[PDF] Chapter 11: The Riemann Integral





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Math 432 - Real Analysis II

(d) Use (c) to show that if f and g are integrable then max{f



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If f;g: [a;b] !R and both f and gare Riemann integrable then fgis Riemann integrable Proof Apply the Composition theorem The function h(x) = x2 is continuous on any nite interval Then h f= h(f) = f2 and h g= h(g) = g2 are Riemann integrable Also (f+g)2 is Riemann integrable (why?) Therefore fg= 1 2 [(f+ g)2 f2 g2] is Riemann integrable



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To prove that fg is integrable when f and g are simply note that fg= (1=2) (f +g)2f2g2!: Property (6) and the linearity of the integral then imply fg is integrable (8) If there exists m;M such that 00 such that m jf(x)j M for all x2[a;b] Note that 1 f(x) 1 f(y) = f(y)f(x) f



Searches related to if f and g are integrable then fg is integrable filetype:pdf

Theorem 1 If f and g are integrable on [a;b] then the product fg is integrable on [a;b] Proof First notice that fg = 1 2 (f +g)2 f2 g2: So it is enough to prove that if f is integrable then f2 is integrable If f is integrable then jfj is integrable Also f2 = jfj2 Let M = supjfj over [a;b] Suppose that P is a partition of [a;b] and I

Is the composite function f g integrable?

    For the composite function f ? g, He presented three cases: 1) both f and g are Riemann integrable; 2) f is continuous and g is Riemann integrable; 3) f is Riemann integrable and g is continuous. For case 1 there is a counterexample using Riemann function. For case 2 the proof of the integrability is straight forward.

Is F integrable?

    Since f ? is a probability density function, it is trivially integrable, so by the dominated convergence theorem, ? S g n + d ? ? 0 as n ? ?. But ? R g n d ? = 0 so ? R g n + d ? = ? R g n ? d ?.

What makes a function integrable?

    Assuming we're talking about Riemann integrals, in order for a function to be Riemann integrable, every sequence of Riemann sums must converge to the same limit. If you sometimes get different limits depending on the partitions of the interval you take, then the function is not integrable.

Is F[A;B]!R Riemann integrable?

    A bounded function f: [a;b] !R is Riemann integrable i fis continuous almost everywhere on [a;b]. Theorem 12 (Composition Theorem). Let f : [a;b] !R be Riemann integrable and f([a;b]) [c;d].

Chapter 11

The Riemann Integral

I know of some universities in England where the Lebesgue integral is taughtin the rst year of a mathematics degree instead of the Riemann integral, but I know of no universities in England where studentslearn the Lebesgue integral in the rst year of a mathematics degree. (Ap- proximate quotation attributed to T. W. Korner) Letf: [a;b]!Rbe a bounded (not necessarily continuous) function on a compact (closed, bounded) interval. We will dene what it means forfto be Riemann integrable on [a;b] and, in that case, dene its Riemann integralRb af. The integral offon [a;b] is a real number whose geometrical interpretation is the signed area under the graphy=f(x) foraxb. This number is also called the denite integral off. By integratingfover an interval [a;x] with varying right end-point, we get a function ofx, called an indenite integral off. The most important result about integration is the fundamental theorem of calculus, which states that integration and dierentiation are inverse operations in an appropriately understood sense. Among other things, this connection enables us to compute many integrals explicitly. We will prove the fundamental theorem in the next chapter. In this chapter, we dene the Riemann integral and prove some of its basic properties. Integrability is a less restrictive condition on a function than dierentiabil- ity. Generally speaking, integration makes functions smoother, while dierentiation makes functions rougher. For example, the indenite integral of every continuous function exists and is dierentiable, whereas the derivative of a continuous function need not exist (and typically doesn't). The Riemann integral is the simplest integral to dene, and it allows one to integrate every continuous function as well as some not-too-badly discontinuous functions. There are, however, many other types of integrals, the most important of which is the Lebesgue integral. The Lebesgue integral allows one to integrate unbounded or highly discontinuous functions whose Riemann integrals do not exist,205

20611. The Riemann Integraland it has better mathematical properties than the Riemann integral. The deni-

tion of the Lebesgue integral is more involved, requiring the use of measure theory, and we will not discuss it here. In any event, the Riemann integral is adequate for many purposes, and even if one needs the Lebesgue integral, it is best to understand the Riemann integral rst.

11.1. The supremum and inmum of functions

In this section we collect some results about the supremum and inmum of functions that we use to study Riemann integration. These results can be referred back to as needed. From Denition 6.11, the supremum or inmum of a function is the supremum or inmum of its range, and results about the supremum or inmum of sets translate immediately to results about functions. There are, however, a few dierences, which come from the fact that we often compare the values of functions at the same point, rather than all of their values simultaneously. Inequalities and operations on functions are dened pointwise as usual; for example, iff;g:A!R, thenfgmeans thatf(x)g(x) for everyx2A, and f+g:A!Ris dened by (f+g)(x) =f(x) +g(x).

Proposition 11.1.Suppose thatf;g:A!Randfg. Then

sup Afsup

Ag;infAfinfAg:

Proof.If supg=1, then supfsupg. Otherwise, iffgandgis bounded from above, then f(x)g(x)sup

Agfor everyx2A:

Thus,fis bounded from above by supAg, so supAfsupAg. Similarly,f g implies that sup

A(f)supA(g), so infAfinfAg.

