78 2. THE LEBESGUE INTEGRATION THEORY 11. Improper Riemann integrals 11.1. Preliminaries. The Riemann integral is defined for a bounded function f and a bounded region Ω. Can the Riemann integral be defined for f that is not bounded in a neighborhood of some point of Ω and (or) not bounded Ω? For example, a continuous function f(x)= e−x can be integrated over an unbounded interval [0, ) using the rule ∞ ∞ b e−x dx = lim e−x dx = lim (1 e−b) = 1 Z0 b→∞ Z0 b→∞ − Note f is integrable on every [0,b] because f is continuous. So, the rule makes sense. Similarly, the function f(x)= x−1/2 is not bounded on [0, 1], but it is continuous on every [a, 1] so it makes sense to define 1 dx 1 dx = lim = lim (2 2√a) = 2 + + Z0 √x a→0 Za √x a→0 − The above Riemann integrals are known as improper Riemann integral. So, a general idea is to reduce the integration domain to a bounded region so that the function in question is integrable on it, and after calculation of the integral over the reduced domain one has to take an appropriate limit with respect to parameters of the reduced domain in which the later becomes the region in question. The procedure is also often called a regularization of the improper Riemann integral. Does the value of the improper integral depend on the regulariza- tion? The answer is not straightforward, especially for higher dimen- sional integrals. Consider the function of two variables: x2 y2 f(x,y)= − (x2 + y2)2 Evidently it is defined everywhere except the origin. One can choose f(0, 0) to be any number. Regardless of the choice, f is not bounded in any neighborhood of the origin. Suppose one wants to integrate this function over Ω= (x,y) x2 + y2 1 , x 0 , y 0 { | ≤ ≥ ≥ } which is the part of the unit disk that lies in the positive quadrant. So, f is not bounded on Ω. In attempt to mimic a one-dimensional improper integral, let us take a subregion Ω Ω a ⊂ Ω = (x,y) a2 x2 + y2 1 , x 0 , y 0 a { | ≤ ≤ ≥ ≥ } 11. IMPROPER RIEMANN INTEGRALS 79 so that Ωa Ω as a 0. Using the polar coordinates it is not difficult to show that→ → f(x,y) dxdy = 0 ZZΩa Alternatively, the result follows from the symmetry argument. The region Ωa is symmetric under the reflection about the line y = x: (x,y) (y,x), whereas the integrand is skew-symmetric, f(x,y) = f(y,x→). Can one conclude that the improper Riemann integral of f over− Ω exists and is equal to zero? It is obvious that Ω can be obtained in many ways as the limit of subregions. For example, consider a collection of expanding subregions Ω Ω which are defined in polar coordinates k ⊂ x = r cos(θ) , y = r sin(θ) by the conditions Ω = (r, θ) a r 1 , β θ π/2 , k = 1, 2,... k { | k ≤ ≤ k ≤ ≤ } where a and β are positive sequences that converge to 0 mono- { k} { k} tonically. These regions coincide with Ωa with a = ak in which a small sector with the angle βk is removed. So, ∞ Ω Ω Ω , Ω = Ω k ⊂ k+1 ⊂ k k[=1 The latter union is a proper mathematical way of saying that Ωk “ap- proaches Ω and coincides with Ω in the limit k ”. Using polar coordinates → ∞ 1 π/2 r2 cos(2θ) 1 f(x,y) dxdy = 4 rdrdθ = sin(2βk)ln(ak) ZZΩk Zak Zβk r 2 The left side is an indeterminate form “0 ”. Its limit may not even exist and, even if it exists, it may have×∞ any value! Indeed, take −c/βk ak = e where c > 0 so that ak 0 monotonically if βk 0 monotonically. Then → → c sin(2β ) lim f(x,y) dxdy = lim k = c k→∞ ZZΩk − k→∞ 2βk − 2 −c/βk If ak = βk, then the limit is 0 and, if ak = e , c> 0, then the limit is . The reader is asked to verify that if the range of the polar angle −∞ in Ωk is restricted to the interval π/2 βk θ π/2, then the limit can be made arbitrary positive number− or≤ + ≤by a suitable choice ∞ of ak. So, the answer really depends on how the improper Riemann integral is regularized! In fact, a similar result can be established for the 80 2. THE LEBESGUE INTEGRATION THEORY integral of f over an unbounded region x2 + y2 1, x 0, y 0 (see Exercises). Naturally, one wants a definitive (or≥ unique)≥ value≥ of the integral, and, for this reason, the definition of the improper Riemann integral requires some amendments in order to be applicable for higher dimensional integrals. 