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Familiarizing Students with Definition of Lebesgue Integral: Examples of Calculation Directly from Its Definition Using Mathemat

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Familiarizing Students with Definition of Lebesgue Integral: Examples of Calculation Directly from Its Definition Using Mathemat Math.Comput.Sci. (2017) 11:363–381 DOI 10.1007/s11786-017-0321-5 Mathematics in Computer Science Familiarizing Students with Definition of Lebesgue Integral: Examples of Calculation Directly from Its Definition Using Mathematica Włodzimierz Wojas · Jan Krupa Received: 7 December 2016 / Revised: 19 April 2017 / Accepted: 20 April 2017 / Published online: 30 May 2017 © The Author(s) 2017. This article is an open access publication Abstract We present in this paper several examples of Lebesgue integral calculated directly from its definitions using Mathematica. Calculation of Riemann integrals directly from its definitions for some elementary functions is standard in higher mathematics education. But it is difficult to find analogical examples for Lebesgue integral in the available literature. The paper contains Mathematica codes which we prepared to calculate symbolically Lebesgue sums and limits of sums. We also visualize the graphs of simple functions used for approximation of the integrals. We also show how to calculate the needed sums and limits by hand (without CAS). We compare our calculations in Mathematica with calculations in some other CAS programs such as wxMaxima, MuPAD and Sage for the same integrals. Keywords Higher education · Lebesgue integral · Application of CAS · Mathematica · Mathematical didactics Mathematics Subject Classification 97R20 · 97I50 · 97B40 1 Introduction Young man, in mathematics you don’t understand things. You just get used to them John von Neumann I hear and I forget. I see and I remember. I do and I understand Chinese Quote Electronic Supplementary material The online version of this article (doi:10.1007/s11786-017-0321-5) contains supplementary material, which is available to authorized users. W. Wojas · J. Krupa (B) Department of Applied Mathematics, Warsaw University of Life Sciences (SGGW), ul. Nowoursynowska 159, 02-776 Warsaw, Poland e-mail: [email protected] W. Wojas e-mail: [email protected] 364 W. Wojas, J. Krupa In popular books of calculus, for example [3,7,12], we can find many examples of Riemann integral calculated directly from its definition. The aim of these examples is to familiarize students with the definition of Riemann integral. But we cannot find analogical examples for Lebesgue integral. In this article, with similar aim but for Lebesgue integral definition, we present the following examples of π/2 ( ) 1 ( ) ( ) π ( − + 2) ( ) > calculation directly from its definition: 0 sin x dm x , 0 exp x dm x , 0 ln 1 2r cos x r dm x (r 1), 1 xk dm(x) (k ∈ N), 1 x2 dm(x) where dm(x) denotes the Lebesgue measure on the real line. The way we 0 −1 1 ( ) 1 2 ( ) 1 3 ( ) presented we may also calculate e.g. 0 x dm x , 0 x dm x , 0 x dm x . We also consider the calculation of 1 2 ( ) 1 1/m ( ) ∈ N b χ ( ) e ( ) integrals 0 x dm x , 0 x dm x (m ), a [a,b]\Q dm x , 1 ln x dm x using “partition the range of f ” Lebesgue philosophy. We calculate sums, limits and plot graphs of needed simple functions using Mathematica. 2 Definitions of Lebesgue Integral The following two definitions of Lebesgue integral are used in the first part of this article: Let (R, M, m) be the measure space, where M is σ -algebra of Lebesgue measurable subsets in R, and m is the Lebesgue measure on R. Let R be extended real numbers R with two more elements adjoined, denoted +∞ and −∞.LetA be Lebesgue measurable subset of R. A real-valued function f on A is called simple if it assumes only finitely many distinct values. Let f : A → R be measurable nonnegative function (we’ve omitted the definition of Lebesgue integral for simple real measurable functions). Definition 1 (See [4,6,8,10,13]) f dm(x) = sup s dm(x) : 0 ≤ s ≤ f, s : A → R is simple measurable function . (2.1) A A Definition 2 (See [1,11,15]) Let sn : A → R be nondecreasing sequence of nonnegative simple measurable functions such that limn→∞ sn(x) = f (x) for every x ∈ A. Then: f dm(x) = lim s dm(x). (2.2) →∞ n A n A We stress the fact that the value of the Lebesgue integral of function f in the Definition 2 is independent of the choice of the sequence sn. The proof of this fact can be found in [1, p. 130, Theorem 17.5.] and in [11, p. 300], [15, p. 289, Theorem 4.6]. The equivalence of the two definitions follows from the Lebesgue’s Monotone convergence theorem (See [13, Theorem 11.28, p. 318], [6]) or it could be proved more elementary (sketch of the proof): ( ) From the basic properties of Lebesgue integral for simple measurable function we have that A sn dm x is ( ) ∞ nondecreasing sequence of real numbers. Hence the limit limn→∞ A sn dm x exists (finte or ). Suppose that ( ) = ≥ limn→∞ A sn dm x a 0 for some nondecreasing sequence sn of nonnegative simple measurable functions such that limn→∞ sn(x) = f (x) for every x ∈ A. Directly from properties of the least upper bound we have that sup s dm(x) : 0 ≤ s ≤ f, s : A → R is simple measurable function ≥ a. A When a =∞then the equivalence is obvious. Suppose that a < ∞ and that sup s dm(x) : 0 ≤ s ≤ f, s : A A → R is simple measurable function > a. That means that there exists simple measurable function s such that ≤ ≤ ( ) ( )> ( ) = { ( ), ( )} ∈ = , ,... 0 s f and A s x dm x a.Lettn x max s x sn x for all x A and for n 1 2 . Familiarizing Students with Definition of Lebesgue Integral... 365 It can be shown that tn are simple measurable functions for n = 1, 2,..., s(x) ≤ tn(x) ≤ tn+1(x) for all x ∈ A, n ∈ N and limn→∞ tn(x) = f (x) for all x ∈ A. < ( ) ( ) ≤ Because the limit in the Definition 2 is independent of the choice of the sequence sn and a A s x dm x ( ) ( ) A tn x dm x we have: a < s(x) dm(x) ≤ lim t (x) dm(x) = lim s (x) dm(x) →∞ n →∞ n A n A n A ( ) = which contradicts the assumption limn→∞ A sn dm x a. So it must be: sup s dm(x) : 0 ≤ s ≤ f, s is simple measurable function = lim s (x) dm(x). →∞ n A n A We will consider six examples of calculating Lebesgue integral directly from its definition. π/2 ( ) 3 Example: 0 sin x dm x Let us consider the function: f (x) = sin x, x ∈[0,π/2). For the rest of this example we will restrict our consideration to x ∈[0,π/2). π/2 ( ) We will calculate 0 sin x dm x applying directly Definition 1. Consider n 2−1 k s (x) = π χ + (x), x ∈[ ,π/ ), n = , ,... n sin n+1 k π, k 1 π for 0 2 1 2 2 n+1 n+1 k=0 2 2 n 2 k s¯ (x) = π χ − (x), x ∈[ ,π/ ), n = , ,.... and n sin n+1 k 1 π, k π for 0 2 1 2 2 n+1 n+1 k=1 2 2 Using Wolfram Mathematica we get the following Figs. 1 and 2: Fig. 1 Graphs of functions f , s1, s2. We can see that s1(x) ≤ s2(x) for x ∈[0,π/2) 366 W. Wojas, J. Krupa Fig. 2 Graphs of functions f , s2, s3. We can see that s2(x) ≤ s3(x) for x ∈[0,π/2) It is clear that sn, s¯n are sequences of nonnegative simple measurable functions and that sn ≤ f and s¯n ≥ f on [0,π/2) for all n = 1, 2,.... Using Wolfram Mathematica we get: Listing 1 Mathematica code: n − π 21 πk In[1]:= Simplify Sin 2n+1 2n+1 k=0 − − − − Out[1]= 2 2 nπ −1 + Cot 2 2 nπ In[2]:= Limit[%, n →∞] Out[2]= 1 So n π/2 2−1 π k 1 −2−n −2−n an = sn dm(x) = sin · π = 2 π(−1 + cot(2 π)) → 1 (3.1) 2n+1 2n+1 0 k=0 Similarly Listing 2 Mathematica code: n π 2 πk In[3]:= Simplify Sin 2n+1 2n+1 k= 1 − − − − Out[3]= 2 2 nπ 1 + Cot 2 2 nπ In[4]:= Limit[%, n →∞] Out[4]= 1 Familiarizing Students with Definition of Lebesgue Integral... 367 So n π/2 2 π k 1 −2−n −2−n a¯n = s¯n dm(x) = sin · π = 2 π 1 + cot(2 π) → 1(3.2) 2n+1 2n+1 0 k=1 + sin n 1 x sin n x n ( ) = 2 2 sin x = Of course we could use the following formulae: k=1 sin kx x and limx→0 1 instead of sin 2 x the code in Listings 1 and 2 to get the results in formulae (3.1) and (3.2). Using formulae (3.1) and (3.2), basic properties of least upper, greatest lower bounds and properties of Lebesgue integral of simple measurable functions we will prove that: π/2 sup s dm(x) : 0 ≤ s ≤ f, s is simple measurable function ≥ 1 (3.3) 0 and π/2 sup s dm(x) : 0 ≤ s ≤ f, s is simple measurable function ≤ 1. (3.4) 0 Inequalities (3.3) and (3.4)give π/2 sup s dm(x) : 0 ≤ s ≤ f, s is simple measurable function = 1, 0 π/ π/ which means that 2 f dm(x) = 2 sin x dm(x) = 1. 0 0 π/2 Because limn→∞ an = 1, so sup sn dm(x) : n = 1, 2,..
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