Mathematical Methods for Physics and Engineering

Mathematical Methods for Physics and Engineering Lecture notes for PDEs Sergei V. Shabanov Department of Mathematics, University of Florida, Gainesville, FL 32611 USA

CHAPTER 1 The integration theory 1. Riemann integral over an interval 1.1. Bounded functions. An open interval is denoted by (a, b), a closed interval is denoted by [a, b]. Intervals [a, b) or (a, b] are sometimes called semi-open or semi-closed. A function f is continuous at x if lim x x f(x) = f(x ). A function f is continuous on an interval (a, b) if it is continuous at every point of the interval. The set of all functions continuous on (a, b) is denoted C ((a, b)). A continuous function of (a, b) is said to have a continuous extension to [a, b] if the limits lim x b f(x) = B, lim x a + f(x) = A exist and, in this case, f is continuously extended to [a, b] by the rule f(a) = A and f(b) = B. The set of all continuous functions on (a, b) that are continuously extendable to [a, b] is denoted by C ([a, b]). For example, the function f(x) = sin(x) on (, π/2) is continuously extendable to the endpoints. In this case, A = and B = 1 in the above notations. However, the functions g(x) = ln(x) and f(x) = sin( 1 x ) on (, 1) are not from C ([, 1]). Indeed, these functions are continuous on (, 1) and continuously extendable to x = 1: g(x) and f(x) sin(1) as x 1, but g(x) (not a number!) as x +, whereas the limit of sin( 1 x ) as x + does not even exist because sin( 1 x ) take values ±1 in any interval (, δ). Therefore no numerical values g() and f() exist to make g and f continuous at x =. A function f is said to have a maximum at c on an interval I if if f(c) f(x) for any x I. A function f is said to have a minimum at c on an interval I if if f(c) f(x) for any x I. The maximal and minimal values are also called extreme values of a function. Theorem 1.1. (Extreme values of a continuous function) Let f C ([a, b]). Then there are points c 1 and c 2 in [a, b] such that f(c 1 ) f(x) f(c 2 ), x [a, b]. In other words, a function continuous on a closed interval always attains its maximal and minimal values. 3

4 1. THE INTEGRATION THEORY The closedness of the interval is essential for this theorem to hold. For example, the function f(x) = 1/x is continuous on (, 1), but it has no maximal value on (, 1) and no minimal value despite that f(x) < 1 = f(1) because x = 1 does not belong to (, 1). Let f : I R be a bounded function, that is, m f(x) M for all x I. The numbers m and M are called a lower bound and an upper bound of f on I. For example, the function f(x) = 1/x if x and f() = is not bounded on any interval that contains. The function f(x) = sin(x)/x if x and f() = 1 is bounded on any interval because f(x) x / x = 1 because sin(x) x. Clearly, if M 1 M, then M 1 is also an upper bound of f and, if m 1 m, then m 1 is also a lower bound of f. The greatest lower bound is denoted by inf I f(x) pronounced as the infimum of f on I, and the least upper bound is denoted by supf(x) I pronounced as the supremum of f on I. In other words, for any positive number ε >, the number sup I f ε is not an upper bound, and inf I f + ε is not a lower bound. Thus, for any bounded function inf I f f(x) sup f, I x I In particular, if f C ([a, b]), then by Theorem 1.1 inf I f = minf = f(c 1 ), sup I for some c 1 and c 2 in [a, b]. I f = max f = f(c 2 ) I 1.2. Riemann integral. Suppose that f is a bounded function on [a, b]. A finite set of points a x x 1 x 2 x n 1 x n b is called a partition P of [a, b]. The intervals I j = [x j 1, x j ] are partition intervals, j = 1, 2,..., n, and δx j = x j x j 1 is the length of I j. Define M j = sup f(x), m j = inf f(x) I j I j and the upper and lower sums, respectively n U(P, f) = M j x j, L(P, f) = j=1 n m j x j j=1

1. RIEMANN INTEGRAL OVER AN INTERVAL 5 If an interval I 1 is contained in an interval I 2, then sup I 1 f sup I 2 f, inf I 1 f inf I 2 f, I 1 I 2. Therefore, the upper and lower sums for any partition P are bounded m(b a) L(P, f) U(P, f) M(b a) Furthermore, let P be a partition obtained from P by adding to the latter a point x such that x k 1 < x < x k for some k. Then the partition P has two intervals [x k 1, x ] and [x, x k ] that are subintervals of the interval [x k 1, x k ] in P. By the stated properties of the supremum and infimum on subintervals, L(P, f) L(P, f) L(P, f) U(P, f) The process of adding points to a partition is called a refinement. In the process of refinement the lower sums are increasing, while they remain bounded from above. Therefore there exists the least upper bound of all lower sums L(P, f) sup L(P, f) P where the supremum is taken over all partitions of [a, b]. Similarly, in the process of refinement the upper sums are decreasing, while they remain bounded from below. Therefore there exists the greatest lower bound for all upper sums inf U(P, f) U(P, f) P where the infimum is taken over all partitions of [a, b]. Definition 1.1. (Riemann integral over an interval) A bounded function f : [a, b] R is said to be Riemann integrable on [a, b] if inf P U(P, f) = sup P L(P, f) and in this case the number inf P U(P, f) = supl(p, f) = P b a f(x)dx is called the Riemann integral of f over [a, b]. The set of Riemann integrable functions on [a, b] is denoted by R([a, b]). It is clear that if the Riemann integral of f over [a, b] exists, then for any partition P L(P, f) b a f(x)dx U(P, f). In other words, the lower and upper sums can be viewed as estimates of the integral from below and above, respectively, and in the process of refinement of partition, the accuracy of the estimates is increasing.

