consists of two disjoint copies of X n, each scaled down by 1,

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1 Homework 4 Solutions, Real Analysis I, Fall, 200. (4) Let be a topological space and M be a σ-algebra on which contains all Borel sets. Let m, µ be two positive measures on M. Assume there is a constant c [0, ) so that µ(e) = c m(e) for each Borel set E. Assume that m is regular. Let A M and m(a) = 0. Prove µ(a) = 0. Solution: Assume A M and m(a) = 0. Let ɛ > 0. Then since m is (outer) regular, there is an open set V A so that m(v ) < ɛ/c. By assumption, then µ(v ) = c m(v ) < ɛ. Then since A V, µ(a) µ(v ) < ɛ. Since ɛ is arbitrary, we must have µ(a) = 0. (6) The Cantor set. (a) Define subsets of the unit interval O n, n inductively by O = ( 3, 2 3 ), O n+ = 3 O n ( 3 O n ), 0 = [0, ], n+ = n O n. Show that n is a disjoint union of closed intervals of each of length 3 n. Solution: We first prove the following inductive definition for n alone: () 0 = [0, ], n+ = 3 n ( 3 n ). To show this, compute n+ = n O n = [0, ] (O O 2 O n ) = ([0, ] O ) (O 2 O n ) = ([0, 3 ] [ 2 3, ]) { 3 O ( 3 O ) 3 O n 3 (O n )} = 3 ([0, ] O O n ) { 3 ([0, ] O O n ) } = 3 n ( 3 n ). Now we prove by induction that n is a disjoint union of 2 n closed subintervals of [0, ], each of length 3 n. This is clearly true for n = 0. Now we assume the statement is true for n. To prove it for n+, note that since n [0, ], n+ = 3 n ( 3 n + 2) 3 consists of two disjoint copies of n, each scaled down by, 3 contained in the intervals [0, ] and [ 2, ]. Thus the number 3 3 of intervals in n+ is double that of n : 2 2 n = 2 n+,

2 and each of these intervals has length of the length of 3 those in n : 3 3 n = 3 (n+). Thus the inductive case is checked and () is proved. (b) Define the Cantor set ( ) K = n = 0 O n. n= n= Show K is a compact subset of R with Lebesgue measure 0. Solution: By (), m( n ) = 2 n 3 n = ( 2 3 )n. Therefore, since n are nested measurable sets of finite measure, m(k) = lim n m( n ) = lim n ( 2 3 )n = 0. As an intersection of closed sets, K is closed. It is also bounded in R. Thus it is compact. (c) Show that K consists of all numbers in the unit interval [0, ] which have a base-3 expansion which only uses 0 s and 2 s. (One special point to recall is that (0. 2) 3 = (. 0) 3, where the bar signifies infinite repetition of the ternary digits. Therefore, the endpoint is legitimately in K.) Solution: We prove by induction that n consists of ternary numbers of the form (0.d d 2 d n d n+ ) 3 so that d i {0, 2} for i n. For the base case, we consider = [0, ] [ 2, ]. The first interval consists of all ternary numbers between 0 = (0.0 0) 3 to = (0.0 2) 3 3, while the second 3 3 interval consists all ternary numbers between 2 = (0.2 0) 3 3 and = (0.2 2) 3. This proves the base case. For the inductive case, assume the statement is true for n. Then if x = (0.d d 2 d n d n+ ) 3 n, then d i {0, 2} for i n. So the elements of n+ are all of the form x = (0.0d 3 d 2 d n d n+ ) 3 and x+ 2 = (0.2d 3 3 d 2 d n d n+ ) 3. This clearly shows that criterion for n+ is true. Thus the inductive proof is complete. Now since K is the intersection of all the n, we have that the elements of K can all be represented by ternary expansions all of whose digits are 0 or 2. (d) Construct a surjective map from K onto [0, ]. (Hint: Relate the base-3 expansions of elements of K to base-2 expansions of elements of [0, ].) Conclude that K is uncountable. 2

