#### Ma635-Lecture Notes 8. Lebesgue measure

Stevens, MA 635
Excerpt: ... Ma 635. Real Analysis I. Lecture Notes VIII. MEASURE 8.1 Denition. Lebesgue measure of interval (a, b) of numerical line is L (a, b) = b a. 8.2 Denition. Lebesgue outer measure of a set E is (E) = inf L n=1 L (In ) : E n=1 In where the inmum is taken over all coverings of E by countable unions of intervals. 8.3 () = 0 L (countable set) = 0 L (a, b) = L (a, b) L (E) (F ) if E F L L (monotonicity) 8.4 Denition If {(ai , bi )} is a collection of pairwise disjoint open intervals then L i=1 (ai , bi ) := i=1 (bi ai ). The sum may be innite. 1 ...

#### Math301HW2

Charleston Law, MATH 301
Excerpt: ... Math 301 - Spring 2008 Homework 2: due 23 April (1) Skim through Chapter 1 in the text to make sure you are familiar with the concepts from todays class. (2) From todays lecture, you should recognize and understand the following vocabulary: set (proper) subset Venn Diagram intersection (relative) complement partition N, Z, R universal set power set disjoint index set ordered pair cardinality intervals union dierence pairwise disjoint sets Cartesian product (3) Your rst hand-in homework assignment is due at the beginning of class next Tuesday, 29 April. The problems are listed below, and all come from your textbook. You may discuss your work with your classmates, but you should neatly write your solution in your own words without consultation from your classmates. The underlined problems must be completed completely on your own; the problems in parentheses must be typeset using LaTeX. (1.7) 1.40 2.17 2.34 1.23 1.52 2.21 2.39 1.25 2.6 2.29 (2.50) 1.31 2.11 2.32 2.55 (4) There is a link on the ...

#### lecture_19

Harvey Mudd College, MATH 171
Excerpt: ... arose only later. With that in mind, let's apply some of our knowledge of group actions to the specific example of the symmetric group. Definition 12. Let Sn be a permutation. The cycle type of is a nonincreasing sequence of positive integers (l1 , l2 , . . . , lr ), l1 l2 lr such that may be decomposed into disjoint cycles length li , = 1 2 r Example 13. The cycle type of (123)(34) S4 is 4, as (123)(34) = (1234). The cycle type of (12)(34) is (2, 2), as this permutation is written as a product of disjoint transpositions. 3 |i | = li . Proposition 14. Every permutation is a product of pairwise disjoint cycles; moreover this decomposition is unique up to the order of the terms. Theorem 15. Two permutations in Sn are conjugate if and only if they are of the same cycle type. Lemma 16. For a permutation Sn , and = (a1 , . . . , ak ) a k-cycle in Sn , -1 = (a1 ), . . . , (ak ). Proof. Note that -1 (ai ) = (ai ) = (ai+1 ). Also, if b = a1 , . . . , ak , then -1 (b) = a1 , . . . , ak . Th ...

#### 5.1 Definitions

University of Illinois, Urbana Champaign, MATH 124
Excerpt: ... et. It is composed of / all the elements not in A. Denition If a set A has nitely many elements n(A) is dened to be the number of elements in A. Denition Sets are called disjoint if their intersection is the empty set. A collection of sets is called pairwise disjoint if you pick any two sets in the collection, the sets are disjoint. Denition A collection of sets is called a partition of another set A if their union is A and they are pairwise disjoint . Denition The power set of a set A is the set of all subsets of A. 1 ...

#### LY

W. Alabama, CO 442
Excerpt: ... Cuts and the Lucchesi-Younger Theorem Let D = (V, E) be a directed graph. For a subset W of V , the coboundary of W is the set (W ) of edges of D having precisely one end in W . The coboundary (W ) is directed if either every edge of (W ) has its tail in W or every edge of (W ) has its head in W . Lemma 1 D is strongly connected if and only if D has no directed coboundary (W ) where W = and W = V (D). Proof: exercise. A directed cut is an inclusion-wise minimal nonempty directed coboundary. (Inclusion-wise minimal means that no proper subset also has this property. Size-wise minimal means that no smaller set at all also has this property.) Let L be the set of directed cuts of D. A transversal of L is a subset t of E such that, for every L, t = . Clearly, if t is a transversal of L and P L is a set of pairwise disjoint directed cuts, then |t| |P|. (This is because t must have at least one edge from every cut in P and, because the cuts in P are pairwise disjoint , no two of ...

