Appendix I Integrating Step Functions In Chapter I, section 1, we were considering step functions of the form (1) where II>"" In were brick functions and AI> ... ,An were real coeffi­ cients. The integral of I was defined by the formula (2) In case of a single real variable, it is almost obvious that the value JI does not depend on the representation (2). In Chapter I, we admitted formula (2) for several variables, relying on analogy only. There are various methods to prove that formula precisely. In this Appendix we are going to give a proof which uses the concept of Heaviside functions. Since the geometrical intuition, above all in multidimensional case, might be decep­ tive, all the given arguments will be purely algebraic. 1. Heaviside functions The Heaviside function H(x) on the real line Rl admits the value 1 for x;;;o °and vanishes elsewhere; it thus can be defined on writing H(x) = {I for x ;;;0O, ° for x<O. By a Heaviside function H(x) in the plane R2, where x actually denotes points x = (~I> ~2) of R2, we understand the function which admits the value 1 for x ;;;0 0, i.e., for ~l ~ 0, ~2;;;o 0, and vanishes elsewhere. In symbols, H(x) = {I for x ;;;0O, (1.1) ° for x to. Note that we cannot use here the notation x < °(which would mean ~l <0, ~2<0) instead of x~O, because then H(x) would not be defined in the whole plane R 2 • Similarly, in Euclidean q-dimensional space R'l, the Heaviside function H(x), where x = (~I>' .. , ~) E R q, is defined on assuming their values equal to 1 for x;;;o 0, i.e., for ~l;;;o 0, ... , ~q ;;;0 0, and to °elsewhere. The notation (1.1) can be used also in this general case. The Heaviside functions belong to the simplest non-constant real valued functions; they admit the values 1 or °only. Note that H(x) = H(~l) ... H(~); 212 Appendix I. Integrating Step Functions this equality can be considered as another, but equivalent, definition of the q-dimensional Heaviside function, provided we already know what a one dimensional Heaviside function is. q If x = (~1. ... ,~) and a = (a1. ... , a q ) are points of R , then the differ­ ence x - a = (~1 - a1. ... , ~ - a q ) is another point of Rq. Evidently, the value of H at x - a equals to the product H(x - a) = H(~l - (1) ... H(~ - a q ). (1.2) This implies that for x~a, H(X-a)={~ for x;;p a. 2. Bricks and brick functions Let a = (a1. ... ,aq) and b = ({31. ... ,(3q) be points of Rq such that a < b, i.e., ai < (3i for i = 1, ... , q. By a brick [a, b) we understand the half closed interval a:so; x < b, i.e., the set of all points x = (~1. ... , ~) of Rq satisfying aj:SO; ~j < {3j for j = 1, ... , q. A point c = ('Yt. ••. , 'Yq) such that, for each j = 1, ... , q, we have 'Yj = aj or 'Yj = (3j is called a vertex of the brick [a, b). A brick has exactly 2q vertices. A vertex is even, if the number of the (3j'S appearing among the coordinates 'Yh ... , 'Yq is even. It is odd, if that number is odd. The point a is always an even vertex of [a, b). The point b is an even vertex, if the number q is even, and an odd vertex, if the number q is odd. A brick has exactly 2q - 1 even vertices and 2q - 1 odd vertices. Let lab denote a function, related to the brick [a, b), such that 1, if x is an even vertex, lab (X) = { -1, if x is an odd vertex, 0, if x is no vertex. It is easy to see that, if q is even and x is an even or odd vertex of [a, b), then -x is an even or odd vertex of [-b,-a), respectively. If q is odd and x is an even or odd vertex of [a, b), then -x is, conversely, an odd or even vertex of [-b, -a). This implies that (2.1) Let Kab denote the characteristic function of the brick [a, b), i.e., a function such that for a:SO;x< b, elsewhere. 