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ANALYSIS 3 PROBLEMS FOR WEEK 5 SOLUTIONS Problem 1. Recall that for a function f : X → R the following three statements are equivalent: (i) f is measurable, (ii) {f < a} is measurable for each a ∈ R, (iii) {f > a} is measurable for each a ∈ R. First of all let us prove that |f | is measurable, indeed this follows from {|f | < a} = ∅, if a < 0 and {|f | < a} = {f < a} ∩ {f > −a}, if a ≥ 0 combined with (ii) and (iii) and the measurability of f . Next assume that p > 0. We have {|f |p > a} = X, and if a ≤ 0 1 {|f |p > a} = {|f | > a p }, if a > 0, and so by (iii) and the measurability of |f | it follows that |f |p is measurable for every p > 0. Finally assume that f (x) 6= 0 for every x ∈ X. Note that o n1 > 0 = {f > 0} f n1 o n 1o >a = f < , if a > 0 f a and n1 o n 1o >a = f < ∪ {f > 0}, if a < 0 f a so that using (iii) again we conclude. Problem 2. You have seen in class that B(R2 ) = B(R) ⊗ B(R), so if we prove that A is open in R2 with the euclidean topology we are done. This is trivial because the function R2 3 (x, y) 7→ f (x, y) := |x − y| ∈ R is continuous, indeed |f (x, y) − f (x0 , y 0 )| = ||x − y| − |x0 − y 0 || ≤ |x − y − x0 + y 0 | ≤ |x − x0 | + |y − y 0 | and the last term is a norm on R2 and so, since all norms are equivalent on a finite dimensional space, it induces the usual Euclidean topology on R2 . Problem 3. You have seen the implication (⇒) in class. Let us prove (⇐). So let (fn )n be a sequence of bounded simple functions and assume (1) lim sup |fn (x) − f (x)| = 0 n→∞ x∈X In particular this implies that fn (x) → f (x) for every x ∈ X, which implies the measurability of f (the pointwise limit of measurable functions is measurable, as you have seen in class). For the boundedness of f simply notice that by (1) there exists N such that sup |fN (x) − f (x)| ≤ ε x∈X and moreover supx∈X |fN (x)| ≤ C < ∞, therefore |f (x)| ≤ |f (x) − fN (x)| + |fN (x)| ≤ ε + C < ∞ 1 2 ANALYSIS 3 PROBLEMS FOR WEEK 5 SOLUTIONS for every x ∈ X. Problem 4. Without loss of generality we can assume that the the decimal representation of every rational number of the form k10−m ends with infinite zeros (rather than infinite nines), because the set of all such rationals has measure zero. With this assumption each real number has a uniquely defined decimal expansion. For each n ≥ 1 let us define En = {x ∈ [0, 1] | the nth decimal place of x is zero} Then we have A= ∞ \ Enc n=1 and claim that m(A) = 0. It clearly suffices to show (by the decreasing monotone convergence theorem for measures) that N \ m( Enc ) → 0 n=1 as N → ∞. We shall prove that N \ m( Enc ) = ( n=1 9 N ) 10 by induction. In fact, given this for N − 1, we consider c EN ∩ N −1 \ Enc . n=1 For any y ∈ TN −1 n=1 Enc we have y= N −1 X −n yn 10 n=1 + ∞ X yn 10−n n=N 0 where each yn 6= 0 for n ≤ N − 1. For each fixed sequence y10 , . . . , yN −1 , the collection of all y in this set with that fixed sequence is in one-to-one correspondence with the interval [0, 1] by the affine map N −1 X yn0 10−n 10−N +1 y 7→ y − n=1 By the scaling property of Lebesgue measure, this set has measure 10−N +1 . But then the interc section of EN and the set of y with this fixed sequence has measure (0.9)10−N +1 (independent of the fixed sequence) so the result follows. Problem 5. There are many possible constructions. For example, let (rn )∞ n=1 be an enumeration of Q ∩ [0, 1] (since the latter set is countable) and let ∞ [ U= (rn − 2−n , rn + 2−n ) n=1 for some small > 0. Then we have m(U ) ≤ ∞ X 2−n+1 = . n=1 Furthermore, U is a dense set in [0, 1] (since it contains the dense set Q ∩ [0, 1]) so the closure U = [0, 1] (in the restriction topology). Thus m(U \ U ) = m([0, 1] \ U ) = 1 − and the result follows for any < 1. ANALYSIS 3 PROBLEMS FOR WEEK 5 SOLUTIONS 3 Problem 6. If f is bounded the result trivially holds by choosing δ < µ(E)εsup f . If f is not bounded consider the sequence fn := min{f, n}, then fn (x) ≤ n for every x ∈ X, (fn )n is an increasing sequence of positive functions and limn→∞ fn (x) = f (x). Therefore we can apply the monotone convergence theorem to conclude Z Z f dµ = lim fn dµ n→∞ E E R for every E ⊂ X. Note that by assumption X f dµ < ∞, hence both sides of the equality above are finite. Now let ε > 0 be given, then there exists N such that Z Z ε f dµ − f dµ < . N 2 E E Take δ = ε/2N , then we have Z Z Z Z ε f dµ ≤ f dµ − fN dµ + fN dµ ≤ + N µ(E) < ε 2 E E E E whenever E ⊂ X with µ(E) < δ.