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Math 7350 Homework 5 Solutions Fall 2004
- Evaluate, with justification, the limit limn→∞
0 (1 +^ x/n)
−n (^) n sin(x/n) dx.
log(1+ xn )−n^ = −n( xn +O(( xn )^2 )) = −x+O(x
2 n )^ → −x^ as^ n^ → ∞, thus (1+^
x n )
−n (^) → e−x as n → ∞. Also n sin( xn ) = n( xn + O(( xn )^3 )) = x + O( (^) nx^32 ) → x as n → ∞. Thus if we set fn(x) = (1 + x/n)−n^ n sin(x/n) then fn → f as n → ∞ where f (x) = xe−x. We would expect the integral to converge to
0 xe
−x (^) dx. To prove this, use Dominated Convergence Theorem. Since sin x ≤ x for x ≥ 0 we have |fn(x)| ≤ x(1 + x/n)−n. If we assume n ≥ 3 then (1+ xn )n^ ≥ 1+
(n 1
) (^) x n +
(n 2
) (^) x 2 n^2 +
(n 3
) (^) x 3 n^3 ≥^ 1+x+^
1 2 (1−^
1 n )x
6 (1−^
1 n )(1−^
2 n )x
(^3) ≥ x+ 1 27 x
Thus |fn(x)| ≤ g(x) = 1/(1 + x
2 27 ), which is integrable. Thus the limit is^
0 xe
−x (^) dx. Now letting gn = xe−xχ[0,n],
gn →
f by MCT. But
∫ (^) n 0 xe
−x (^) dx = [−(x+1)e−x]n 0 = 1 − (n + 1)e−n^ → 1 as n → ∞. Thus the limit is 1.
- Let S be a measurable set with finite measure. Show that f (x) = λ(S ∩ (S + x)) is a continuous function and limx→+∞ f (x) = 0. If I = (a, b) is an interval and z ∈ S∩(I+x) but z /∈ S∩(I+y) then z ∈ (I+x)(I+y). Now λ((I + x) \ (I + y)) = λ((a + x, b + x) \ (a + y, b + y)) ≤ |x − y|. Thus λ(S∩(I +x))−λ(S∩(I +y)) ≤ |x−y|. If U =
⋃N
i=1 Ii^ is a disjoint union of^ N^ intervals, then λ(S∩(U +x)) =
λ(S∩(Ii +x)). Thus λ(S∩(U +x))−λ(S∩(U +y)) ≤ N |x−y|. If λ(S) < ∞, then there is such a U with λ(S 4 U ) < ε. Then |λ(S ∩ (S + x)) − λ(S ∩ (U + x))| < ε. Thus f (x) − f (y) = λ(S ∩ (S + x)) − λ(S ∩ (S + y)) < 2 ε + N |x − y|. Similarly f (y) − f (x) < 2 ε + N |y − x|, so |f (y) − f (x)| < 2 ε + N |x − y|. Since this is < 3 ε for |x − y| sufficiently small, and ε > 0 is arbitrary, f (x) is continuous. Let K be the maximum distance between any two points in the (bounded) set U. Let V = S \U. Since S ⊆ U ∪V , S ∩(S +x) ⊆ (U ∩(U +x))∪(U ∩(V +x))∪(V ∩(S +x)). If x > K then U ∩ (U + x) = ∅, so λ(S ∩ (S + x)) ≤ λ(V + x) + λ(V ) < 2 ε. Hence f (x) → 0 as x → +∞.
