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15. Caratheodory's Theorem; Nonmeasurable and Non-Borel Sets Theorem 15.1 (Carathodory). A RN is measurable if and only if e for each E RN . (E) = (E A) + (E Ac )
Lemma 15.2. For every measurable set B and every A B, Exercise 15.1. Prove Lemma
ASSIGNMENT 5 SOLUTIONS
MAT 473 SPRING 2012
Exercise 8.1. Suppose U Rn is open and f, g : U Rm are continuously differentiable on U . Prove that the function : U R defined by (x) = f (x), g(x) is continuously differentiable on U . Proof. By the Product Rul
ASSIGNMENT 4 SOLUTIONS
MAT 473 SPRING 2012
Exercise 4.2. Consider the function f : R2 R defined by 2 2 xy x - y if (x, y) = (0, 0) f (x, y) = x2 + y 2 0 if (x, y) = (0, 0).
Prove that all second-order partial derivatives of f exist, but the conclusion to
ASSIGNMENT 9 SOLUTIONS
MAT 473 SPRING 2012
Exercise 16.2. Prove that each of the following collections generates the -algebra B of Borel sets in R: (ii) cfw_(a, b] | a < b in R, and (vii) cfw_[t, +) | t R. Proof. Let C be the collection in part (ii), and
ASSIGNMENT 8 SOLUTIONS
MAT 473 SPRING 2012
Exercise 14.1. For A Rn and x Rn , prove that A + x L if and only if A L, in which case (A + x) = (A). Proof. For any x Rn , translation by x is continuous on Rn with continuous inverse (namely, translation by -x
ASSIGNMENT 1 SOLUTIONS
MAT 473 SPRING 2012
Problem 1 (1.1). Given that R is complete, prove that Rn is complete (as a metric space). Proof. First note that for any y = (y1 , y2 , . . . , yn ) Rn , and for any 1 i n, we have 2 2 2 |yi | y1 + y2 + + yn = y
ASSIGNMENT 3 SOLUTIONS
MAT 473 SPRING 2012
Exercise 4.2. Suppose f : R2 R is defined by 2 xy if (x, y) = (0, 0) f (x, y) = x2 + y 2 0 if (x, y) = (0, 0).
Show that all partial derivatives of f exist everywhere, and f is continuous everywhere, but f (0, 0)
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1. Euclidean Space, Linear Maps. f (x + h) - f (x) h0 h exists. When it does, we call the value of the limit the derivative of f at x, and denote it by f (x). Equivalently, f is differentiable at x if there exists c R such that lim lim Recall th
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2. Linear Maps, Operator Norm It's not obvious from the definition that T is finite for all T L(Rn , Rm ). If [T ] = (aij ) is the matrix representation of T with respect to the standard bases, define the 2-norm of T to be m n 2 T 2 = aij .
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14. Measurability Criteria Theorem 14.1. L is a -algebra of subsets of Rn , and is a measure on L. Proof. L0 , so L. If A L, then Ac M = M \A = M \(AM ) L0 for each M L0 , so Ac L. Finally, if Ai L for each i N and A = i Ai , then (A M ) (M
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4. Chain Rule. Partial Derivatives. Proposition 4.1 (Chain Rule). Suppose f : U Rn Rm is differentiable at x U and g : V Rm R is differentiable at f (x) V . Then g f is differentiable at x, with (g f ) (x) = g (f (x) f (x). Proof. Put y = f (x)
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16. Measurable Functions Recall that the -algebra B of Borel sets in R is the smallest -algebra of subsets of R which contains the open sets. Definition 16.1. Let M be a -algebra of subsets of a set X. A function f : X R is said to be M-measura
ASSIGNMENT 6 SOLUTIONS
MAT 473 SPRING 2012
Exercise 10.3. The equation x3 + y 3 + z 3 - 3xyz = 0 can be solved for z as a function of (x, y) near (-1, -1, 2). If z = g(x, y) denotes such a function, find the degree 2 Taylor polynomial of g for (x, y) near
ASSIGNMENT 7 SOLUTIONS
MAT 473 SPRING 2012
Exercise 12.2. Prove that if x Rn and P is a special polygon in Rn , then (P ) = (P +x). Conclude that (G) = (G + x) for every open set G in Rn . Solution. It is clear that I +x is a closed box if and only if I i
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17. Measurable Functions, Simple Functions Proposition 17.1. If f, g : X R are M-measurable, then f +g and f g are M-measurable. Proof. For each t R, cfw_x | f (x) + g(x) < t = cfw_x | f (x) < r cfw_x | g(x) < t - r .
rQ
Since f and g are M
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7. Taylor's Theorem; The Inverse Function Theorem. Theorem 7.1 (Taylor's Theorem). Let U Rn be open and convex, and let f : U R be C N . For each a U and each h such that a + h U , there exists c on the line segment from a to a + h such that
f
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8. Proof of The Inverse Function Theorem Proof. By Exercises 7.3 and 7.4, we may assume that a = 0, f (0) = 0, and f (0) = Id. Now, since f is continuous on U , (f - Id) = f - Id is, so there exists > 0 such that (f - Id) (x) - (f - Id) (0)
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Consider f : R2 R defined by f (x, y) = x2 + y 2 - 1, which is C 1 with f (x, y) = 2x 2y for all (x, y). The level set cfw_(x, y) | f (x, y) = 0 is a nice curve in R2 , namely the unit circle x2 + y 2 = 1. Moreover, if we look near the poi
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12. Lebesgue Measure: Open Sets and Compact Sets So far we have defined (A) for A empty, for any closed box A, and more generally for any special polygon A Rn . Definition 12.1. If G Rn is a nonempty open set, let (G) = supcfw_(P ) | P is a
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3. Invertible Operators, Derivatives For each T L(Rn ), let det T be the determinant of the n n matrix representing T with respect to the standard basis; this defines a map det : L(Rn ) R. In fact, det is continuous. To see this, note that t
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10. Proof of the Implicit Function Theorem. Proof. The formula for g (x) will follow from the rest of the theorem, as in Exercise 9.1, by defining : Rn Rn Rm by (x) = (x, g(x), so that (f )(x) = f (x, g(x) = 0 for all x I, and then using th
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13. Lebesgue measure: inner and outer measures Proposition 13.1. Let Ai be subsets of RN . (i) If A2 , then (A1 ) (A2 ) and (A1 ) (A2 ). A1 (ii) i=1 Ai i=1 (Ai ). (iii) If the Ai are disjoint, then Ai (Ai ). i=1 i=1
Proof. (i) is routine. For (
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5. Partial Derivatives, continued The extra condition on f that provides a converse for Proposition 4.5 is the continuity of its partial derivatives, in the sense that x Dj fi (x) should be continuous from U Rn into R for each i, j. (See Pr
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6. Mean Value Theorem. Clairaut's Theorem Recall the Mean Value Theorem from Calculus: if f : [a, b] R is continuous on [a, b] and differentiable on (a, b), then there exists c between a and b such that Disappointingly, we can have no such
ASSIGNMENT 2 SOLUTIONS
MAT 473 SPRING 2012
Exercise 3.2.
then T GL(n). (ii) Explain how it follows from this that GL(n) is an open subset of L(Rn ). Proof. For (i), apply Proposition 3.2 to T S -1 : T S -1 - Id = (T - S) S -1 T - SS -1 < 1, cfw_T L(Rn ) |
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11. Lebesgue Measure: Closed Boxes and Special Polygons Recall that relatively few functions f : R R are Riemann integrable: the continuous functions, and in a sense not too many more. Also, the Riemann integral behaves badly with respect to li