3. Step functions 213 Characteristic functions of bricks are called brick functions. Evidently, eve:r:y q-dimensional brick function can be expressed as a product of one-dimensional brick functions Kab(x) = K"',I3,(gl)' .. K"'ql3q(~)' On the other hand, each of the one dimensional brick functions equals to a difference of shifted Heaviside functions Hence q Kab(x) = II [H(~j - elj) - H(~j - f3J]' j~l On expanding this product, we can write (2.2) where the sum is extended over all c E R q • Since the number of c such that lab(c) +0 is finite (equal to 2q), the sum consists, in fact, of a finite number of summands. It is interesting that, beside (2.2), also the formula (2.3) c holds. In fact, Kab(-x) = K-b,-a(x) = L Lb,-a(c)H(x-c) and, by (2.1), Kab(-X) = (-l)q L lab(-c)H(x - c) = (-l)q L lab (c)H(x + c). c c Substituting -x instead of x, we obtain hence formula (2.3). 3. Step functions By step functions we shall understand linear combinations AlH(x - al) + ... + AnH(X - an), (3.1) where the coefficients Ai are real numbers, and x, ai are points of Rq. The sum (3.1) can also be written in the form f AH(x-ai)' (3.2) i=1 It is almost obvious that a linear combination of step functions is another step function. However, it is awkward to write down an exact proof of this 214 Appendix I. Integrating Step Functions fact in symbols like (3.1) or (3.2). A more convenient notation is c where the summation runs over all points c E R'I. But we assume that there is only a finite number of points c for which Ac:f: O. This assumption will be kept throughout the whole Appendix 1. In particular, from (2.2) follows that each brick function Kab is a step function. Now, if is a linear combination of step functions c we can write f(x) = L P-b L AbcH(X - c) = L L P-bAbcH(x - c), b c c b because the number of non-vanishing summands is finite in each sum, and the order of summation can therefore be interchanged. Letting "c = L P-bAbc, we get b f(x) = L "cH(x - c). c This proves that a linear combination of step functions is a step function. Theorem 3. For each step function f there is only one representation in the form f(x) = L AcH(x - c). c Proof. Assume there is another representation f(x) = L P-cH(x - c). Let- c ting Pc = Ac - P-c, we get L PcH(x - c) = 0 for each x E R'I. If [a, b) is a c brick and lab its vertex function defined as in section 2 we can write x c or, interchanging the order of summation, L Pc L lab (x)H(x - c) = O. c x 4. Step functions of bounded carrier 215 If we apply now formula (2.3), with interchanged roles of x and c, we get after cancelling the factor (-I)q (3.3) c where Kab is the characteristic function of the brick [a, b). Let d be an arbitrary fixed point of Rq. Since there is only a finite number of points c such that Pc =l= 0, we can choose the brick [a, b) such that d is the only point in [a, b) with Pd = O. Then Kab(c) = 0 for each point c =l= d such that Pc =l= 0 and, subsequently, the sum in (3.3) reduces to the only summand PdKab(d) = Pd. Hence Pd = O. Since d is arbitrary, we have Pd = 0 and, in turn, Ad = ILd for all dE Rq. 4. Step functions of bounded carrier A function is said to be of bounded carrier, iff it vanishes outside a bounded set. Since a briCk is a bounded set and, moreover, each bounded set is included in a brick, we can also say that a function is of bounded carrier, iff it vanishes outside a brick. Evidently, each brick function is of bounded carrier. A linear combination of functions of bounded carriers is a function of bounded carrier. Theorem 4. A step function is of bounded carrier, iff it can be represented in the form (4.1) a where the Kaa * denote characteristic functions of pairwise disjoint bricks [a, a*). Proof. Since brick functions Kaa* are step functions, so is their linear combination (4.1). Moreover, since the number of the Kaa* in (4.1) is finite, and each of the Kaa* vanishes outside a bounded interval, there is a common bounded interval which all the Kaa* vanish outside. Subse­ quently, also the function f vanishes outisde that interval. In other words, f is of bounded carrier. It remains to prove that, conversely, if a step function f(x) = L ILcH(X - c) c is of bounded carrier, then it can be represented in the form (4.1). Denote by A the set of all points a such that, for each j = 1, ... ,q, the jth coordinate of a equals to the jth coordinate of some point c such that ILc =l= O.