Math 563 Measure Theory Project 1 (Funky Functions Group) Luis Zerón, Sergey Dyachenko

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1 Math 563 Measure Theory Project (Funky Functions Group) Luis Zerón, Sergey Dyachenko 34 Let C and C be any two Cantor sets (constructed in Exercise 3) Show that there exists a function F: [,] [,] with the following properties: i) F is continuous and bijective ii) F is monotically increasing iii) F maps surjectively onto C C [ Hint: Copy the construction of the standard Cantor-Lebesgue function] Proof Let I, I, be the intervals removed from [,] in the construction of and let J, J, be the intervals deleted from [,] in the construction of C, arranged in the same order That is, I and J are the middle intervals removed in the first step; I and J are the left middles and I 3, J 3 are the right middles removed in the second step; and so on Then map the interval I n onto the interval J n linearly and increasingly, for n=,, (see graph below) C J 3 J J I I I 3 Note that [,] \ C is dense in [,] Indeed, take x [,] ; if x [,] \ C there is nothing to prove If x C then by Exercise 4 there is a sequence { x n } n such that x n [,] \ C and x n x Thus any neighborhood of x contains one element of [,] \ C (in fact infinitely many) different from x Hence x is a limit point of [,] \ C This proves that [,] \ C is dense in [,] A similar argument shows that [,] \ C is also dense in [,]

2 Hence F: [,] \ C [,] \ C is defined and strictly increasing on a dense subset of [,] Since the range of F is also dense on [,], the domain of F can be extended to [,] so that F is increasing and continuous on [,] with range [,] We extend the domain of definition of F to all of [,] by putting and F() =, F() =, F(x ) = Sup{ F(: x [,] \ C, x < x} We see that F is now defined on all of [,] since F is bounded on [,] \ C Property (ii) F is monotonically increasing Let x [,] such that x < We must consider several cases: x, x a) Both and are in [,] \ C Then, by construction, F x ) < F( ) x x x C b) is in [,] \ and x is in C Then F(x ) Sup{ F(: x [,] \ C, x < x} = F(x ) ( x c) x is in C and x is in [,] \ C Then x is in some interval Ik and therefore for all x in [,] \ C such that x < x we have that x < x < x and x is in some interval I j at the left of I k and thus, by construction, F ( < F( for all x in [,] \ C with x < x But this implies that Sup{ F(: x [,] \ C, x < x} F(x ), ie, F(x ) F(x ) Thus, in any case, F is monotonically increasing In fact F is strictly increasing, since in cases (a) and (b) above, if F(x ) = F(x ), then this value is in some interval J k, but F is injective there which implies that x = x ; but this contradicts that x < x Property (i) F is continuous and bijective Let us show that F is continuous Since F is monotonically increasing on [,], then by a well known theorem of real analysis (see for example Rudin s Principle of Mathematical Analysis, Theorem 49, p95) we have that for any x (,), + F( x ) : lim F( = inf{ F( : x [,] \ C, x > x} = + x x = lim x x F( x ) : F( = sup{ F( : x [,] \ C, x < x}

3 We claim that sup{ F( : x [,] \ C, x < x} = inf{ F ( : x [,] \ C, x > x} Notice that if for any x < x and any t > x, such that x, t [,] \ C, we have F( F( t) and F sup{ F( : x [,] \ C, x < } ( x which implies that sup{ F( : x [,] \ C, x < x} F( t) for any t > x and therefore sup{ F( : x [,] \ C, x < x } inf{ F( t) : t [,] \ C, t > x = inf{ F( : x [,] \ C, x > x For the reverse inequality, suppose that sup{ F ( : x < x} < inf{ F( : x > x} Since [,] \ C is dense in [,] then any sub-interval of [,] contains points of [,] \ C, ie, values of F in [,] \ C Then the sub-interval (sup{ F ( : x < x}, inf{ F( : x > x}) must contain values of F Let y = F( x ) be one of such values If x >, then y F x ) inf{ F( : x > }, a contradiction If x = ( x x < x F( sup{ F( : x x}, again a contradiction y = < Thus we conclude that sup{ F( : x [,] \ C, x < x} = inf{ F ( : x [,] \ C, x > x} ie, lim F( = lim F( = F( x) x x x x + This proves that F is continuous at x } }, then By a similar argument we can prove that F is continuous at and at Therefore F is continuous on [,] Let us now show that F is bijective Since F is increasing and continuous on [,] then F is injective For the surjectivity, take any y in [,] Since f is continuous on [,], then by the intermediate value theorem there is an x in [,] such that F( = y Hence F maps [,] onto [,] and therefore F: [,] [,] is bijective