6 1. THE INTEGRATION THEORY Definition 1.2. (Riemann sums) Let f be a bounded function on [a, b] and P is a partition of [a, b]. Choose a point x j [x j 1, x j ] in each partition interval. Points x j are called sample points. The sum n R(P, f) = f(x j) x j j=1 is called a Riemann sum of f for the partition P. By definition, the value of the Riemann sum depends on the choice of sample points. It follows from the inequality m j = inf I j f f(x j) sup f = M j I j m j x j f(x j) x j M j x j that for any partition P and any choice of sample points, Riemann sums are bounded L(P, f) R(P, f) U(P, f) Therefore if f is integrable on [a, b], then in the process of refinement the lower and upper sums converge to the integral of f and, hence, by the squeeze principle, so should do the Riemann sums. The limit is independent of the choice of sample points and the very process of refienment. For example, consider a uniform partition: x j = a + j x, j =, 1,..., n, x = b a n so that x j = x (all partition interval have the same length). Then n b f(x j) x = f(x)dx lim n j=1 for any integrable f and any choice of sample points. 1.3. Integrability and continuity. Theorem 1.2. (Integrability of continuous functions) A bounded function that is not continuous at finitely many points in [a, b] is Riemann integrable on [a, b]. In particular, every continuous function on [a, b] is integrable on [a, b] (note every continuous function is bounded on a closed interval: minf f(x) maxf). Furthermore, if f(x) = in [a, b] except a

1. RIEMANN INTEGRAL OVER AN INTERVAL 7 possibly for finitely many points where it has some non-zero values, then f is integrable on [a, b] and b a f(x)dx = if f(x) at finitely many points. A simple consequence of this observation f(x) g(x) at finitely many points for any integrable f and g on [a, b] b a f(x)dx = b a g(x)dx Theorem 1.3. (Integrability of composition) Suppose that f is Riemann integrable on [a, b] and m f(x) M. Suppose that g is a continuous function on [m, M]. Then the composition function h(x) = g(f(x)) is Riemann integrable on [a, b]. A further relation between continuity and integrability is studied in Section 1.6. 1.4. Properties of the Riemann integral. Suppose that f and g are Riemann integrable on [a, b] and m f(x) M. The following properties of the Riemann integral can be proved. It follows from the properties of the lower and upper sums that m(b a) b a f(x)dx M(b a). For any constants c 1 and c 2, the linear combination of integrable functions c 1 f(x) + c 2 g(x) is Riemann integrable and b ( b b c 1 f(x) + c 2 g(x) )dx = c 1 f(x)dx + c 2 g(x)dx a If f(x) g(x) in [a, b], then b a f(x)dx a b a g(x)dx Note that for any partition interval sup Ij f sup Ij g and, hence, U(P, f) U(P, g) from which the stated property follows. In particular, f(x) b a f(x)dx. Let f(x) M on [a, b]. Then b f(x)dx M(b a) a a

8 1. THE INTEGRATION THEORY The product of two Riemann integrable functions on [a, b] is also Riemann integrable on [a, b]: f R([a, b]), g R([a, b]) fg R([a, b]). The assertion is a consequence of Theorem 1.2. Indeed, take g(x) = x 2 which is continuous on any interval. Then the composition h(x) = (f(x)) 2 R([a, b]) if f R([a, b]). A linear combination of integrable functions is integrable and, therefore, the functions f ± g are integrable and so are their squares (f ±g) 2 R([a, b]). It follows from the identity fg = 1 ) ((f + g) 2 (f g) 2 4 that the product is a linear combination of integrable functions and, hence, is also integrable. The absolute value of a Riemann integrable function is Riemann integrable: and f R([a, b]) f R([a, b]) b a b f(x)dx f(x) dx Note that the converse is not true: the integrability of the absolute value of f does not imply integrability of f. An example is given in Section 1.5. For any c [a, b], b a f(x)dx = c a a f(x)dx + b c f(x)dx the relation is known as the additivity property of the Riemann integral. 1.5. An example of a Riemann non-integrable function. Let Q R be the set of all rational numbers. Consider the function { m, x Q f(x) = M, x / Q where M m. In any partition interval I j of non-zero length x j for a partition P of an interval [a, b], there exist rational and irrational numbers. Therefore m j = inf I j f = m, M j = sup I j f = M.

1. RIEMANN INTEGRAL OVER AN INTERVAL 9 Hence, L(P, f) = U(P, f) = n n m j x j = m x j = m(b a) j=1 n M j x j = M j=1 j=1 j=1 n x j = M(b a) for any partition P supl(p, f) = m(b a) M(b a) = inf U(P, f) P P which means that f is not Riemann integrable on any [a, b]. Note that f(x) is nowhere continuous because any interval (c ε, c + ε) contains rational and irrational numbers for any ε > and, therefore, the limit of f(x) as x c does not exist for any number c. Put m = M and M > in the above example. Then f(x) = M and therefore the absolute value of f is Riemann integrable on any interval b a f(x) dx = M(b a) where as f(x) is not integrable on any [a, b]. This example illustrates the assertion that the Riemann integrability of the absolute value of a function does not imply integrability of the function. 1.6. Riemann integrable functions. It is natural to ask what functions are Riemann integrable? This question is answered in this section. By definition, the length of an open interval (a, b) is b a. Definition 1.3. (Sets of measure zero in R) A set of real numbers is said to have measure if it can be covered by a union of open intervals of total length less than any preassigned positive number ε >. A point is evidently a set of measure zero. Indeed, a point can be covered by an interval (a, b) whose length b a cab be made arbitrary small. In particular, the open interval (a, b), the closed interval [a, b], and semi-open intervals (a, b] and [a, b) differ from one another by at most two points, that is, by a set of measure zero. So, the length of all these intervals should be b a (should coincide with the length of (a, b)).