Note thatfgdoes not imply that supAfinfAg; to get that conclusion, we need to know thatf(x)g(y) for allx;y2Aand use Proposition 2.24. Example 11.2.Denef;g: [0;1]!Rbyf(x) = 2x,g(x) = 2x+ 1. Thenf < g and sup [0;1]f= 2;inf[0;1]f= 0;sup [0;1]g= 3;inf[0;1]g= 1:

Thus, supf >infgeven thoughf < g.

As for sets, the supremum and inmum of functions do not, in general, preserve strict inequalities, and a function need not attain its supremum or inmum even if it exists.

Example 11.3.Denef: [0;1]!Rby

f(x) =( xif 0x <1,

0 ifx= 1.

Thenf <1 on [0;1] but sup[0;1]f= 1, and there is no pointx2[0;1] such that f(x) = 1.

11.1. The supremum and inmum of functions207Next, we consider the supremum and inmum of linear combinations of func-

tions. Multiplication of a function by a positive constant multiplies the inf or sup, while multiplication by a negative constant switches the inf and sup, Proposition 11.4.Suppose thatf:A!Ris a bounded function andc2R. If c0, then sup

Acf=csup

Af;infAcf=cinfAf:

Ifc <0, then

sup

Acf=cinfAf;infAcf=csup

Af: Proof.Apply Proposition 2.23 to the setfcf(x) :x2Ag=cff(x) :x2Ag.

For sums of functions, we get an inequality.

Proposition 11.5.Iff;g:A!Rare bounded functions, then sup

A(f+g)sup

Af+ sup

Ag;infA(f+g)infAf+ infAg:

Proof.Sincef(x)supAfandg(x)supAgfor everyx2[a;b], we have f(x) +g(x)sup

Af+ sup

Ag:

Thus,f+gis bounded from above by supAf+ supAg, so

sup

A(f+g)sup

Af+ sup

Ag: The proof for the inmum is analogous (or apply the result for the supremum to the functionsf,g). We may have strict inequality in Proposition 11.5 becausefandgmay take values close to their suprema (or inma) at dierent points. Example 11.6.Denef;g: [0;1]!Rbyf(x) =x,g(x) = 1x. Then sup [0;1]f= sup [0;1]g= sup [0;1](f+g) = 1; so sup(f+g) = 1 but supf+ supg= 2. Here,fattains its supremum at 1, while gattains its supremum at 0. Finally, we prove some inequalities that involve the absolute value. Proposition 11.7.Iff;g:A!Rare bounded functions, thensup Afsup Agsup

Ajfgj;infAfinfAgsup

Ajfgj:

Proof.Sincef=fg+gandfg jfgj, we get from Proposition 11.5 and

Proposition 11.1 that

sup Afsup

A(fg) + sup

Agsup

Ajfgj+ sup

Ag; so sup Afsup Agsup

Ajfgj:

20811. The Riemann IntegralExchangingfandgin this inequality, we get

sup Agsup Afsup

Ajfgj;

which implies that sup Afsup Agsup

Ajfgj:

Replacingfbyfandgbygin this inequality, we getinfAfinfAgsup

Ajfgj;

where we use the fact that sup(f) =inff. Proposition 11.8.Iff;g:A!Rare bounded functions such that jf(x)f(y)j jg(x)g(y)jfor allx;y2A; then sup

AfinfAfsup

AginfAg:

Proof.The condition implies that for allx;y2A, we have f(x)f(y) jg(x)g(y)j= max[g(x);g(y)]min[g(x);g(y)]sup

AginfAg;

which implies that supff(x)f(y) :x;y2Ag sup

AginfAg:

From Proposition 2.24, we have

supff(x)f(y) :x;y2Ag= sup

AfinfAf;

so the result follows.

11.2. Denition of the integral

The denition of the integral is more involved than the denition of the derivative. The derivative is approximated by dierence quotients, whereas the integral is approximated by upper and lower sums based on a partition of an interval. We say that two intervals are almost disjoint if they are disjoint or intersect only at a common endpoint. For example, the intervals [0;1] and [1;3] are almost disjoint, whereas the intervals [0;2] and [1;3] are not. Denition 11.9.LetIbe a nonempty, compact interval. A partition ofIis a nite collectionfI1;I2;:::;Ingof almost disjoint, nonempty, compact subintervals whose union isI. A partition of [a;b] with subintervalsIk= [xk1;xk] is determined by the set of endpoints of the intervals a=x0< x1< x2<< xn1< xn=b: Abusing notation, we will denote a partitionPeither by its intervals

P=fI1;I2;:::;Ing

11.2. Denition of the integral209or by the set of endpoints of the intervals

P=fx0;x1;x2;:::;xn1;xng:

We'll adopt either notation as convenient; the context should make it clear which one is being used. There is always one more endpoint than interval.

Example 11.10.The set of intervals

is a partition of [0;1]. The corresponding set of endpoints is f0;1=5;1=4;1=3;1=2;1g:

We denote the length of an intervalI= [a;b] by

jIj=ba: Note that the sum of the lengthsjIkj=xkxk1of the almost disjoint subintervals in a partitionfI1;I2;:::;Ingof an intervalIis equal to length of the whole interval. This is obvious geometrically; algebraically, it follows from the telescoping series nX k=1jIkj=nX k=1(xkxk1) =xnxn1+xn1xn2++x2x1+x1x0 =xnx0 =jIj: Suppose thatf: [a;b]!Ris a bounded function on the compact interval

I= [a;b] with

M= sup

If; m= infIf:

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