Real Analysis HW 7 Solutions Problem 44: Let f be integrable over R and > 0. Establish the following three approximation properties. R (i) There is a simple function η on R which has finite support and R |f − η| < (ii) RThere is a step function s on R which vanishes outside a closed, bounded interval and |f − s| < . R (iii) RThere is a continuous function g on R which vanishes outside a bounded set and |f − g| < . R Solution: (i) As shown in Problem 24, if f is non-negative we may find an increasing sequence of non-negative simple functions {φn } with finite support such that φn → f pointwise. It follows by monotone convergence that we can find a φ such that Z Z |f − φ| = f − φ < . R R For generalR f we write f = f + − fR− , and find η1 and η2 simple and of finite support such that R |f + − η1 | < /2 and R |f − − η2 | < /2. Since f + and f − have disjoint support, we see that η1 and η2 must also have disjoint support, therefore η = η1 − η2 is also simple with finite support and it follows that Z Z Z + |f − η| ≤ |f − η1 | + |f − − η2 | < . R R R (ii) By part (i), since we can approximate by simple functions, by the triangle inequality it suffices to show that the characteristic function χE of a bounded measurable set E can be approximated by step functions. Note that since E isSmeasurable, we can ∞ find a disjoint collection of open intervals {Ik }∞ k=1 such that O = k=1 Ik , and m(O ∼ E) S < /2. Since O must have finite measure we can find an N large enough such that m( ∞ k=N +1 Ik ) < /2. Therefore N X s= χIn k=1 1 is a step function and Z |χE − s| ≤ R N Z X k=1 ≤m Z ∞ X |χE∩Ik − χIk | + R N [ ! Ik ∼ E +m k=1 χE∩Ik R k=N +1 ∞ [ ! Ik ∩ E k=N +1 ∞ [ ≤ m(O ∼ E) + m ! Ik < . k=N +1 (iii) Using part (ii), once again we see by the triangle inequality that it suffices to show that any characteristic function of a bounded interval χ[a,b] can be approximated by a continuous function. Let g be the continuous function which is 1 on [a + /2, b − /2] and linearly interpolated to 0 outside of [a, b], then Z |χ[a,b] − g| < m([a, a + /2) ∪ (b − /2, b]) = . R Problem 46: (Riemann-Lebesgue) Let f be integrable over (−∞, ∞). Show that Z ∞ lim f (x) cos nxdx = 0. n→∞ −∞ Solution: Using the result from problem 44, we Rknow there exists a step functon s, vanishing outside of a closed bounded interval such that R |f − s| < /2. Let s be the step function which we write in cannonical form as s= K X sk χ(ak ,bk ) , k=1 where {(ak , bk )} are a disjoint collection of bounded open intervals and {sk } are distinct. Note that it doesn’t matter that we don’t define s at the end points of the intervals since they are a set of measure 0. We see that Z ∞ −∞ Z K X s(x) cos nxdx ≤ |sk | = k=1 K X k=1 ≤ bk ak cos nxdx |sk | | sin nbk − sin nak | n 2K max{|si |} . n 2 Therefore if n > N ≡ 4K max{si }/, we conclude Z ∞ Z Z ∞ ≤ f (x) cos nxdx |f (x) − s(x)|dx + −∞ −∞ ∞ −∞ s(x) cos nxdx < /2 + /2 = . Problem 47: Let f be integrable over (−∞, ∞). (i) Show that for each t, ∞ Z Z ∞ f (x + t) dx. f (x) dx = −∞ −∞ (ii) Let g be a bounded measurable function on R. Show that Z ∞ g(x) · [f (x) − f (x + t)]dx = 0. lim t→0 −∞ Solution: P (i) Note that this result is true for simple functions. Let ϕ = nk=1 ci χEk be simple, then since χE (x + t) = χE−t (x), we haven by the translation invariance of Lebesgue Z ϕ(x + t) dx = R n X ck m(Ek − t) = n X Z ck m(Ek ) = ϕ(x) dx. R k=1 k=1 For the case of integrable f , we may restrict ourselves to f non-negative, since we may always write any integrable f as f = f + − f − , where f + (x) = max{f (x), 0} and f − (x) = max{−f (x), 0}. For such a non-negative f , we know that there is a sequence of simple functions with finite support {ϕn }∞ n=1 such that ϕn ≤ f with ϕn converging to f pointwise and monotonically. In particular ϕ(· + t) → f (· + t) monotonically for every fixed t. It follows from the monotone convergence theorem that Z Z Z Z f (x) dx. f (x + t) dx = lim ϕn (x + t) dx = lim ϕn (x) dx = R n→∞ n→∞ R R R (ii) Let M < ∞ be a constant such that |g| ≤ M . Since f is integrable, we know by Problem 44 R that for every > 0 there exists a h, continuous and of bounded support such that R |f − h| < /2M . It follows that h uniformly continuous and bounded by some constant N . Let {tn } be any sequence tn → 0 and define hn (x) = |h(x)−h(x+tn )|. Then hn → 0 pointwise and 0 ≤ hn (x) ≤ 2N. It follows by the bounded convergence theorem that Z lim |f (x) − f (x + tn )| = 0. n→∞ R 3 Therefore Z Z g(x) · [f (x) − f (x + tn )] dx ≤ M |f (x) − h(x)| dx R R Z Z +M |f (x + tn ) − h(x + tn )| dx + M |f (x) − f (x + tn )| dx R R By part i) and the way that h was chosen, the sum of the first two terms on the righthand side above are bounded by . Taking the lim supn of both sides, we obtain Z Z lim sup g(x) · [f (x) − f (x + t)] dx < + M lim |f (x) − f (x + tn )| dx = . n n→∞ R R Since this is true for every > 0 and every {tn }, tn → 0, we conclude Z lim g(x) · [f (x) − f (x + t)] dx. t→0 R Problem 50: Let F be a family of functions, each which is integrable over E if and only if for each > 0, there is a δ > 0 such that for each f ∈ F, Z if A ⊆ E is measurable and m(A) < δ, then f < . A R R Solution: Clearly by the triangle inequality if A |f | < , then A f and so uniform integrability R easily implies this condition. However, for any A measurable we may write R R + f + − f where A+ = {x ∈ A : f ≥ 0} and A− = {x ∈ A : f < 0} are |f | = A A A disjoint. It follows that there exists δ + and δ − such that if m(A+ ) < δ + and m(A− ) < δ − , then Z Z f < /2 and f < /2. A− A+ Choosing δ = min{δ + , δ − }, we conclude that if m(A) < δ, then Z |f | < . A Therefore this condition implies uniform integrability. Problem 51: Let F be a family of functions, each of which is integrable over E. Show that F is uniformly integrable over E if and only if for each > 0, there is a δ > 0 such that for each f ∈ F, if U is open and m(E ∩ U) < δ, Z |f | < . E∩U 4 Proof: Clearly if F is uniformly integrable, it follows trivially tha there exists a δ > 0 such that if U is open and m(E ∩ U) < δ, then Z |f | < . E∩U Now assume the other property, and let δ > 0 be such that if U is open and m(E ∩ U) < δ, R then E∩U |f | < . If A ⊆ E is measurable such that m(A) < δ, then we know that there is an open set O containing A such that m(O ∼ A) < δ − m(A). Therefore m(E ∩ O) ≤ m(O) = m(A) + m(O ∼ A) < δ, and since A ⊆ E ∩ O, it follows Z Z |f | ≤ A |f | < . E∩O 5