1 1. THE INTEGRATION THEORY A countable union of sets of measure zero is also a set of measure zero. Let G = G n, G n R n=1 and all G n are sets of measure zero. Fix ε. Let I n the union of open intervals that covers G n. Since G n is of measure zero, the total length of intervals in I n can made arbitrary small. In particular, let the total length of intervals in I n be less than 2 n ε. The union of intervals in all I n covers the set G and the total length of these intervals does not exceed L 2 n ε = ε ( 1 + 1 ( 1 ) 2 ( 1 3 ) 2 2 + + + = 2 2) ε 2 1 1 1 = ε 2 n=1 Therefore the total length of intervals covering G can be made arbitrary small and, hence, G is a set of measure zero. Are sets consisting of at most countably many points all sets of measure zero? The answer is negative. There are uncountable collections of numbers which contain no interval. One of the most famous examples is the Cantor set. Let G = [, 1]. Remove the open interval ( 1 3, 2 3 ) and put G 1 = [, 1 3 ] [2 3, 1] Next, the middle thirds are removed from each of the intervals to obtain G 2 = [, 1 9 ] [2 9, 3 9 ] [6 9, 7 9 ] [8 9, 1] Continuing the procedure, a sequence of closed sets G n is obtained such that G 1 G 2 G n G n+1 and G n is the union of 2 n intervals, each of length 3 n, so that total length is ( 2 3 )n tends to zero as n. The set G = n=1 is called the Cantor set. It is clear that no interval of the form I k,m = (3 m (3k + 1), 3 m (3k + 2)), where k and m are positive integers, has a common point with G. Since every segment (a, b) contains a segment I k,m if 3 m < (b a)/6, the set G cannot contain any interval (a, b). So the Cantor set is a set of measure zero. One can also prove that G contains uncountably many elements. So, sets of measure zero in R have a more complex structure than merely a countable collection of numbers. G n

1. RIEMANN INTEGRAL OVER AN INTERVAL 11 Definition 1.4. (Limit point of a set) A point x is called a limit point of a set A R if an open interval (x ε, x + ε) contains a point y A and y x for any ε >. Definition 1.5. (Closed set) A set is closed if it contains all its limit points. Definition 1.6. (Closure of a set) The union of a set A with the set of its limits points is called the closure of A and denoted A. For example, any point of A = (, 1) is a limit point of A, but the points x = and x = 1 are also limit points of A. So the closure of A is A = [, 1]. If A is the collection of all integers, then A has no limit point (there is only one integer is a sufficiently small interval centered at that integer). So, A is a closed set and A = A. Let A = {x n } 1 be a converging sequence of numbers such that lim x n = x, x n x, n = 1, 2,.... n Then, by definition of the limit, for any ε > there exists an integer N such that x x n < ε for all n > N. This implies that any open interval centered at x contains infinitely many elements of the sequence and, hence, x is the limit point of A. The closure of A is A = A {x}. Let A = Q (the set of all rational numbers). Recall that R is the collection of elements that are limits of all converging sequences of rational numbers. Therefore real numbers are limit points of the set of rational numbers. Any positive rational number q is the ration of two positive integers q = n/m. Therefore the set E 2 of pairs of positive integers (n, m) contains no less elements than the set of all positive rational numbers. But the set E 2 is countable. In the positive quadrant of the coordinate plane, the elements of E 2 are points with integer-valued coordinates, or one can think of E 2 as a table in which n enumerates the columns, while m enumerates the rows. The grid points or elements of the table can be counted, for example, as (1, ) (, 1) (, 2) (1, 1) (2, ) (3, ) (2, 1) (1, 2) (, 3) (, 4) (1, 3) This procedure establishes one-to-one correspondence between positive integers and elements of E 2. Any subset of a countable set is at most countable (meaning that it can also be finite). So, the set of positive integers is countable. In other words, all positive rational numbers form a sequence {q n } 1. Similarly all negative rational numbers also

12 1. THE INTEGRATION THEORY form a sequence { q n } 1. Then all all rational numbers can be counted as follows: q 1 q 1 q 2 q 2 q 3 q 2 This shows that the set of rational numbers Q is countable and, therefore, has measure zero in R. In particular, rational numbers in an interval [a, b] form a subset of measure zero. However, if all limit points of this set are added, the whole interval [a, b] is recovered and its length is b a. This implies that the closure of Q is the whole real line, Q = R. Definition 1.7. (Dense subsets) A subset B of a set A R is said dense in A if for any element x A is a limit point of B. Rational numbers form a dense subset in R. Definition 1.8. (An interior point of a set) A point x of a set A R is said to be an interior point of A there exists an open interval (a, b) containing x and (a, b) A Definition 1.9. (An open set) A set A R is said to be open if all points of A are interior points. The following assertion is true. An open set is not a set measure zero. Indeed, a point of an open set is covered by an interval (a, b) that is contained in the set A. Any cover of A by open intervals must contain the interval (a, b) and, hence, the total length of all intervals of the cover cannot be made arbitrary small because because the interval (a, b) has finite length b a. A property is said hold almost everywhere (indicated by a.e.) in R if the set where this property does not hold has measure zero. For example, if M = in the example of the function given in Section 1.5, then f(x) =, a.e. meaning that, the function f is not zero only on a set of measure in R (on the set of rational numbers in this particular case). The function { 1, x [m, M] χ I (x) =, otherwise is called the characteristic function of the interval I = [m, M]. Clearly, it is bounded χ I (x) 1 on any interval. One can also say that χ I (x) is continuous a.e.. Indeed, this function has two jumpdiscontinuities at x = m and x = M. Two points form a set of measure

1. RIEMANN INTEGRAL OVER AN INTERVAL 13 zero. By Theorem 1.2, the characteristic function is integrable on any interval [a, b] and b a χ I (x)dx = L where L is the length of the interval of intersection of [a, b] with [m, M]. This can proved either directly from the definition of the Riemann integral (by studying the limits of the lower and upper sums under the process of refinement) or by calculating the limit of the Riemann sum for a uniform partition (in this case, the integrability is established first by Theorem 1.2). On the other hand, the function defined in Section 1.5 is continuous nowhere and not Riemann integrable on any interval. The following theorem establishes the most precise relation between Riemann integrability and continuity and, yet, provide a necessary and sufficient condition for Riemann integrability. Theorem 1.4. (Riemann integrable functions) Let f be a bounded function on [a, b]. Then f is Riemann integrable on [a, b] if and only if it is continuous almost everywhere on [a, b]. The value of the integral of a bounded function that is continuous a.e. in [a, b] does not depend on the values of f the set A on which it is not continuous. Indeed, the set A is of measure zero and can be covered by open intervals of total lengths less than any preassigned positive number. Therefore the contribution of the values of f on A to a Riemann sum is proportional to the total length of partition intervals covering the set A and, hence, should tend to zero in the process of refinement of the partition. It is important to stress that by altering values of a continuous function on a set of measure zero, the resulting function is not necessarily a function continuous almost everywhere. For example, let f(x) = 1 for all x R. A constant function is continuous everywhere. Let us alter the value of f at the set Q by setting f(x) = if x Q. Although Q is a set of measure zero, the resulting function is nowhere continuous and, by Theorem 1.4, is not Riemann integrable (see also Section 1.5). Consider a different set of measure zero on which f is altered. Let [a, b] = [, 1] and the function f is altered at the points x n = 1/n, n = 1, 2, 3,..., by setting f(x n ) =. The points A = {x n } form a countable set of points and any such set has measure zero. So a continuous function was altered again on a set of measure zero. However, in this case, the resulting function is Riemann integrable. Indeed, For any partition P of [a, b] L(P, f) U(P, f) 1