- Suppose f : [0, 1] → R and f n^ is integrable for all n ≥ 1. If
0 f^
n (^) = ∫^1 0 f^ for all n ≥ 1, show that f = χS a.e. for some measurable set S ⊆ [0, 1]. We start be showing∫ f 2 = f a.e.. Consider g = (f 2 − f )^2. Then g ≥ 0 and 1 0 g^ =^
0 (f^
(^4) − 2 f 3 + f 2 ) = ∫^1 0 f^ −^2
0 f^ +^
0 f^ = 0. Let^ Ek^ =^ {x^ :^ g(x)^ >^1 /k}. Then 0 =
g ≥ λ(Ek)/k. Thus λ(Ek) = 0. But then λ(
Ek) = 0 and
Ek = {x : g(x) > 0 } = {x : f (x)^2 6 = f (x)}. Thus f 2 = f a.e.. Since f is measurable, S = f −^1 [{ 1 }] is measurable. If f (x) 6 = χS (x) then f (x) 6 = 0, 1, so f (x)^2 6 = f (x). Thus f = χS a.e..
- (a) Show that if S is measurable with finite measure then limn→∞
S cos^ nx dx^ = 0. [Hint: Recall that there exists a finite union U of intervals with λ(S 4 U ) < ε.] Assume first that S = (a, b) is an open interval. Then |
S cos^ nx dx|^ =^ |^
∫ (^) b a cos^ nx dx|^ =^ |[^
1 n sin(nx)]
ba| ≤ 2 n. Fix ε > 0 and let U be a union of k open intervals with λ(S 4 U ) < ε. Now |
S cos^ nx dx−
U cos^ nx dx|^ =^ |^
(χS −χU ) cos nx dx| ≤
|χS −χU | = λ(S 4 U ). Thus |
S cos^ nx dx^ −^
U cos^ nx dx|^ < ε^ and^ |^
U cos^ nx dx| ≤^ k(2/n).^ Hence |
S cos^ nx dx|^ < ε^ +^ k(2/n) which is^ <^2 ε^ for sufficiently large^ n.^ Since this holds for any ε > 0, limn→∞
S cos^ nx dx^ = 0. (b) Deduce the Riemann-Lebesgue Theorem: If f is integrable then limn→∞
f (x) cos nx dx = 0. By writing f = f+ −f− it is enough to prove the result when f ≥ 0 and
f < ∞. Let φ be a simple function with 0 ≤ φ ≤ f , φ =
∑k i=1 aiχSi , and^
φ ≥
f −ε/2. Then
φ cos nx dx =
∑k i=1 ai
Si cos^ nx dx^ →^ 0 as^ n^ → ∞. Thus for sufficiently large n, |
φ cos nx dx| < ε/2. But |
(f (x) − φ(x)) cos nx dx| ≤
(f − φ) ≤ ε/2, so |
f (x) cos nx dx| ≤ |
φ(x) cos nx dx| + |
(f − φ) cos nx dx| < ε for sufficiently large n.
- Suppose that for all n,
0 fn(x)
(^2) dx ≤ 1 n^4. Show that^ fn^ →^ 0 a.e. on [0,^ 1]. Let En = {x : |fn(x)| > 1 /n}. Then
0 fn(x)
(^2) dx ≥ λ(En)/n (^2). Thus λ(En) ≤ 1 /n (^2). If fn(x) 6 → 0 then there is some ε = 1/k with fn(x) > 1 /k for infinitely many n. But then fn(x) > 1 /n for infinitely many n. Thus x ∈
n 0 Ek. But the measure of this set is at most
n 0 1 /n
(^2) → 0 as n 0 → ∞. Thus the set of x where fn 6 → 0 is of measure zero. Alternative solution: Let g(x) =
n=1 fn(x)
(^2) ∈ [0, ∞]. Now ∑N n=1 f^ n^2 is increasing in^ N^ , so by MCT, ∫ g =
limN
∑N
n=1 f^ n^2 = limN^ ∫ ∑N n=1 f^ n^2 = limN^ ∑N n=
f (^) n^2 ≤
n=
1 n^4 <^ ∞. But this implies {x : g(x) = +∞} has measure 0. Thus g(x) =
n=1 fn(x)
(^2) con- verges a.e., but this implies fn(x)^2 → 0 a.e., and so fn(x) → 0 a.e..
