Math630-13 - Real Analysis Math 630 Homework Set#13 Chapter...

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Unformatted text preview: Real Analysis - Math 630 Homework Set #13 - Chapter 12 by Bobby Rohde 12-11-00 Problem 12.19 Let X Y be the set of positive integers, X , and let be the measure defined by setting E equal to the number of points in E if E is finite and if E is an infinite set. (This measure is called the counting measure.) State the Fubini and Tonelli Theorems explicitly for this case. Answer While I am unclear on exactly what suffices as response to this question, I am going to say when we can apply the Fubini and Tonelli theorems and then simpify the conclusions of the theorems as applicable to this specific case. Let f x, y be a real-valued function from X Y . First considering the Tonelli theorem, we know that the set of positive integers is may apply the Tonelli theorem provided that f is nonnegative everywhere. -finite, so we In the case of Fubini theorem we need that X Y f . However since X Y is a set of cardinality 0 we know that an enumeration of the points in X Y . Let ri i 1 be such an enumeration. Then we can equate f x, y = i 1 f ri If we further define ri x, y . n n x, y ri x, y , then the sequence of functions are each simple and clearly i 1 f ri converge to f. Suppose we consider the positive and negative parts of f. Then f n , and by application of the monotone convergence theorem, f limn n . However the right hand n side is equivalent to limn ri = i 1 f ri , since the counting measure over a i 1 f ri set of one element is 1. Thus saying that X Y f is equivalent to asking that and i 1 f ri . i 1 f ri MATH630-13.nb 2 Let us now consider the implication of the theorems which are of course exactly the same. For simplicity I will not state the results generate by interchange of x and y. (i) for almost all x the function fx defined by fx y Y. Well fx y i1 f x, y is a measurable function on i1 f ri ri x, y , where x is fixed. Which is equal to f x, i i y. = A . We need only show that A is well defined for all . However A Consider y : fx y is some subset of Y, and thus A Y , so A is measurable, since it is in the collection of measurable sets, and hence fx y is a measurable function by Proposition 11.5. (ii) Y f x, y y is a measurable function of X . Let us explicitly compute Y f x, y y = Y fx y = Y i 1 f x, i . Using the iy same caveat as above we may consider this as positive and negative pieces of f. Since integration is a linear operation, we will only consider nonnegative functions f. In the nonnegative case we may exploit the fact that the sum is the limit of a sequence of increasing simple function and apply the monotone convergence theorem to say that Y fx y = i1 f x, i i= i1 f x, i , since i 1, i. X , and = That the function i 1 f x, i is measurable follows immediately form the fact that x (X). The argument is the same as the 2nd paragraph in (i). (iii) XY f x, y XY f x, y Working from the conclusion of (ii), that Y fx y i 1 f x, i , we will then integrate with respect to , to find that X Y f x, y = i 1 j i f i, j i . (Note that technically we msut first consider this as the limit of finite sums, but the functions in the finite sums are increasing nonnegative functions whose limit is the sum, and the integrals are thus the same by Monotone convergence theorem.) However i 1 since is the counting measure, and thus = i 1 j i f i, j . However we also know that X Y f x, y = X Y f x, y = ri = since ri x, y i 1 f ri i 1 f ri , X Y i 1 f ri ri 1, i . However since the ri 's range over ever pair j, k of integers exactly once, this is identical to the above result. Hence XY f x, y XY f x, y = i1 ji f i, j . MATH630-13.nb 3 Problem 12.22 Let h and g be integrable functions on X and Y, and define f x, y h x g y . Then f is integrable on X Y and XY f Xh Yg Proof Without loss of generality we may suppose that f , g and h are nonnegative functions. If not we may rewrite them as the differance of their negative and positive parts and by the linearity of integration we only need to show the proof for nonnegative functions. Consider inf f XY XY f . If the integral exists then by definition of integration n i 1 ci XY Ei f is = a Ei XY, X Y. so Yg where x, y n i 1 ci x, y nonnegative simple function and each Ei However Xh h h g, inf m1 i 1 di h h X Hi h are integrable m1 i 1 di g functions, X and Gi they inf g each g Y g admit a representation Gi Y , with Hi m2 i 1 fi m2 i 1 fi x x and y y. g h Since f x, y h x g y m1 n i 1 ci Ei x, y i 1 di Hi x Hi x G j y = Hi G j x, y . that every m2 i 1 fi Gi y = m1 i1 is m2 j 1 di a possible function for f j Hi x G j y . However This implies that f admits a representation as a sum of n m1 m2 elements, with each Ei H j i Gk i and ci d j i fk i , where the j, k depend on i and each pair of j, k are used exactly once over the m1 m2 terms. m This that n 1 ci Ei X Y = i 11 m2 d j i fk i H j i Gk i X Y . We know i that each G j and Hk is in X and Y respectively since these are the basis sets, so this devolves to m1 m2 fk i H j i Gk i . However H j Gk is a rectangle in X Y , so i 1 dj i H j i Gk i = H j i Gk i fk i H j i Gk i = m the j i and k i this is just equal to i 11 di m1 m2 i 1 dj i m1 m2 fk i i 1 dj i m2 Hi i 1 fi Gi Hj i =X h Gk i , but by definition of . Yg MATH630-13.nb 4 Thus inf f XY XY inf h h X h inf g g Yg . Evaluating the infs we arrive at f , assuming the integral over f exists. However there is nothing Xh Yg specific in this argument to the fact we choose inf, except the last inequality, hence by the sup over simple functions less than the given function approach we can arrive at , if the integral over f exists. X Yf Xh Yg However Xh Yg what sup f XY we have inf f XY in fact Xh shown Yg is the integrals. XY that middle , inequality Xh is Yg always