14 1. THE INTEGRATION THEORY because f(x) 1 and b a = 1. Fix ε > and consider a partition P whose two first points are x = and x 1 = ε. Since x n as n, the partition interval [, ε] contains points of A for any ε >. The interval [ε, 1] contains finitely many elements of A so that one can find a partition such that the partition intervals containing the elements of A in [ε, 1] have the total length less than ε. For any such partition U(P, f) = 1, L(P, f) = 1 2ε because inf f = on [, ε] and any partition interval containing elements of A in [ε, 1], while sup f = 1 on any partition interval. Therefore sup L(P, f) = 1 because ε > is arbitrary small. It follows from the inequality L(P, f) U(P, f) 1 that sup L(P, f) = inf U(P, f) = 1 and the function f is integrable on [, 1]. A continuous function f altered on the set {x n = 1/n} is continuous a.e. on [, 1]. These examples show that the Riemann integrability is sensitive to the structure of a set of measure zero on which a continuous function was altered. More generally, an alteration of values of a Riemann integrable function on a set of measure zero does not generally give a Riemann integrable function. This is a main deficiency of the Riemann integration theory, which is eliminated in the Lebesgue integration theory. 1.7. Improper Riemann integrals. The Riemann integrability is defined only for bounded functions on a finite interval [a, b]. A point c is a singular point of a function f if f is not bounded on any closed interval containing c. For example, the function f defined by the rule f(x) = 1/x if x and f() = has a singular point x =. Let c [a, b] be a singular point of f. Suppose that f is Riemann integrable on any closed subinterval of [a, b] that does not contain the point c. The limits lim b α a + α β lim β b a α lim α c + a f(x)dx if f(x)dx if f(x)dx + lim β c c = a c = b b β f(x)dx if a < c < b are called an improper Riemann integral of f over [a, b], provided the limits exist. If the limits do not exist, the improper Riemann integral

1. RIEMANN INTEGRAL OVER AN INTERVAL 15 is said to diverge. The same notation b a f(x)dx will be used for improper integrals. For example, let f(x) = x n and [a, b] = [, b]. If n <, then x = is singular point of f (the value of f at x = is not relevant). In this case, the improper Riemann integral is b lim α + lim α + α b x n dx = lim α n + 1 dx ( x = lim ln(b) ln(α) α + α + 1 The limit exists only if n > 1. So b x n dx = lim α + b α (b n+1 α n+1 ), n 1, ) x n dx = bn+1 n + 1, n > 1 and the improper Riemann integral diverges if n 1. Suppose that a function f is Riemann integrable on an interval [a, b] for any b > a. Then, if the limit lim b b a f(x)dx exists, it is called an improper Riemann integral of f over [a, ) and is denoted as lim b b a f(x)dx = a f(x)dx The Riemann integral over the real line is defined similarly: lim lim a b b a f(x)dx = f(x)dx provided the limits exist. Note that the limits are taken independently of one another. For example, let f be continuous on R and F be an antiderivative of f, that is F (x) = f(x), then by the fundamental theorem of calculus f(x)dx = lim a lim b b a f(x)dx = lim lim b a [ ] F(b) F(a)

16 1. THE INTEGRATION THEORY For instance, dx 1 + x 2 = lim a lim b tan 1 (x) = lim tan 1 (b) lim b a tan 1 (a) = π ( 2 π ) = π 2 The following theorems provide sufficient conditions for improper Riemann integrals to exist. Theorem 1.5. (Integrability with a singular point) Let f be Riemann integrable on [a, b] for any < a < b. Suppose that exist constants M and ν such that f(x) M, M >, n < 1, x (, b). xν Then f is Riemann integrable over [, b]. Proof. The point x = is possibly a singular point. So the integral of f over [, b] is understood as the improper integral. Let {a n } (, b) be a sequence of strictly positive numbers (not exceeding b) that converges to. Recall that the Cauchy criterion for sequences in R: A sequence {a n } in R converges if and only if for any ε > there exists an integer N such that b a a n a m < ε, n, m > N. In other words, the difference a n a m can be made arbitrary small for all large enough n and m (such sequences are also called Cauchy sequences). So, the objective is to show that the sequence of integrals I n = b a n f(x)dx is a Cauchy sequence and, hence, by the Cauchy criterion converges. Owing to an arbitrary choice of the sequence, it can then be concluded that the limit lim a + b a f(x)dx exists, which means that f is Riemann integrable on [, b]. Let a n a m. Using the properties of the Riemann integral, the following chain of

1. RIEMANN INTEGRAL OVER AN INTERVAL 17 inequalities is proved to hold I n I m b an a n f(x)dx a m = M 1 ν b f(x) dx M ( a 1 ν n f(x)dx = a m a 1 ν m an ) a m dx x ν an a m f(x)dx The case a m a n is studied similarly so that in either case I n I m M 1 ν a 1 ν n a 1 ν m Note that x 1 ν as x + because ν < 1. Therefore an 1 ν as n by continuity of the power function. Any numerical converging sequence is a Cauchy sequence and so is the sequence an 1 ν. Therefore the right side of the above inequality can be made arbitrary small for all sufficiently large n and m. Thus, the inequality implies that I n is a Cauchy sequence and, hence, converges. The theorem shows that the function should grow not too fast at a singular point in order to be Riemann integrability. It should be stressed that the theorem provides only a sufficient condition for integrability. For example f(x) = 1 x sin ( 1 x ) satisfies the inequality f(x) 1, x (, b) x which follows from sin(u) 1, but f is integrable on [, b]. Indeed, using the change of integration variable y = 1/x, dx = dy/y 2, b sin( 1 lim ) 1/a x sin(y) sin(y) dx = lim dy = dy a + a x a + 1/b y 1/b y