11.2. Improper Riemann integrals. Let Ω RN bounded or unbounded. An exhaustion of Ω is a sequence of regions⊂ Ω ∞ such that { k}1 each Ωk is bounded, closed, and contained in Ω; • Ω contains Ω ; • k+1 k the union of all Ωk coincides with Ω except possibly a set of • measure zero. An example of an exhaustion was given in the previous section. Ifa bounded function f is Riemann integrable on a closed bounded region Ω, then it can be proved that lim f(x) dN x = f(x) dN x k→∞ ZΩk ZΩ which is nothing but the continuity property of the Riemann integral. If f is not bounded and/or Ω is not bounded, the improper integral is defined by this property. Definition 11.1. Let Ω ∞ be an exhaustion of Ω. Suppose that { k}1 a function f on Ω is Riemann integrable on each Ωk. Then the function f is said to be Riemann integrable on Ω if the limit lim f(x) dN x = f(x) dN x k→∞ ZΩk ZΩ exists and is independent of the choice of Ωk. In this case, the value of the limit is called an improper Riemann integral of f over Ω. 11.3. Improper integral of non-negative functions. It is not difficult to find a particular exhaustion and investigate the limit of the sequence of the integrals. Even if the limit is found to exist, one still has to show that it is independent of the choice of the exhaustion in order to prove that the found limit is the value of the improper integral. It is not possible to do so by constructing all possible exhaustions. How then can this be done? Suppose that f(x) is non-negative on Ω. For any exhaustion, the sequence of integrals is monotonically increasing 0 f(x) dN x f(x) dN x ≤ ZΩk ≤ ZΩk+1 11. IMPROPER RIEMANN INTEGRALS 81 by the positivity property of the Riemann integral and that Ωk Ωk+1. The sequence converges only if it is bounded. So, there are on⊂ly two possibilities: either the limit is a number N lim f(x) d x = If < k→∞ ZΩk ∞ or it is infinite, If = . Suppose that this limit exists, If < . Let 0 ∞ ∞ Ωk be another exhaustion of Ω. Then the sequence of the integrals is{ bounded:} N f(x) d x If 0 ZΩk ≤ 0 because f(x) 0 and Ωk Ω. Since the sequence is also monotonic, it converges ≥ ⊂ N 0 lim f(x) d x = If If k→∞ 0 ZΩk ≤ and its limit cannot exceed If . On the other hand, one can swap the roles of the exhaustions and use the same argument: N 0 0 f(x) d x If If If ZΩk | | ≤ ⇒ ≤ 0 because Ωk Ω and f(x) 0. Therefore If = If and the value of the limit does not⊂ depend on the≥ choice of the exhaustion. Theorem 11.1. (Improper integral for non-negative functions) Suppose that (i) f(x) 0 , x Ω; ≥ ∀0 ∈ (ii) Ωn and Ωn are exhaustions of Ω; { } { } 0 (iii) f is Riemann integrable on each Ωn and Ωn Then lim f(x) dN x = lim f(x) dN x n→∞ n→∞ 0 ZΩn ZΩn where the limit can also be + . ∞ Suppose that a bounded function f is Riemann integrable on a bounded, closed region Ω, then the non-negative functions 1 f±(x)= f(x) f(x) 0 2| |± ≥ are also integrable (because the absolute value of an integrable func- tion is integrable). The function f+(x) coincides with f(x) whenever f(x) 0 and vanishes otherwise, whereas f (x) coincides with f(x) ≥ − − 82 2. THE LEBESGUE INTEGRATION THEORY whenever f(x) 0 and vanishes otherwise. They can alternatively be written as ≤ f (x) = max f(x), 0 , f (x)= min f(x), 0 + { } − − { } Thus, any Riemann integrable function can be written as the difference of two non-negative integrable functions: f(x) = f (x) f (x) , + − − N N N f(x) d x = f+(x) d x f−(x) d x ZΩ ZΩ − ZΩ and vice versa (integrability of f implies integrability of f and f = ± | | f+ + f− by the linearity of the integral). Using the limit laws, the following theorem can easily be established from the above representation. Theorem 11.2. Suppose that the improper integrals of f± over Ω exist. Then the improper integral of f over Ω exists and can be com- puted in any exhaustion Ω of Ω. { n} Indeed, by the limit laws and the existence of the improper integral of f±, N N f(x) d x = lim f+(x) f−(x) d x n→∞ ZΩ ZΩn − N N = lim f+(x) d x lim f−(x) d x n→∞ n→∞ ZΩn − ZΩn N N = f+(x) d x f−(x) d x ZΩ − ZΩ and, by Theorem 11.1 the values of the limits in the right side of the equation do not depend on the choice of the exhaustion (or regulariza- tion) of the integrals.