sup f true XY of the sups inf f XY and infs used to define , which implies that f exists and its value is Xh Yg . QED. Author's note: I am all but sure that there is a better way to do this, but I don't know what it is so I decided to do it based on first principles. MATH630-13.nb 5 Problem 12.24 The following example shows that we cannot remove the hypothesis that f be nonnegative from the Tonelli Theorem or that f be integrable from the Fubini Theorem. Let X Y be the positive integers and be the counting measure. Let f x, y 2 2 x, if x y 2 2 x, if x y 1 0, otherwise 22 x A Contradiction First it is neccesary to notice that f x, y A x, y : x y and B x, y : x y 1 . x, y 22 x B x, y , where In order to disprove the Fubini and Tonelli thereoms it suffices to show that property (iii) in each fails to hold. In other words I wish to show that X Y f x, y . Y X f x, y . So computing Y f x, y at a particular x, we have that Let us first consider X Y f x, y x x 22 Ax 22 Bx . Furthermore we know that for a fixed x, Ax has only one point determined by x y, and similarly Bx has only one point. Hence the counting measure evaluated on these sets is just 1. Thus Y f x, y = 2 2x 2 2 x 0. Hence 0. X Y f x, y Now let us consider Y X f x, y that we first evaluate X f x, y for a fixed y. Thus y y1 Ay 22 B y , note the substitution in the exponent in gives the integral as 2 2 terms of the fixed coordinate. Again each of these has exactly one item in its set, with the exception of B y at y 1, which is empty because x 0 is not in the domain of positive integers. So for y 2, X f x, y 2 2y 2 2y1 = 2y 1 1 2 y 1 . Also at y 1, 2 3 =2 21 X f x, y 2. Now in order to compute the integral over Y, it is not hard to see that by restricing the integral to the first N integers we are in fact integrating over a simple function. Thus Y X f x, y = N 3 3 y1 i1 2 =2 1 i Since we are using the 1 1 i2 2 1,2,... ,N 2 counting measure, each evaluation is in fact one. Simplifying the geometric series we have MATH630-13.nb 6 3 2 1 1 2 Fubini nor the Tonelli theoreoms will hold for this particular fucntion and choice of measure. QED. 1 2 = 1 2 . Thus YX f x, y 1 2 0 XY f x, y . Hence neither the Problem 12.25 The following example shows that we cannot remove the hypothesis that f be integrable from the Fubini Theorem or that and are -finite from the Tonelli Theorem: Let X Y be the interval 0, 1 , with the class of Borel sets. Let be the Lebesgue measure and the the counting measure. Then the diagonal x, y X Y : x y is measurable (is an , in fact), but its characteristic function fails to satisfy any of the equalities in condition (iii) of the Fubini and Tonelli Theorems. Contradiction Again we will seek an exception of the Fubini and Tonelli Theorems by computing the differant halves of part (iii) in each result and showing that they are not equal. Consider X Y x, y . First we wish to compute Y x, y , for a particular choice of x. x, y = y is only non-zero for the one value where x y, This means that Y x , but x and so since is the counting measure, we have that Y x, y = 1. So X Y x, y = =X 1. X1 Now consider Y X x, y . First we start with X x, y , for a particular value of y. y. Which evaluates the integral to y , which again is a set containing only the point where x However since is the Lesbegue measure this evaluates to 0. So we have that Y X x, y = Y0 0. Since 0 1, we have thus shown that fails to satisfy the (iii) property of the Fubini and Tunelli Theorems. QED. MATH630-13.nb 7 Problem 12.31 Let f be a nonnegative measurable function on two-dimensional Lebesgue measure on 2. Then m2 x, y t t t , y , and let m2 be the fx fx x :0 y fx m2 x, y :0 Let 0 m x: f x fx x. t . Then is a decreasing function and Proof m2 x, y :0 y fx m2 x, y :0 y fx fx x Essentially in the first part we are attempting to show that under the curve (with and without its boundary). fx x is the same thing as the area Let A x, y : 0 y f x and B x, y : 0 y f x . Let denote the Lesbegue measure. Consider , since is -finite and is everywhere nonnegative, we may A apply Tonelli Theorem to conclude that Ax . For any A A particular x we have that Ax is the interval 0, f x , and thus that Ax f x . Thus A = fx , but the integral with respect to Lesbegue measure is exactly what is intended by the = A by the definition of an integral notation f x x. Finally we know that A of a characteristic function, but m2 . Hence we have that m2 A = f x x. If we then note that B behaves exactly the same as A in the above proof since Bx = 0, f x , and 0, f x = f x . Thus the conclusion is equally valid that m2 B f x x. hence Bx MATH630-13.nb 8 0 t t fx x While in the above proof we made a construction that showed the area was the same as that under the curve f x , by exploiting a series of vertical bars, in the proof that follows we will attempt to demonstrate that it is equivalent to a series of horizontal bars. On the question that is a decreasing function, this follows immediately from that fact that mA mB, whenever A ! B, and the observation that if x x : f x t1 then x x : f x t2 , t2 t1 . Thus for t1 t2 , we have that x : f x t1 " x : f x t2 . So m x : f x t1 m x : f x t2 . Let x, t :f x t 0 Consider 0 t t = 0 m x : f x t t = 0 m t t = 0 x, t x t. The last equality coming from the fact that the measure of a cross-section is the same as the integral of its characteristic function. Since characteristic functions are bounded and both measure are taken with respect to the -finite real numbers, we may apply Tonelli thoerem to conclude that x, t x t = x, t t x. From here we need only observe that is the same 0 0 as A in the first section of this problem, upto the labelling of variables. Thus from above we may conclude that x, t t x = f x x. tt f x x. QED. 0 0 ...
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