18 1. THE INTEGRATION THEORY The convergence of the improper integral over the unbounded interval is equivalent to the convergence of the series of positive terms 2π sin(x) x dx = = = = = n=1 2π(n+1) 2πn ( π(2n+1) + n=1 2πn π n=1 π n=1 n=1 n=1 sin(x) dx x ( sin(x) x + 2πn 2π(n+1) π(2n+1) ) sin(x) x sin(x) x + π(2n + 1) dx π sin(x) (x + 2πn)(x + π(2n + 1)) dx 1 2πn(2n + 1) π 1 πn(2n + 1) < sin(x)dx ) dx where the inequality is obtained by taking the minimal value of the denominator (x+2πn)(x+π(2n+1)) in the interval [, π] which occurs at x = (to maximize the fraction). The series converges by the comparison test (recall that the p series 1 n p converges for p > 1). Theorem 1.6. (Integrability over an unbounded interval) Let f be Riemann integrable on [a, b] for any a < b. Suppose that there exist constants M and ν such that f(x) M, M >, ν > 1, x > c xν for some c > a. Then the improper Riemann integral of f over [a, ) converges. Proof. Let {b n } [a, ) be a monotonic sequence, b n+1 b n, such that b n as n. Consider the sequence of integrals I n = bn a f(x)dx

1. RIEMANN INTEGRAL OVER AN INTERVAL 19 Let n > m. Then by the properties of the Riemann integral and hypotheses of the theorem: bn bm bn I n I m = f(x)dx f(x)dx = f(x)dx a bn b m = M(ν 1) a f(x) dx M ( 1 b ν 1 m bn b m 1 b ν 1 n For any n and m, the inequality can be written in the form I n I m M(ν 1) 1 1 dx x ν ) b ν 1 m bn ν 1 This inequality shows that the sequence I n is a Cauchy sequence and, hence, converges. Indeed, the sequence c n = 1/bn ν 1 converges to because ν > 1 and therefore c n is a Cauchy sequence. The latter implies that the right side of the above inequality can be made arbitrary small for all large enough n and m. Owing to the arbitrariness of the sequence b n, it is concluded that the improper Riemann integral in question converges. In particular, Theorem 1.6 implies that any continuous function that decreases faster than the inverse power function as x is integrable over an unbounded interval [a, ) (in the sense of the improper Riemann integral). 1.8. Fresnel integrals. It has to be stressed that Theorem 1.6 provides only a sufficient condition for the convergence of improper integrals. For example, the following improper integrals exist sin(x 2 )dx = cos(x 2 )dx = 1 π 2 2 while the integrands do not even decrease in the asymptotic region x, rather they oscillate between 1 and 1. These integrals are known as Fresnel integrals. The convergence of the Fresnel integrals can be established by means of Abel s test for alternating series, according to which an alternating series ( 1) n b n, b n, n converges if the sequence of non-negative terms b n converges to monotonically, that is, b n+1 b n for all n and lim n b n =. b m

2 1. THE INTEGRATION THEORY Put π(n+1) a n = πn sin(x 2 )dx = = ( 1) n π sin(y) 2 πn + y dy π(n+1) πn sin(u) 2 u du where the change of integration variables has been made, u = x 2 and u = y + πn. The sequence b n = π sin(y) 2 πn + y dy is non-negative and monotonically decreasing to with increasing n, that is, < b n+1 < b n and b n as n. Therefore by Abel s test for the alternating series ( 1) n b n converges. On the other hand, sin(x 2 )dx = lim N πn sin(x 2 )dx = lim N N a n = n= ( 1) n b n The other Fresnel integral can be treated similarly. The actual numerical value of the integrals can be found, e.g., by converting the complex integral a e ix2 dx = to the Gaussian integral lim a a a cos(x 2 )dx + i e ix2 dx = e iπ/4 lim a a a e t2 dt = sin(x 2 )dx n= π 2 eiπ/4 by means of the residue theorem for the Cauchy line integral of the analytic function e iz2 over the boundary of the wedge of the disk z a in which Rez Imz in the complex plane spanned by z (the integral over the arc part of the boundary, z = a, vanishes in the limit a ).

2. LEBESGUE INTEGRATION THEORY IN R 21 2. Lebesgue integration theory in R 2.1. Motivations. As has been already noted, the Riemann integrability depends on the structure of a set of measure zero on which the function is not continuous. Changing a function on a set of measure zero may destroy its integrability. The latter feature of the Riemann integral is rather awkward in some applications. For example, suppose that a quantity f depends on a quantity x and, hence, this dependence can be modeled by a function f(x). However, only mean values of f over sets of values of x are relevant for modeling f(x) (e.g., they are available from some data base). The mean value of a function on an interval is defined by 1 b f(x)dx, b a a Here b a is the length of the interval. Let S be a bounded set, S [ R, R] for some R > and χ S (x) be the characteristic function of the set S: { 1, x S χ S (x) =, x / S Then one can define the integral of f over S by R f(x)dx = χ S (x)f(x)dx S R and the length of S can be defined as L(S) = R R χ S (x)dx Then, by analogy with the case S = [a, b], the mean value value of f on S is the ratio of the above two quantities. Clearly if S = [a, b], then the length defined by the integral of the characteristic function is b a. If a set of measure zero is removed from the interval [a, b], then the length of the resulting set should not change because a set of measure zero can be covered by intervals whose total length is less than any preassigned positive number. However, the above definition of the length fails to reproduce this natural result because the very existence of the integral of the characteristic function depends on the structure of a set of measure removed from [a, b]. For example, rational numbers in [a, b] form a set of measure zero, but the function that is equal zero for a rational number and to 1 otherwise is shown to be non-integrable over any interval. Similarly, the product χ S f may not be integrable (because it amounts to changing values of an integrable f on a set of measure zero and the resulting function