#### mat473notes_lecture12

ASU, MAT 473
Excerpt: ... Lecture 12, 2/23 Denition. Let X be a set. A non-empty collection of subsets of X is called an algebra (of sets) if it is closed under the formation of complements and intersections. An algebra of subsets of X is called a -algebra if it is closed under the formation of countable intersections. Note that it follows from DeMorgans laws that algebras are closed under unions and dierences, and contain X and , while -algebras are closed under countable unions. Denition. We let A denote the collection of all subsets of Rn that can be expressed as a nite disjoint union of boxes: A = {k Ei : E1 , . . . , Ek E are pairwise disjoint }. i=1 Lemma. A is an algebra. Proof. Let A, B A. Then A = k Ei and B = j=1 Fj , where the Ei are pairwise i=1 disjoint boxes, as are the Fj . Then E F = (i Ei ) (j Fj ) = i,j Ei Fj . It is clear that the collection {Ei Fj : 1 i k, 1 j } is pairwise disjoint . The sets in the collection are boxes by part (i) of th ...

#### 051019

Berkeley, MATH 202a
Excerpt: ... MATH 202A LECTURE NOTES FOR OCT 19, 2005 PROFESSOR DONALD SARASON 1. Semirings Denition 1.1 (Semiring). A semiring (Folland: an elementary family) on X is a family S P(X) such that (1) S (2) If A, B S, then A B S (3) If A, B S, then A \ B is a nite disjoint union of sets in S Example 1.2. A cell in R is [a, b) or . The cells in R form a semiring. Example 1.3. A cell in RN (N > 1) is or a product of cells RN form a semiring. N 1 [ai , bi ). Cells in Proposition 1.4. Let S be a semiring of X and let R be the family of nite disjoint unions of sets in S. Then R is the ring generated by S (i.e. in particular R is a ring). 2. Measure Denition 2.1 (Measure). Let S P(X) be a semiring. A measure on S is a function : S [0, ] (if maps to [0, ), we say is a nite measure) such that (1) () = 0 (2) If A1 , A2 , S are pairwise disjoint , then i=1 Ai = i=1 (Ai ) N Example 2.2. Let S be a semiring of cells in RN . ...

#### set1

USC, MATH 525
Excerpt: ... AB = . What is the set B? distributes over : A (BC) = (A B)(A C). And in fact, the ring is commutative: A B = B A. Note: rings need not have a multiplicative unit. What would it take for R to be a ring with unit? Problem 3. Is it possible for a -ring to be infinite yet countable? Give a careful proof of your statement-this problem is harder than it looks. Hint: If you could find countably many pairwise disjoint sets E1 , E2 , . . . in the -ring, you'd be home free: for each possible sequence of 0's and 1's would lead to a different countable union (a 0 in the i-th position meaning don't include Ei in the union, a 1 meaning do include it). Ah, but how to get them to be pairwise disjoint ? Problem 4. Let X be an uncountable set (say R for specificity) and let A be the -algebra of subsets of X which are either countable (finite or denumerable), or whose complements are countable. Does there exist a topology T on X such that A is the Borel field on X generated by T ? Hint: None. I don't know the answer ...

#### 24-RecDecAndLR-0328

University of Illinois, Urbana Champaign, CS 421
Excerpt: ... gain, you get the idea. So, along with that issue, we need a grammar that will allow us to gure out what rule we are going to use to parse with after only having had seen one token. One way that we saw for dealing with this was to extend the grammar and not make a choice until it was absolutely obvious which rule we needed to be using. But we need to test our grammar to see if it is even possible to write a parser for the grammar. 2.2 Pairwise Disjoint Test (slides 34-36 The test we are going to use is called the Pairwise disjoint edness Test. And note, that this is an approximate test, so we may not get perfect results every time. And this test simply checks to see if it can make a decision on which rule to use based on the rst token seen and not extending the grammar like we mentioned before. So this test is a little strict because it would have told us that the grammar we looked at last time would not have a parser, but in fact, we did have one. So for this test, we are going to calculate what we w ...

