Download Math 7350 Homework Solutions: Convergence, Fσδ-Sets, and Lebesgue Measurable Sets and more Assignments Mathematics in PDF only on Docsity!
Math 7350 Homework 4 Solutions Fall 2004
- Let fn : R → R be a sequence of continuous functions. Show that the set of points C = {x ∈ R : fn(x) converges as n → ∞ } is an Fσδ-set. The sequence fn(x) converges iff fn(x) is a Cauchy sequence: ∀ε > 0 : ∃n 0 > 0 : ∀n, m ≥ n 0 : |fn(x) − fm(x)| < ε. The ‘σδ’ should take care of the quantifiers, but we need {x : |fn(x) − fm(x)| < ε} to be closed. It is not closed as it stands, but the set En,m,ε = {x : |fn(x) − fm(x)| ≤ ε} is closed since it is the complement of the set {x : |fn(x) − fm(x)| > ε} which is the inverse image of the open set (−∞, −ε) ∪ (ε, ∞) under the continuous function g(x) = fn(x) − fm(x). Now
Cn 0 ,ε =
n,m≥n 0
En,m,ε = {x : ∀n, m ≥ n 0 : |fn(x) − fm(x)| ≤ ε}
is an intersection of closed sets, so is closed. Thus Aε =
n 0
Cn 0 ,ε = {x : ∃n 0 : ∀n, m ≥ n 0 : |fn(x) − fm(x)| ≤ ε}
is an Fσ-set, and B =
k
A 1 /k =
ε> 0
Aε = {x : fn(x) converges}
is an Fσδ-set.
- If E ⊂ R is a Lebesgue measurable set with a finite measure, prove that for any given ≤ > 0, there is a set U which is a finite union of open intervals such that λ(U 4 E) < ≤. Here U 4 E = (U \ E) ∪ (E \ U ). Since λ(E) = λ∗(E) < ∞, we can find a countable sequence of open intervals I 1 , I 2 ,... such that E ⊆
i=1 Ii^ and^
i=1 λ(Ii) =^
i=1 l(Ii)^ < λ(E) +^ ε/2.^ If we set U∞ =
i=1 Ii, then^ λ(U∞)^ ≤^
λ(Ii) < λ(E) + ε/2, so λ(U∞ \ E) < ε/2. Set Un =
⋃n i=1 Ii. Then^ U^1 ⊆^ U^2 ⊆^...^ and^
n=1 Un^ =^ U∞. Hence limn→∞^ λ(Un) = λ(U∞). Thus there is some n for which λ(Un) ≥ λ(U∞)−ε/2, and so λ(U∞\Un) < ε/2. Now Un is a finite union of intervals and Un 4 E = (U∞ \ E) 4 (U∞ \ Un) ⊆ (U∞ \ E) ∪ (U∞ \ Un), so λ(Un 4 E) < ε/2 + ε/2 = ε.
- Show that if U is a finite union of open intervals and Q ∩ [0, 1] ⊆ U , then λ(U ) ≥ 1. Suppose U =
⋃n i=1 Ii^ where^ Ii^ are open intervals and^ Q^ ∩^ [0,^ 1]^ ⊆^ U^.^ We need to show λ(U ) ≥ 1. Without loss of generality, and by induction of n, we may assume the Ii’s are pairwise disjoint. Otherwise, if Ii ∩Ij 6 = ∅, then Ii ∪Ij is also an open interval, and we can write U as a union of a smaller number of intervals. Similarly, without loss of generality,
each Ii intersects [0, 1], otherwise we could remove Ii from the union to get a set U ′ which is the union of n − 1 intervals, Q ∩ [0, 1] ⊆ U ′, and λ(U ) ≥ λ(U ′) ≥ 1. Now write the intervals as Ii = (ai, bi) and order the intervals so that a 1 < a 2 < · · · < an (if ai = aj then Ii and Ij intersect, so we may assume the ai are distinct). Now if bi > ai+1 then Ii and Ii+1 intersect. Hence we may assume bi ≤ ai+1. In particular we may assume U ∩ (ai, ai+1) = Ii. On the other hand, if bi < ai+1 then either bi ≥ 1 (so Ii+1 ∩ [0, 1] = ∅) or ai+1 ≤ 0 (so Ii ∩ [0, 1] = ∅) or (bi, ai+1) ∩ [0, 1] 6 = ∅ (so (bi, ai+1)∩[0, 1] contains a rational that is not in U ). Since none of these are possible, we must have bi = ai+1 for all i with 1 ≤ i < n. Thus U = (a 1 , bn) \ {a 2 ,... , an}. But 0 ∈ U so a 1 < 0, and 1 ∈ U so bn > 1. Now λ(U ) = bn − a 1 > 1.
- The first Borel-Cantelli Lemma states that if the sets∑ B 1 , B 2 ,... are measurable and ∞ i=1 λ(Bi)^ <^ ∞, then the set of points that belong to infinitely many^ Bi^ is a set of measure 0. Prove the first Borel-Cantelli Lemma. Let E be the set of points that belong to infinitely many Bi. Since
i=1 λ(Bi) converges, for any ε > 0 there is an n 0 such that
i=n 0 λ(Bi)^ < ε.^ But then λ(
i=n 0 Bi)^ ≤^
i=n 0 λ(Bi)^ < ε. Now if^ x^ ∈^ E^ then^ x^ is in infinitely many^ Bi, and so lies in some Bi with i ≥ n 0. Thus E ⊆
i=n 0 Bi. Hence^ λ(E)^ ≤^ λ(
i=n 0 Bi)^ < ε. Since this holds for all ε > 0, λ(E) = 0.
- Let A be a subset of R such that λ(A) > 0. Denote by A−A the set {x−y : x, y ∈ A}.
(a) Prove that there is an interval [a, b] such that λ(A ∩ [a, b]) > 34 (b − a). First by setting An = A∩[n, n+1) and noting that and
λ(An) = λ(A), we may assume λ(An) > 0 for some n. Replacing A by An, we may assume λ(A) < ∞. Now A ⊆ U with U open and λ(U ) < 43 λ(A). Writing U as a countable disjoint union of intervals Ii, we get
λ(Ii) = λ(U ) < 43 λ(A) = (^43)
λ(Ii ∩ A). If λ(Ii) ≥ 43 λ(Ii ∩ A) for all i we obtain a contradiction. Hence there is an interval Ii with λ(Ii ∩ A) > 34 λ(Ii). (b) Show that if 0 ≤ δ ≤ 14 (b − a) then A ∩ (A + δ) ∩ [a, b] is non-empty. Let A′^ = A ∩ [a, b]. Now if A′^ ∩ (A′^ + δ) = ∅ then λ(A′) + λ(A′^ + δ) = λ(A′^ ∪ (A′^ + δ)) ≤ λ([a, b + δ]) ≤ 54 (b − a). But λ(A′) = λ(A′^ + δ) > 34 (b − a), a contradiction. Hence A′^ ∩ (A′^ + δ) ⊆ A ∩ (A + δ) ∩ [a, b] is non-empty. (c) Deduce that A − A ⊇ [−^14 (b − a), 14 (b − a)]. If 0 ≤ δ ≤ 14 (b − a) then A ∩ (A + δ) 6 = ∅, so x ∈ A ∩ (A + δ) for some x. But then x ∈ A and x − δ ∈ A, so ±x ∈ A − A. Thus A − A ⊇ [−^14 (b − a), 14 (b − a)].