3 Solution: First of all we show that every element x of K can be uniquely determined by x = (0.d d 2 ) 3, where each d i {0, 2}. The only case of non-uniqueness in a ternary expansion is that (0.d d 2 d n d n 2) 3 = (0.d d 2 d n e n 0) 3, where d n 2 and e n = d n +. But when all the d i s are 0 or 2, this case can only happen if d n = 0 and so e n =, which is not allowed in our ternary expansions of elements of K. Thus the expansions involving only 0 s and 2 s are unique for elements of K. Now we define our surjective map from φ: K [0, ] by φ: (0.d d 2 ) 3 (0.f f 2 ) 2, where f i = 0 if d i = 0 and f i = if d i = 2. Since every element of [0, ] can be represented by such a base-2 representation, φ is surjective. (φ is not injective, since binary representations are not unique, for the same reasons as above.) Using the fact that [0, ] is uncountable, since φ: K [0, ] is surjective, then K must be uncountable as well. (Since if K were countable, there would be a one-to-one correspondence with [0, ] with a subset of K, which contradicts the fact that [0, ] is uncountable.) (7) Let f be an L function with respect to Lebesgue measure on R. Use the result of Chapter, Problem 2, to show that g(x) = x 0 f(t) dt is continuous. Solution: Let ɛ > 0. Then by Chapter, Problem 2, in Rudin, there is a δ > 0 so that if E is measurable and µ(e) < δ, then f dµ < ɛ. E For x y < δ, let I be the interval from x to y. Then g(x) g(y) = f(t) dt f(t) dt < ɛ. I This is exactly the condition that g is (uniformly) continuous. (9) Rudin, chapter 3, problem 4. Suppose f is a complex measurable function on, µ is a positive measure on, and ϕ(p) = f p dµ = f p p (0 < p < ). I 3

4 Let E = {p : ϕ(p) < }. Assume f > 0. (a) If r < p < s, r E, and s E, prove that p E. (b) Prove that if log ϕ is convex in the interior of E and that ϕ is continuous on E. Solution: The main idea is to use Hölder s inequality on integrals of the form f p dµ = f γ f p γ dµ. So for < α < and + =, we have α β ϕ(p) = f p dµ = f γ f p γ dµ = f γ α f p γ β ( ) ( f γα α dµ = ϕ(γα) α ϕ((p γ)β) β, ) f (p γ)β β dµ log ϕ(p) log ϕ(γα) + log ϕ((p γ)β). α β Now put r = γα and s = (p γ)β. (Notice given r < p < s, we can solve for α (, ) and γ (0, p).) In this case, Thus we have r + s = γ + (p γ) = p. α β log ϕ( r + s) log ϕ(r) + log ϕ(s) α β α β for (0, ) and + =. This is the definition of the α α β convexity of ϕ. ϕ is continuous on the interior of E by Theorem 3.2 Note that we have proved (a) along the way, since if φ(r) and φ(s) are finite, then the inequality implies φ(p) φ(r) α φ(s) β <. (c) By (a), E is connected. Is E necessarily open? Closed? Can E consist of a single point? Can E be any connected subset of (0, )? Solution: For Lebesgue measure on R, E can be any connected subset of (0, ). The basic functions to consider are of the form x l and x l log x m near x = 0 and x =. 4

5 Thus we will multiply them by characteristic functions of the form χ (0, 2 ] and χ [2, ). So define g l (x) = x l χ (0, 2 h l (x) = x l χ [2, ) (x), g l,m (x) = x l log x m χ (0, 2 h l,m (x) = x l log x m χ [2, ) (x). It is easy to check that g R l < if and only if l >. Similarly, h R l < if and only if l <. By comparing as x 0 + log x x ɛ for all ɛ > 0, we find that g R l,m is finite for l > and infinite for l <. For l =, compute by substituting u = log x R g,m = 2 0 x log x m dx = ln 2 u m du, which is finite if and only if m <. Similarly, we can show h R l,m is finite if l < and infinite for l >. If l =, this integral is finite if and only if m <. Note that g p l = g lp, h p l = h lp, g p l,m = g lp,mp, h p l,m = h lp,mp. For f = g, 2 + h, 2, then E = {}. Replacing this by f p gives E = {p}. To achieve E =, one may take f = g + h. To achieve E = [, p), one may take f = g + h, 2. To p get an interval of the form E = [a, b) for b = pa, take f a instead. To get E = (, p), take f = g p 5 + h, and f a gives E = (a, pa). To get E = [, p], take f = g p, 2 + h, 2. Then f a gives p E = [a, pa]. E = (, ) comes from f = h. E = [, ) comes from f = h, 2. E = (0, ) comes from f = g. E = (0, ] comes from f = g, 2. All other intervals with 0 or as one endpoint can be achieved by taking f a as above. To get E = (0, ) for f > 0, we may take f = e x. This exhausts all the connected subsets of (0, ). (For different measure spaces, the same is not true. For example, if µ() <, then Hölder s inequality applied to f r = f r shows that s E = r E for all r < s.)

6 (d) If r < p < s, prove that f p max( f r, f s ). Show that this implies the inclusion L r (µ) L s (µ) L p (µ). Solution: By the solution to (b) above, we have f p p = ϕ(p) ϕ(r) r α ϕ(s) = f α r f s β s 6 and so f p f r pα r f s pβ s. ( f p = Compute by our definition of r, s that r pα + s pβ = γ p + p γ p =. So if M = max( f r, f s ), we have f p M r pα M s pβ = M. So if f L r (µ) L s (µ), then f r and f s are finite. So for r < p < s, we have f p max( f r, f s ) <, and f L p (µ). Therefore, L r (µ) L s (µ) L p (µ). (e) Assume that f r < for some r < and prove that f p f as p. Solution: First of all, let α < f. We ll show that there is P so that if p P, then f p > α. First of all, choose β (α, f ). By the definition of f as the essential supremum, if E = {x f(x) > β}, we have µ(e) > 0. Then ) ( ) ( ) f p p dµ f p p dµ β p p dµ = β(µ(e)) p. E E But as p, β(µ(e)) p β > α, and so there is a P so that if p > P, we have f p > α. This allows us to settle the case of f = immediately, as the definition of lim p f p = = f is satisfied.