#### wk10

Allan Hancock College, MATH 2902
Excerpt: ... 4 are precisely 36, 18, 12, 9, 6, and 4, we conclude that these are the only possibilities for the number of right translates of X. Furthermore, if the number of right translates is 4 then, as we proved in the previous 4 lemma, the translates are pairwise disjoint . And by the proposition we stated above, any nonempty subset whose translate are pairwise disjoint must be a coset of a subgroup of G; so the result is proved. Each equivalence class for the relation on S consists of the right translates of an 9-element subset, and so the number of elements in the equivalence class must be 36, 18, 12, 9, 6, or 4. The total number of elements in S is the sum of the numbers of elements in the various equivalence classes; so #S = 36n1 + 18n2 + 12n3 + 9n4 + 6n5 + 4n6 for some nonnegative integers n1 , n2 , n3 , n4 , n5 and n6 . But the right hand side above can be written as 3(12n1 + 6n2 + 4n3 + 3n4 + 2n5 + n6 ) + n6 , which differs from n6 by a multiple of 3. So #S n6 (mod 3). But we have seen that #S = 94143280 ...

#### homework1sol

Kentucky, STA 320
Excerpt: ... ition of S. Prove that each outcome x S is contained in EXACTLY ONE partition set. Start with an arbitrary x S. We need to show that x is in exactly one Bi set. To do this, will will show that x cannot be in MORE than one Bi and that x cannot be in LESS than one Bi set. If this is true, that x must be in exactly one Bi set. To show x is in no more than one Bi set, let's argue by contradiction. Suppose we could find i and j such that x Bi and x Bj . Then x would be a member of Bi Bj . However, the Bi sets are pairwise disjoint (part of the definition of a partition) and thus Bi Bj = for all i and j. Thus, x CANNOT be a member of Bi Bj , resulting in a contradiction. Thus, x cannot be in any more than one Bi set. To show x is in at least one Bi set, again let's argue by contradiction. Suppose x were in NO Bi set. Then x would not be in n Bi by the definition of collective union. But by i=1 the definition of partition, n Bi = S, and x was assumed to be in S (S contains ALL i=1 outcomes). Therefore, we again ...

#### probability set up

Cornell, BTRY 4080
Excerpt: ... n Ai . i=1 An exercise is to show that (n Ai )c = n Ac and (n Ai )c = n Ac . These are called DeMori=1 i=1 i i=1 i=1 i gan's laws. There are no restrictions on S. The collection of events, F, must be a -field, which means that it satisfies the following: (i) , S is in F; (ii) if A is in F, then Ac is in F; (iii) if A1 , A2 , . . . are in F, then Ai and Ai are in F. i=1 i=1 Typically we will take F to be all subsets of S, and so (i)-(iii) are automatically satisfied. The only times we won't have F be all subsets is for technical reasons or when we talk about conditional expectation. So now we have a space S, a -field F, and we need to talk about what a probability is. There are three axioms: (1) 0 P(E) 1 for all events E. (2) P(S) = 1. (3) If the Ei are pairwise disjoint , P( Ei ) = i=1 i=1 P(Ei ). Pairwise disjoint means that Ei Ej = unless i = j. Note that probabilities are probabilities of subsets of S, not of points of S. However it is common to write P(x) for P({x}). Intuitively, the probability ...

#### hw5

UNL, SHARTKE 2
Excerpt: ... or two adjacent vertices. (Hint: Study s(u) - s(v) when u and v are adjacent.) b) Determine the maximum distance between the center and the barycenter in a tree of diameter d. (Example: In the tree below, the center is {x, y}, the barycenter is {z}, and the distance between them is 1.) x y z 3. Let G be the 3-regular graph with 4m vertices formed from m pairwise disjoint kites by adding m edges to link them in a ring, as shown on the right above for m = 6. Prove that (G) = 2m8m . 4. Count the following sets of trees with vertex set [n], giving two proofs for each: one using the Prfer correspondence and one by direct counting arguments. u a) trees that have 2 leaves. b) trees that have n - 2 leaves. 5. Prove that if the Graceful Tree Conjecture is true and T is a tree with m edges, then K2m decomposes into 2m - 1 copies of T . (Hint: If every tree T with m - 1 edges is graceful, then K2m-1 has a cyclically invariant decomposition into copie ...