22 1. THE INTEGRATION THEORY may not be integrable). Thus, the Riemann integral does not look like a suitable mathematical tool for modeling. Furthermore, suppose one wants to approximate a true dependence of f on x that has mean values on intervals (rather than on general sets to avoid the aforementioned length problem). Approximations of f form a sequence of functions f n (x) such that it converges to f(x) for every x as n. If f n (x) is integrable for every n, can one expect that the limit function f(x) is integrable, too? For example, consider f(x) that is obtained by taking the double limit: f(x) = lim n ( lim f nm(x) m ), f nm (x) = [cos(πn!x)] 2m, where n and m are positive integers. For every n and m, the function f nm (x) is a non-negative, continuous, and bounded function, f nm (x) 1 on any interval [a, b], and, hence, it is integrable and b a f nm (x)dx b a. If x is a rational number, then x = p/q for some integers p and q. Therefore n!x = pn!/q is an integer if n q so that a = [cos(πn!x)] 2 = 1 and, hence, f(x) = 1 (because a n = 1). If x is irrational, then n!x is not an integer for any n and therefore [cos(πn!x)] 2 < 1 so that f(x) = because a n as n if a < 1. Thus, the limit function has only two values 1 or if the argument is either rational or irrational, respectively. This function is not Riemann integrable (it is nowhere continuous). The integration theory developed by Lebesgue eliminates the dependence of the integral on values of a function on set of measure zero. In other words, in the Lebesgue integration theory, a function g obtained from an integrable function f by changing values of the latter on a set of measure zero is integrable and the Lebesgue integrals of f and g coincide. There are some additional advantages of the Lebesgue integral over Riemann integral. In particular, conditions under which a sequence of integrable functions converges to an integrable function are simpler in the Lebesgue theory. 2.2. Measurable functions and sets. Consider a collection of ordered numbers c n, c n < c n+1, which is finite or countable. Suppose that any interval (a, b) contains only finitely many numbers c n. Consider the intervals I n = (c n, c n+1 ). If the set of numbers c n contains the smallest number m, then the interval I = (, m) is added to the set of intervals I n. If the set of numbers c n contains the greatest number

2. LEBESGUE INTEGRATION THEORY IN R 23 M, then the interval I + = (M, ) is added to the set of intervals I n. The union of the closures I n = [c n, c n+1 ] (and possibly I = (, m] and I + = [M, )) is the whole real line R. If the collection of numbers c n has both the smallest and greatest numbers, m and M then this collection has finitely many elements. Indeed, m c n M for all n, but any interval (a, b) can contain only finitely many numbers c n. In particular, take a and b such that [m, M] (a, b). Then all c n are in (a, b) and, hence, there are only finitely many of them. For example, let c n = n where n is an integer. Then any interval (a, b) contains only finitely many integers. The real line R is the union of intervals [n, n+1] because every real x either lies between two integers or coincides with an integer. If c n is the collection of all non-negative integers, then R is the union of I = (, ] and all [n, n + 1], n =, 1,... However, the collection c n = 1, n = 1, 2,..., does not have n the property that any interval (a, b) can contains only finitely many elements c n because < c n < 2 for all n. Piecewise continuous functions. The constructed set of open intervals I n = (c n, c n+1 ) and possibly I ± has the following characteristic properties: it contains at most countably many elements; any bounded interval is covered by finitely many closed intervals I n ; intervals I n have no common points; the union of all closed intervals I n is the whole real line R Definition 2.1. (A piecewise continuous function) A function f : R R is said to be piecewise continuous on R if there exist a collection of finitely many or countably many open intervals I n with no common points such that any bounded interval is covered by finitely many closed intervals I n, and f C (I n ). Recall that f C (I n ) means that f is a continuous function on an open interval I n that is continuously extendable to the endpoints of I n. Let A n and B n, n = 1, 2,..., be two numerical sequences. The function, x < 1 (2.1) f(x) = B n (x n), n < x < n + 1 A n, x = n

24 1. THE INTEGRATION THEORY is a piecewise continuous function. The function is continuous on the open intervals I = (, 1) and I n = (n, n+1), n = 1, 2,..., because f is a constant function on I and a linear function on I n. The collection of intervals has the characteristic properties stated in the definition of a piecewise continuous function (the intervals have no common points and any bounded interval is covered by finitely many closed intervals). For example, let a < and b > 1. Then [a, b] is covered by the union of I = (, 1] and I n = [n, n + 1], n = 1, 2,..., N for some N > b. On every I n, the function B n (x n) is continuously extended to x = n and x = n + 1 by the limit values and B n, respectively. So, f C ([n, n + 1]) and C ((, 1]) (on (, 1), f(x) = and, hence, is continuously extended to x = 1 by. Instead of the linear functions B n (x n), one can take any continuous functions g n (x) C ([n, n + 1]) so that the function (2.1) is piecewise continuous. In particular, one can choose g n (x) = B n to obtain a piecewise constant function. A piecewise continuous function is not continuous on a set measure zero, or it is continuous almost everywhere. Indeed, the set of points c n is a countable set of points, that is, it is the countable union of sets of measure zero and such a set has measure zero. By Theorem 1.4, any piecewise continuous function is Riemann integrable on any [a, b]. The value of the Riemann integral does not depend on the values of a piecewise continuous function at the points where it is not continuous. For example, the Riemann integral of the function 2.1 over any [a, b] does not depend on the numbers A n = f(n). In particular, for [a, b] = [, 3] the integral is 3 f(x)dx = 1 dx + 2 1 B 1 (x 1)dx + = + 1 2 B 1 + 1 2 B 2 = 1 2 (B 1 + B 2 ) 3 2 B 2 (x 2)dx Note that in any interval [a, b] there only finitely many points at which a piecewise continuous function is not continuous. Therefore these points are contained in finitely many partition intervals. If f is not continuous at m points in [a, b], then in any partition at most 2m partition intervals contain these points (the maximal number of such intervals occurs when c n are in (a, b) and belong to the partition). Under a refinement process the contribution of these partition intervals into the upper and lower sums tends to zero because the length of partition intervals tends to zero. Therefore the limit of the upper and lower sums is independent of the values of the function at the points where it is not continuous.