7 In the case f <, we need to use Hölder s inequality as above. Compute f p = = ( ( ( ) f p p dµ ) f r f p r p dµ = f r p r p p r f ) f r dµ ess sup( f p r p ) Now as p, f r p r p p r f f. So for any ɛ > 0, there is a P so that if p > P, f p < f +ɛ. The opposite inequality f p > f ɛ for p > P is proved above. Therefore, lim p f p = f in the case f < as well. (20) Rudin, chapter 3, problem 0. Suppose f n L p (µ) for n =, 2, 3,..., and f n f p 0 and f n g a.e., as n. What relation exists between f and g? Solution: f = g a.e. To prove this, note that Theorem 3.2, together with the assumption that f n f p 0, shows that there is a subsequence f ni so that f ni f a.e. as i. But we also assume that f n g a.e., and so the subsequence f ni g a.e. also. Therefore, f = g a.e. (22) Rudin, chapter 3, problem 2. Suppose µ() = and h: [0, ] is measurable. If A = h dµ, prove that + A2 + h2 dµ + A. If µ is Lebesgue measure on [0, ] and if h is continuous, h = f, the above inequalities have a simple geometric interpretation. From this, conjecture (for general ) under what conditions on h equality can hold in either of the above inequalities, and prove your conjecture. 7

8 Solution: To prove the right-hand inequality, note since µ() = that + A = ( + h) dµ. Since h 0, we have + h 2 + 2h + h 2 = + h. Thus integrating this inequality, we have + h2 dµ ( + h) dµ = + A. Thus, as long as the integrals are finite, they are equal if and only if [( + h) + h 2 ] dµ = 0. Since ( + h) + h 2 0, the inequality is true if and only if ( + h) + h 2 = 0 a.e., which is equivalent by the computation above to h = 0 a.e. If h is assumed to be continuous, equality is satisfied if and only if h = 0 everywhere. So if h = f, this equality is satisfied if and only if f is constant. For the other inequality, we will use Jensen s inequality. If φ(t) = + t 2, compute φ (t) = 2 ( + t2 ) 2 (2t) = t( + t 2 ) 2, φ (t) = ( + t 2 ) 2 + t( 2 )( + t2 ) 3 2 (2t) = ( + t 2 ) 3 2. Thus φ (t) > 0 for all real t and so φ is strictly convex on (, ). Thus, in the case h() [0, ), Jensen s inequality implies ( ) + A2 = φ h dµ φ(h) dµ = + h2 dµ. (If h has infinite values, then there are two cases: if h ( ) has positive measure, then all the relevant integrals are infinite and the inequalities hold. On the other hand, if h ( ) has measure zero, then we may redefine h to be 0 on this set without affecting the integrals. Then Jensen s inequality applies as above.) The proof of Jensen s inequality relies simply on integrating the pointwise inequality φ(h(x)) φ(t) β(h(x) t) 0, where t = h dµ and β = sup s<t 8 φ(t) φ(s). t s

9 Thus, when the integrals in Jensen s inequality are both finite, they are equal if and only if the pointwise statement φ(h(x)) φ(t) β(h(x) t) = 0 a.e. In our case φ is differentiable and β = φ (t). The condition is then that φ(h(x)) = φ(t) + φ (t)(h(x) t) a.e. This is the condition that φ(h(x)) is equal to the linear approximation of φ at t, evaluated at h(x). Since φ is strictly convex, this is satisfied only when h(x) = t = h dµ a.e. Thus h is constant almost everywhere. For h continuous, the condition on equality is just that h is constant, and if h = f, this is equivalent to f being a linear function. For the geometric interpretation of the inequality if = [0, ] and h = f is continuous, then + h2 dµ is the arc-length of the graph of f, A = f() f(0) is the height of a right triangle with hypotenuse from (0, f(0)) to (, f()). + A is the sum of the lengths of the two sides of this right triangle. + A2 is the length of the hypotenuse. Thus the left inequality is that the arc-length of the graph is the length of the secant line segment. We have proved above that this inequality is an equality if and only if f is a linear function. The right inequality is that the sum of the lengths of the two sides of the triangle from (0, f(0)) to (, f(0)) to (, f()) is the arc-length of the graph of f. This is an equality if and only if f is constant. 9