#### mat473notes_lecture21

ASU, MAT 473
Excerpt: ... f . Then is countably additive on the -algebra of Lebesgue measurable sets. Proof. Let E1 , E2 , . . . be pairwise disjoint measurable sets. (i Ei ) = i Ei f f i Ei 1 = = () If f 0 then () = f L1 , i f i Ei , by pairwise disjoint ness, f Ei ). i = ( i f Ei = i (Ei ) by the monotone convergence theorem. If |f |i Ei |f | < , |f Ei | = i |f |Ei = by the monotone convergence theorem, and disjointness of the Ei . Then by the previous theorem, () = i Ei f = i (Ei ). Lebesgue integral vs. Riemann integral We want to see the relation between the two types of integrals we are now familiar b with. For a Riemann integrable function f : [a, b] R we will write a f dx for the Riemann integral, to distinguish it from the Lebesgue integral. Lets briey recall the denition of the Riemann integral. First recall the notion of step function: f : [a, b] R is a step function if there is a nite partition of [a, b] consisting o ...

#### Ch04

Mines, CSCI 400

Berkeley, MATH 105
Excerpt: ... ficulty than I'd anticipated; I expected problem (5) to be the tricky one but on the average people had more trouble with number (4). Let's fix firmly in our minds the fact that this result is false if it is only given that 0 fj (x) A (x) for all x. Indeed consider fj = [j,j+1] and A = (with (E, A, ) = (R, B, ). Then 0 fj (x) A (x) for all x, yet fj d = 1 while |A| = 0. Another example is fj (x) = -1/j for all x R. Now all hypotheses are satisfied with A = , except that fj are not nonnegative. Again the conclusion is false. Many solutions introduced a bit of notation and then simply claimed the conclusion without offering any analysis at all. In particular, if you didn't use the monotonicity hypothesis fj fj+1 and the hypothesis that fj 0, then you can't possibly have writtenwritten a correct proof! A more sophisticated fallacy, to which several victims fell prey, is to write fj out in the form Nj fj = k=1 cj,k Bj,k where cj,k are nonnegative scalars and the sets Bj,k are pairwise disjoint for each ...

#### Lecture3

Oregon, MATH 232
Excerpt: ... is a sample space and E is an event in this space, then the complement E c of E is the set E c = \ E, so E and E c are disjoint and = E E c . We are now going to restate the Theorem given on page 191 of the text, but we will omit the proof. However, we expect you to expend a little eort to follow and understand its proof it is not only fun in its own right, but it illustrates some very important properties of sets and probability functions. Theorem 3.1. Let P be a probability function on a nite sample space . Then (a) P () = 0; (b) P (E c ) = 1 P (E) for every event E in ; (c) P (E F ) = P (E) + P (F ) P (E F ) for every pair of events E and F ; (d) For every pairwise disjoint set of events E1 , E2 , . . . , En P (E1 E2 En ) = P (E1 ) + P (E2 ) + + P (En ). Here is one of the great advantages of having only a nite sample space . For then each single outcome is an event and each event E is the disjoint uni ...

#### M502F06test3

Boise State, M 502
Excerpt: ... Math 502 Test III Dr. Holmes December 8, 2006 This is a take-home exam. Do not consult anyone but me (I will be in my office MWTh 9 am - 3 pm at least, and you are encouraged to consult me) and do not consult any written resource but your textbook, the notes I have distributed and your own notes and papers. The exam is due at my office door (slide it under if I am not there), Thursday at 5 pm. 1 1. If real numbers are defined as Dedekind left sets in the rationals, and r and s are real numbers, then r s and r s are also real numbers. Describe them in numerical terms (without reference to set theory). (Every exam should have an easy question. . .) 2 2. The statement (A B) = A B is true. Explain why, in detail (show that any x in the first set is also in the second set, and vice versa). The statement (A B) = is false. Give a counterexample. A B 3 3. Prove by mathematical induction (on the finite size of the family P ) that any finite pairwise disjoint family P of nonempty sets has a choice set (i.e ...

