a rectangle with S ⊇ Int Q. Then � � fT ≤ fS . Q Q (3.102) Claim. Let S1 , S2 be bounded subsets of Rn , and let f : S1 ∪ S2 → R be a bounded function. Suppose that f is R. integrable over both S1 and S2 . Then f is R. integrable over S1 ∩ S2 and over S1 ∪ S2 , and � � � � f= f+ f− f. (3.103) S1 ∪ S2 S1 S2 S1 ∩ S2 Proof. Use the following trick. Notice that fS1 ∪S2 = max(fS1 , fS2 ), fS1 ∩S2 = min(fS1 , fS2 ). Now, choose Q such that Int Q ⊃ S1 ∪ S2 , so fS1 ∪S2 and fS1 ∩S2 are integrable over Q. Note the identity fS1 ∪S2 = fS1 + fS2 − fS1 ∩S2 . So, � fS1 ∪S2 = Q Q (3.104) (3.105) (3.106) (3.107) (3.108) � fS1 + � fS2 − Q � fS1 ∩S2 , Q from which it follows that � f= S1 ∪ S2 � f+ S1 � f− S2 � f. S1 ∩ S2 (3.109) 4 ...
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This note was uploaded on 10/02/2010 for the course MAT unknown taught by Professor Unknown during the Fall '04 term at MIT.

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