2. LEBESGUE INTEGRATION THEORY IN R 25 Measurable functions and measurable sets. Consider a sequence {f n }, n = 1, 2,..., whose terms are real-valued functions on some subset Ω of real numbers: f n : Ω R R. Definition 2.11. (Pointwise convergence) A sequence {f n } of real-valued functions on Ω R is said to converge pointwise on a set Ω R to a function f if for any x Ω. lim f n(x) = f(x), n x Ω The pointwise convergence of a functional sequence means convergence of each numerical sequence {f n (x)} whose terms are values of the functions f n at each x from Ω. Example 2.1. Find the largest set on which the functional sequence f n (x) = x n, n = 1, 2,...,, converges pointwise. Solution: The sequence x n converges to if x < 1 and diverges if x > 1. If x = 1, then the sequence ( 1) n has no limit. If x = 1, the sequence of identical terms 1 n = 1 has the limit 1. Therefore lim f n(x) = f(x) = n {, 1 < x < 1 1, x = 1, x Ω = ( 1, 1] Definition 2.12. (Convergence almost everywhere) A sequence {f n } of real-valued functions on Ω R is said to converge almost everywhere on a set Ω R to a function f if lim f n(x) = f(x) n for almost all x Ω, that is, for all x from Ω except possibly for a subset of measure zero. To indicate that a functional sequence converges to a function almost everywhere, one writes lim f n(x) = f(x) a.e. n As before, a.e. means that the indicated property holds for all x except possibly for a set of measure zero. Example 2.2. Show that the functional sequence f n (x) = x 2 + x2 1 + x + x 2 2 (1 + x 2 ) + + x 2, n = 1, 2,... 2 (1 + x 2 ) n 1 converges almost everywhere in R to f(x) = 1 + x 2.

26 1. THE INTEGRATION THEORY Solution: Let g(x) be the limit function, that is, the sequence f n (x) converges to g(x) pointwise. Since f n () = for all n, g() =. For x, the number g(x) is the sum of the geometric series: Therefore g(x) = x 2 ( 1 + q + q 2 + ), < q = 1 1 + x 2 < 1 lim f n(x) = n x2 1 q = 1 + x2 = g(x), x The limit function g(x) does not coincide with f(x) = 1 + x 2 only at x = because g() = 1 = f(). But a single point is a set of measure zero, which implies that g(x) = f(x) a.e. and, hence, lim f n(x) = 1 + x 2 n a.e. The convergence almost every does not generally imply that a functional sequence converges for all values of the argument. It may not have a limit on a set of measure zero. Example 2.3. Show that the sequence f n (x) = [cos(πx)] n, n = 1, 2,..., converges to zero almost everywhere. Solution: Let g(x) be the limit function. If x m where m is an integer, then cos(πx) < 1 and f n (x) as n. So, g(x) = if x is not an integer. If x = 2m is an even integer, then f n (2m) = 1 or g(2m) = 1. If x = 2m + 1 is an odd integer then the sequence f n (2m+1) = ( 1) n has no limit and the limit function g is not defined at x = 2m + 1. However, the set of all odd integers is a set of measure zero (as a countable collection of points). Therefore the limit exists almost everywhere. Furthermore, the set of all even integers is also a set of measure zero. Therefore the value of the limit function at even integers is not relevant for the convergence almost everywhere. Thus, lim f n(x) = lim [cos(πx)] n = a.e. n n In what follows, it is always assumed that all functions are defined in the whole R. If f(x) is given by some algebraic rule that does not make sense for some x, then one can always extend f to the whole R by setting f to zero for all x for which the rule f(x) is not defined.

2. LEBESGUE INTEGRATION THEORY IN R 27 Definition 2.13. (A measurable function) A function f is called measurable if it coincides almost everywhere with the limit of an almost everywhere convergent sequence of piecewise continuous functions. Recall that the characteristic function of a set of real numbers is the function on R that takes the value 1 on the set and vanishes otherwise. Definition 2.14. (A measurable set) A subset S in R is called measurable if its characteristic function is measurable. Properties of measurable functions. Evidently, every piecewise continuous function f is measurable because one can take a sequence of piecewise continuous functions f n (x) = f(x) of identical terms which obviously converges to f(x). Suppose that f is a measurable function and g coincides with f almost everywhere. Then g is also measurable. Indeed, Let f n be a sequence of piecewise continuous functions that converges to f almost everywhere. Since f and g differ only on a set of measure zero, f n converges to g almost everywhere, too: f(x) is measurable f(x) = g(x) a.e. } g(x) is measurable More generally, one can prove that a function that is not continuous on a set of measure zero is measurable. Therefore every Riemann integrable function is measurable by Theorem 1.4. Furthermore, every function for which the improper Riemann integral exists is also measurable. So, the set of measurable functions contains all Riemann integrable functions (either in the proper or improper sense). There are measurable functions that are not Riemann integrable. For example, the Dirichlet function { 1, x Q f D (x) =, x / Q is not Riemann integrable on any interval. However, the set Q of rational numbers has measure zero in R. Therefore f D (x) = a.e., but any constant function and, in particular, g(x) = is measurable and, hence, so is the Dirichlet function. The latter implies that the set of rational numbers and any its subset are measurable sets. Using the basis limit laws, it is not difficult to show that } f(x) + g(x) is measurable f(x) is measurable f(x)g(x) is measurable g(x) is measurable f(x)/g(x), g(x), is measurable