#### 514s08hw3

Wesleyan, MATH 514
Excerpt: ... Math 514 Solution to HW #3 Spring, 2008 Chapter 1.4, Exercise #1 1a. Fix a line L R2 : We are to show that (L) = 0: We write L as a union of countably many intervals. For example, if we describe L as follows: for some xed non-zero vector ~ 2 R2 and some w 2 R2 ; v ~ L = fw + t~ j t 2 Rg ~ v then we can write L= [ fw + t~ j t 2 [n; n + 1]g ~ v n2Z By countable subadditivity, it is enough to show that for each n 2 Z; (fw + t~ j t 2 [n; n + 1)g) = ~ v 0: So we x n 2 Z. We denote fw + t~ j t 2 [n; n + 1)g by An : First we note ~ v that An is Lebesgue measurable, since it is closed. Now if ~ is a vector in R2 u that is not a multiple of ~ ; then the sets v fAn + s~ gs2(0;1) u are pairwise disjoint , and are all contained in a bounded set, namely the parallelogram P = fw + t~ + s~ j s; t 2 [0; 1]g : ~ v u Since P is bounded (and closed), we know that (P ) < 1: However, since is invariant under translation in R2 , each of the sets An +s~ u has the same measure, namely (An ). So if we choose a sequence of dist ...

#### 10_22

Oregon State, MTH 631
Excerpt: ... Section 22 Quotient Topology Def. A surjective map p : X Y is a quotient map if: 1 (U ) is open in X . U is open in Y iff p Def: Given a surjective map p : X saturated if p1 (p(C ) = C. Y , a subset C of X is Quotient Topology Def. Given p : X space Aset , the quotient topology on A is the topology that makes p a quotient map. Def. Let X be a partition of X into pairwise disjoint subsets. Let [x ] be the element of the partition containing x. Let p : X X be given by p(x ) = [x ]. If X is given the quotient topology induced by p, X is called a quotient space of X . Notation: X /G if G is the partition. Examples: 1/4 Mth 631 Fall 2008 2/4 Note: Being a quotient map is equivalent to being continuous and taking each saturated open set to an open set. Note: Open and closed maps are quotient maps. Examples: Mth 631 Fall 2008 Induced Maps Theorem: (Induced maps on Quotient Spaces) Let p : X Y be a quotient map. Let Z be a space and g : X Z be continuous and constant on each ...

#### hw6

Wisconsin, CS 787
Excerpt: ... CS787: Homework 6 Due date: Tuesday, April 21, 2009 Please note this is a short homework, due on Tuesday. 4/21 1. Give an approximation algorithm for the following problem: Given a directed graph G = (V, E), nd a subset of edges of maximum cardinalit ...

#### Final-205-09

UCSC, MATH 205
Excerpt: ... Final 205 Winter 2009 March 15, 2009 This is a closed book exam 1. Let A be an arbitrary collection of pairwise disjoint subsets of the real line I Suppose each subset has positive measure. Show that A is at most R. countable. 2. Let (Y, M, ) be a measure space, with a positive measure. Let fn be a sequence of measurable functions and let f : X I Suppose that R. n lim [{y Y : |fn (y) - f (y) }] = 0 holds for every > 0. Show that f is measurable. 3. Can you find a counterexample to the following statement: Suppose {fn } L , fn f, a.e and fn f , then limn fn - f = 0. 4. Let and be two measures on a -algebra M . If is a finite measure show that the following statements are equivalent. a. holds. b. For each sequence {Bm } of sets in M with limm (Bm ) = 0, it follows that limm (Bm ) = 0. c. For each > 0 there exists a > 0 (where depends on ), such that whenever B M satisfies (B) < then (B) < . 5. Suppose that and are sigma-finite measures on a -algebra M in a set X. Suppose tha ...

#### mid_2

Kentucky, MA 551
Excerpt: ... [Topology I, Second Midterm Exam, November 14, 2008] The exam consists of 5 questions, the last question being a bonus. Please try to solve as many questions in class as you can (no books, no notes). Your papers are due on Monday, November 17, 11 AM. Good luck! I hope you enjoy the problems. Problem 1 Let A = A0 A1 A 2 be a nested sequence of subspaces of X such that (i) X = n An and (ii) for each n 0 An interior of An+1 . Suppose for each n 0, there is a retraction rn : An+1 An . Prove that there is a retraction r : X A. we have connected. Problem 2 Let A Rn (n > 1) be a convex bounded set. Prove that Rn A is Problem 3 Let A0 , A1 , . . . , An be pairwise disjoint closed subsets of a normal space X . Prove that there is a continuous function f : X [0, n] satisfying f (Ai ) = {i} for all i = 0, 1, . . . , n. Problem 4 Let X be a normal space, let A, B be two non-empty disjoint closed subsets, and suppose f : A R and g : B R are co ...