4 Property (iii) F maps surjectively onto C C Notice that: - by construction F: [,] \ C [,] \ C is bijective - by property (i), F: [,] [,] is bijective Therefore F: C C is bijective, ie, F maps C surjectively onto C Remark Since F is a continuous one-one mapping of the compact metric space [,] onto the metric space [,], then the inverse mapping F defined by F ( F( ) = x is a continuous mapping (see for example Rudin s Principles of Mathematical Analysis, Theorem 47, p9) This implies that any two Cantor sets are homeomorphic 35 Give an example of a measurable function f and a continuous function Φ so that f o Φ is non-measurable [Hint: Let Φ : C C as in Exercise 34, with m( C ) > and m( C ) = Let N C be non-measurable, and take f = χ Φ(N ) ] Use the construction in the hint to show that there exists a Lebesgue measurable set that is not a Borel set Proof Following the hint, let Φ : C C as in Exercise 34, with m( C ) > and m( C ) = Since ( C ) >, there exists a non-measurable set N C Now take f m =, x Φ( N) χ = Φ ( N (, x Φ( N ) ( ) Since Φ( N) C ( by 34(iii) ) and m ( C ) =, then m ( Φ( N)) =, so Φ(N) is measurable and therefore f is measurable Now consider f o Φ Notice that f Φ ( o ) ((, + )) = N is non-measurable and therefore f o Φ is nonmeasurable ** f Φ ( o ) ((, + )) = N Indeed: ** x N Φ( Φ( N) x ( f o Φ) f ( Φ( ) = (, + ) x ( f ((, + )) ( f o Φ)( (, + ) o Φ) f ( Φ( ) (, + ) f ( Φ( ) = Φ( Φ( N) x Φ ((, + )) ( Φ( N)) = N

5 For the second statement, let Φ : C C as in Exercise 34, with m( C ) = and m( C ) > Since m( C ) >, there exists a non-measurable set V C Then A = Φ ( V ) C and since m( C ) = then m( A) = and therefore A is measurable Now suppose that A is a Borel set Then Φ( A ) = Φ( Φ ( V )) = V is also a Borel set since Φ is a continuous one-one function and continuous one-one functions map Borel sets onto Borel sets (see proof of this below) But if V is a Borel set, then V is measurable, a contradiction ** Theorem If f is a one-one continuous mapping of R onto R, then f maps Borel sets onto Borel sets Proof Recall that if f is a one-one map of X onto Y, then for A, B X, f ( A B) = f ( A) f ( B) and f ( A \ B) = f ( A) \ f ( B) Now set U ={ A R : f ( A) B}, where B denotes the collection of Borel sets We claim that U is a σ-algebra Indeed, if A U, then A c c U, because f ( A ) = f ( R \ A) = R \ f ( A) Also, if A } is a sequence of U U n= n A n= n U n = sets in U, then f ( A ) = f ( ), which shows that A U Now since f is strictly monotonic (because it is one-one and continuous), we have that f([a,b]) = [f(a),f(b)] or f([a,b]) = [f(b),f(a)] Hence U contains all closed intervals and therefore all Borel sets n { n