28 1. THE INTEGRATION THEORY Note that if f n (x) and g n (x) are sequences of piecewise continuous functions, then the functions f n (x) + g n (x), f n (x)g n (x), and f n (x)/g n (x), g n (x), are also sequences of piecewise continuous functions, and the above assertion follows from the basic laws of limits. A set of measurable function is complete relative to addition and multiplication by a number. In other words, a linear combination of measurable functions is measurable. Sets with this property are called a linear space. Thus, the set of measurable functions is a linear space. Given two functions f and g, define the following functions max(f, g)(x) = min(f, g)(x) = { f(x), f(x) > g(x) g(x), f(x) g(x) { g(x), f(x) > g(x) f(x), f(x) g(x) One can prove that the functions max(f, g) and min(f, g) are measurable, if f and g are measurable. It follows that the absolute value f(x) = max(f, )(x) + min(f, )(x) of a measurable function f is measurable. A set of functions continuous on an interval [a, b] is a linear space because the sum of two continuous functions and a continuous function multiplied by a number are continuous. However a sequence of continuous functions can converge pointwise to a function that is not continuous (see Examples in the previous subsection). In contrast, the linear space of measurable functions has a different property: Theorem 2.7. (Completeness of the set of measurable functions) A function coinciding almost everywhere with the limit of an almost everywhere convergent sequence of measurable functions is measurable. Thus, the set of measurable functions (and sets) is quite large. It seems that every imaginable function is measurable. So, the question of interest: Are there non-measurable functions and sets? It appears that one can prove that they exist (using the axiom of choice), but no explicit example has been constructed so far! This suggests that all functions and sets that can possibly be used in applications or otherwise are measurable. For this reason, in what follows all sets are assumed to be measurable and all functions are assumed to be measurable and bounded almost everywhere.

2. LEBESGUE INTEGRATION THEORY IN R 29 2.3. Definition of the Lebesgue integral. In what follows, the following notation will be used: b f(x)dx = f(x)dx = lim lim f(x) dx a b assuming that the improper Riemann integral exists. To avoid confusion between Riemann and Lebesgue integrals, the Riemann integral will be denoted as b R f(x) dx or a R f(x) dx where the latter is the improper Riemann integral over the whole real line. Definition 2.15. (The space L + ) Let a real function f(x) be the limit of a non-decreasing sequence of piecewise continuous functions f n (x) such that the sequence of Riemann integrals is bounded: f n (x) f n+1 (x), n = 1, 2,...,, x R, R f n (x)dx M, n = 1, 2,..., for some number M. The limit of the non-decreasing sequence of Riemann integrals is called the Lebesgue integral of f and is denoted by the symbol f(x)dx so that f(x)dx = lim R f n (x)dx. n The set of all such functions is denoted by L +. Note that by the basic law for limits, any linear combination of functions from L + belongs to L +. So, L + is a linear space. One can prove that the Lebesgue integral of f L + does not depend on the choice of the sequence {f n }. So, in order to establish whether or not f is Lebesgue integrable, it is sufficient to find at least one non-decreasing sequence of piecewise continuous functions {f n } that converges to f almost everywhere and has a bounded sequence of Riemann integrals. The value of the Lebesgue integral is given by the limit of the sequence of Riemann integrals which always exists. Recall that a non-decreasing bounded numerical sequence always has the limit. Definition 2.16. (Lebesgue integral) A function f is called Lebesgue integrable if it can be represented as a

3 1. THE INTEGRATION THEORY the difference of two functions from the set L + : f(x) = f 1 (x) f 2 (x), f 1 L +, f 2 L + The number f 1 (x)dx f 2 (x)dx = f(x) dx is called the Lebesgue integral of the function f. The set of all Lebesgue integrable functions is denoted by L. In order to establish that a given function belongs to L or not, one has to find at least one pair of functions f 1 and f 2 from L + whose difference coincides with f. The value of the Lebesgue integral (if it exists) does not depend on the choice of f 1 and f 2. Indeed, suppose that f 1 (x) f 2 (x) = f(x) = g 1 (x) g 2 (x), f i L +, g i L +, i = 1, 2. It follows from the basic laws for limits, that the Lebesgue integral is additive: f L +, g L + f + g L + (f(x) + g(x))dx = f(x)dx + g(x)dx By the additivity of the Lebesgue integral for functions from L + and the relation f 1 + g 2 = g 1 + f 2, one infers that f 1 (x)dx f 2 (x)dx = g 1 (x)dx g 2 (x)dx Thus, the Lebesgue integral of f does not depend on the decomposition of f into the difference of two functions from L +. The Lebesgue integral of a complex-valued function f is the complex number f(x)dx = Re f(x)dx + i Im f(x)dx, provided the real and imaginary parts of f are Lebesgue integrable. Definition 2.17. (The Lebesgue integral over a set) A function f is said Lebesgue integrable on a measurable set S, f L(S), if fχ S L, where χ S is the characteristic function of S. The number f(x)χ S (x)dx = f(x) dx called the Lebesgue integral of f over S. S

2. LEBESGUE INTEGRATION THEORY IN R 31 If S is the union of non-intersecting sets S 1 and S 2, then S = S 1 S 2, S 1 S 2 = S f(x)dx = f(x)dx+ f(x)dx. S 1 S 2 Lebesgue integral of a continuous function. Let f be a continuous function. Then it is Lebesgue integrable on any interval (a, b), f L(a, b), and its Lebesgue integral coincides with the Riemann integral: f C (S), S = [a, b] f L(S), S b f(x)dx = R f(x) dx a Indeed, the sequence χ n (x) of the characteristic functions of the open intervals S n = (a + 1, b 1 ), n = 1, 2,..., (assuming that b a > 2, n n otherwise change n to n + N in S n for some large enough number N) converges to the characteristic function χ S of the interval S = [a, b] almost everywhere (except the end points). Suppose first that f(x) on [a, b]. Then the sequence of piecewise continuous functions f n (x) = χ n (x)f(x) converges to χ S (x)f(x) almost everywhere, it is non-decreasing: lim f n(x) = lim χ n (x)f(x) = χ n n S (x)f(x) a.e. χ n+1 (x) χ n (x) f n+1 (x) f n (x) because f(x). Furthermore the sequence of the Riemann integrals is non-decreasing and converges: lim R n χ n (x)f(x)dx = lim n R b 1/n a+1/n f(x)dx = R b a f(x) dx Thus, χ S f L + for any non-negative continuous f. Now recall that if f(x) is continuous, then its absolute value f(x) is also continuous. Define two non-negative continuous functions f ± (x) = 1 2 ( ) f(x) ± f(x). The function f + (x) vanishes if f(x) and f + (x) = f(x) if f(x) >. Similarly, f (x) = f(x) if f(x) < and f (x) = otherwise. It follows that f is the difference of two non-negative continuous functions and, hence, χ S f is the difference of two functions from L + and,