Lecture 21: The Riemann Integral of a Continuous Function

18.100A: Complete Lecture Notes Lecture 21: The Riemann Integral of a Continuous Function Theorem 1 (Riemann Integral) Rb Let f ∈ C([a, b]). Then, there exists a unique number denoted a f (x) dx ∈ R with the following property: for all sequences of tagged partitions {(xr , ξ r )} such that kxr k → 0, we have r Z r b f (x) dx. lim Sf (x , ξ ) = r→∞ a Remark 2. Uniqueness follows immediately from uniqueness of limits of sequences of real numbers. All we need Rb to prove is existence of a f (x) dx. Before giving the proof of the theorem, we first prove some useful facts. Note that we number the next few theorems to use them in our proof of the above theorem. Definition 3 (Modulus of Continuity) For f ∈ C([a, b]), η > 0, we define the modulus of continuous wf (η) = sup{|f (x) − f (y)| | |x − y| ≤ η}. Remark 4. Note that we number the following theorems to reference later. Theorem 5 (Theorem I) For all f ∈ C([a, b]), limη→0 wf (η) = 0. In other words, for all  > 0, there exists a δ > 0 such that ∀η < δ, wf (η) < . Proof : Let  > 0. Since f ∈ C([a, b]), f is uniformly continuous on [a, b] (as continuity on a bounded interval is equivalent to uniform continuity on the same interval). Thus, ∃δ0 > 0 such that if |x − y| < δ9 , then |f (x) − f (y)| < /2. Choose δ = δ0 and let η < δ. Then, if |x − y| ≤ η < δ = δ0 , then |f (x) − f (y)| <  . 2 Therefore, /2 is an upper bound for {|f (x) − f (y)| | |x − y| ≤ η}. Hence, wf (η) ≤ /2 < . 1 Theorem 6 (Theorem II) If (x, ξ) and (x0 , ξ 0 ) are tagged partitions of [a, b] such that x ⊂ x0 , then if f ∈ C([a, b]) then |Sf (x, ξ) − Sf (x0 , ξ 0 )| ≤ wf (kxk)(b − a). Definition 7 (Refinement) If (x, ξ) and (x0 , ξ 0 ) are tagged partitions of [a, b] such that x ⊂ x0 , we say x0 is a refinement of x. Remark 8. Refinements of x are obtained by adding more partition points. Proof : For k = 1, . . . , n, let y(k) = {xk−1 = x0` , x0`+1 , . . . , x0m = xk } 0 0 0 η(k) = {ξ`+1 , ξ`+2 , . . . , ξm }. Then, |f (ξk )(xk − xk−1 ) − Sf (y(k), η(k))| = f (ξk ) − m X f (ξj0 )(x0j − x0j−1 ) j=`+1 m X = (f (ξk ) − f (ξj0 ))(x0j − x0j−1 ) j=`+1 since Pm j=1 x0j − x0j−1 = xm − x0` = xk − xk−1 . Hence, |f (ξk )(xk − xk−1 ) − Sf (y(k), η(k))| ≤ ≤ m X j=`+1 m X |f (ξk ) − f (ξj0 )|(x0j − x0j−1 ) wf (|xk − xk−1 |)(x0j − x0j−1 ) j=`+1 ≤ wf (kxk(xk − xk−1 ). Thus, |Sf (x, ξ) − Sf (x0 , ξ 0 )| = ≤ m X (f (ξk )(xk − xk−1 ) − Sf (y(k), η(k))) k=1 m X |f (ξk )(xk − xk−1 ) − Sf (y(k), η(k))| k=1 ≤ wf (kxk) n X xk − x1 = wf (kxk)(b − a). k=1 Theorem 9 (Theorem III) If (x, ξ) and (x0 , ξ 0 ) are any two tagged partitions of [a, b] and f ∈ C([a, b]), then |Sf (x, ξ) − Sf (x0 , ξ 0 )| ≤ (wf (kx) + wf (kx0 k))(b − a). 2 Proof : Let x00 = x ∪ x0 (i.e. a common refinement), and ξ 00 be a tag of x00 . Then, by Theorem II, |Sf (x, ξ) − Sf (x0 , ξ 0 )| ≤ |Sf (x, ξ) − Sf (x00 , ξ 00 )| − |Sf (x0 , ξ 0 ) − Sf (x00 , ξ 00 )| ≤ wf (kxk)(b − a) + wf (kx0 k)(b − a). We now have the theorems necessary to prove the big theorem at the beginning of this section. Proof : Let {y(r), ζ(r)}r be a sequence of tagged partitions with ky(r)k → 0 as r → ∞. Claim 1: {Sf (y(r), ζ(r)}r is a Cauchy sequence. Proof: Let  > 0. By Theorem I, ∃δ > 0 such that ∀η < δ, wf (η) <  . 2(b − a) Since ky(r)k → 0, ∃M0 ∈ N such that ∀r ≥ M0 , ky(r)k < δ. Choose M = M0 . Then, if r, s ≥ M = M0 , |Sf (y(r), ζ(r)) − Sf (y(s), ζ(r))| ≤ (w(ky(r)k + w(ky(s)k))(b − a) by Theorem III. Hence, by the above inequalities, it follows that     |Sf (y(r), ζ(r)) − Sf (y(s), ζ(r))| < + (b − a) = . 2(b − a) 2(b − a) This proves Claim 1. Let L = limr→∞ Sf (y(r), ζ(r)), which exists as Cauchy sequences of real numbers are convergent sequences. Claim 2: Let {(x(r), ξ(r))}r be any sequence of partitions with kx(r)k → 0. Then, lim Sf (x(r), ξ(r)) = L. r→∞ With (y(r), ζ(r)) as before, we have by the Triangle Inequality and Theorem III that |Sf (x(r), ξ(r)) − L| ≤ |Sf (x(r), ξ(r)) − Sf (y(r), ζ(r))| + |Sf (y(r), ζ(r)) − L| ≤ (wf (kx(r)k) + wf (ky(r)k))(b − a) + |Sf (y(r), ζ(r)) − L| → 0 as r → ∞ by Theorem I and by the definition of L. Thus, by the Squeeze Theorem, |Sf (x(r), ξ(r)) − L| → 0 as r → 0. Properties of the Riemann Integral Notation 10 We will often abbreviate Rb a f (x) dx to Rb a f. Z b Theorem 11 (Linearity) If f, g ∈ C([a, b]) and α ∈ R, then Z (αf + g) = α a Z f+ a 3 b b g. a Proof : Let {(x(r), ξ(r))}r be a sequence of tagged partitions such that kx(r)k → 0. Then, Sαf +g (x(r), ξ(r)) = αSf (x(r), ξ(r)) + Sg (x(r), ξ(r)). Therefore, Z b (αf + g) = lim Sαf +g (x(r), ξ(r)) a r→∞ = lim (αSf (x(r), ξ(r)) + Sg (x(r), ξ(r)) r→∞ Z b Z b g. f+ =α a a 4