6 Exercise 36 This exercise provides an example of a measurable function f on [, ] such that every function g equivalent to f (in the sense that f and g differ only on a set of measure zero) is discontinuous at every point (a) Construct a measurable set E [, ] such that for any non-empty open sub-interval I in [, ], both sets E I and E c I have positive measure (b) Show that f = χ E has the property that whenever g( = f( ae x, then g must be discontinuous at every point in [, ] Solution to part (a) Let {u k } k= be a sequence of all intervals with rational endpoints We will denote u = (α, β ), where α, β are rational numbers in [, ] Because u is an open interval, we can construct a Cantor-like set A of positive measure entirely inside u Since Cantor (and Cantor-like) sets are nowhere dense, we can always find an open interval (a, b) inside u, such that (a, b) is disjoint from A Therefore we can construct another Cantor-like set A, such that A (a, b) u Now let s take an open interval u = (α, β ) We can construct Cantor-like set A 3 entirely inside u such that A 3 is disjoint from all previosly constructed sets and, by the same argument we construct A 4 inside u such that it is disjoint from all previous Cantor-like sets Notice that sets A k and A j satisfy the following conditions: (i)a k and A j are disjoint and nowhere dense, (ii)a k u k and A k u k Repeating this process indefinitely we get a sequence A k that has the properties: (i) A k and A k are disjoint and nowhere dense k N, (ii) A k u k and A k u k Let s take E = A k then I k : u k I k= E I = ( A k ) I A k I = Ak () k= To show the second inequality let s notice that m(e I) m(a k ) > () A k is not a subset of E k=

7 This means that A k E c k= E c I ( A k ) I A k I = Ak (3) k= This concludes the proof of part (a) Solution to part (b) m(e c I) m(a k ) > (4) To show that the claim is valid, we will need to use the following Theorem(without proof): The Baire Cathegory Theorem: Let X be a complete metric space, (a) If {U n } is a sequence of open dense subsets of X, then U n is dense in X, (b) X is not a countable union of nowhere dense sets Let f( = χ E and let g( = f( almost everywhere By definition of continuity g( is continuous if and only if: lim g( = g( (5) x x Without loss of generality, let s assume that x I, where I is an open interval in [, ] Let F be the set where f( is different from g(, so that m(f ) = We see that x has no other option, but to satisfy one of the following: () x (E I F c ) () x (E c I F c ) (3) x F As we already proved m(e I) > and m(e c I) > Using the result of part (a) we know that m(e I) > and m(e c I) > therefore g( = f( on E I F c and E c I F c

8 First, let s show that if case () is true, then g is discontinuos: Suppose x E F c Because E c = ( A k ) c = A c k, where Ac k are dense in [, ], according to Baire Cathegory Theorem, E c is dense in [, ] and E c F c is also dense in [, ] Because E c F c is dense we can find a sequence {x n } E c F c converging to x, then g(x n ) = f(x n ) = χ E (x n ), but g( = f( = χ E ( = This implies discontinuity of g Now, let s show that if case () is true, then g is discontinuos, too: Suppose x (E c I F c ) Then g( = f( = χ E ( Now let s find a sequence {x n } n= convergent to x, such that g(x n ) does not converge to g( Among the sequence of intervals with rational endpoints {u k }, let s choose a subsequence of shrinking, nested intervals {u kj }, so that: α kj < α kj+ x β kj+ < β kj, (6) this can be done, because as we know that, any number can be approximated arbitrarily close by a rational number Constructing Cantor-like sets A kj in each of the open intervals {u kj } as suggested in part (a), and by choosing a sequence {x kj } j A kj u kj, we obtain a sequence, converging to x, that is entirely inside E F c Note g(x kj ) = χ E ( j and so g(x kj ) does not converge to g(, this implies that g is discontinuous Therefore g( = f( ae x is discontinuous almost everywhere Using case () and case () we notice for case (3) the following fact: Suppose x F, then in any neiborhood of x we have points from both E and E c and thus, the sequence constructed by taking points from each of the sets E and E c such that x n E and x n E c we construct a sequence convergent to x, that has no limit This contradicts the definition of continuity The conclusion for these three cases gives us the result that g has to be discontinuous everywhere 3

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