Learning Dashboard
1. Abstract Integration
Exercise 1
Does there exist an infinite σ-algebra which has only countably many members?
Solution
No. Let X be a measurable set with an infinite σ-algebra M. Since M is infinite, there exists nonempty E ∈ M properly contained in X. Both E and Ec are measurable spaces by letting the measurable subsets of E (resp. Ec) be the intersections of measurable subsets of X with E (resp. Ec). Since M is infinite, at least one of these two σ-algebras must be infinite.
Now we define a rooted binary tree inductively as follows. The root is our set X. Given a vertex which is a measurable subset E of X, if it contains a proper measurable subset E0, pick one such subset, and let its two successors be E0 and E \ E0. The remarks above guarantee that this tree is infinite, and hence has infinite depth. So pick an infinite path consisting of subsets E0 ⊇ E1 ⊇ E2 ⊇ .... Then the sets Fi = Ei \ Ei+1 form an infinite collection of disjoint nonempty measurable subsets of X by construction. At the very least, M needs to contain every union of such sets, and this is in bijection with the set of subsets of N, which is uncountable. Thus, M must be uncountable.
Exercise 2
Prove an analogue of Theorem 1.8 for n functions.
Solution
We need to prove the following: if u1, ..., un are real measurable functions on a measurable space X, and Φ is a continuous map of Rn into a topological space Y, then h(x) = Φ(u1(x), ..., un(x)) is a measurable function from X to Y.
Define f : X → Rn by x ↦ (u1(x), ..., un(x)). By Theorem 1.7(b), to prove that h is measurable, it is enough to prove that f is measurable. If R is any open rectangle in Rn which is the Cartesian product of n segments I1, ..., In, then f−1(R) = u−1(I1) ∩ ... ∩ u−1(In), which is measurable since u1, ..., un is measurable. Finally, every open set of Rn is the countable union of such rectangles, so we are done.
Exercise 3
Prove that if f is a real function on a measurable space X such that {x | f(x) ≥ r} is measurable for every rational r, then f is measurable.
Solution
Let U ⊆ R be an open set. First, U can be written as a union of countably many open balls with rational radii that are centered at rational points. So to prove that f−1(U) is measurable, it is enough to prove this when U is an open ball of this form, say with radius r and center c. Since the set of measurable sets is closed under complements and finite intersections, every set of the form {x | r1 > f(x) ≥ r2} is measurable for rational r1, r2. Now note that {x | c + r > f(x) > c − r} can be written as the countable union ∪n≥1{x | c + r > f(x) ≥ c − r + 1/n}, so f−1(U) is measurable.
Exercise 4
Let {an} and {bn} be sequences in [-1, 1], and prove the following assertions:
- (a) lim sup (−an) = − lim inf an
- (b) lim sup (an + bn) ≤ lim sup an + lim sup bn provided none of the sums is of the form 1 - 1
- (c) If an ≤ bn for all n, then lim inf an ≤ lim inf bn
Show by an example that strict inequality can hold in (b).
Solution
(a) The supremum Ak of the set {-ak, -ak+1, ...} is the negative of the infimum A'k of the set {ak, ak+1, ...}. Hence inf_k{Ak} = −sup_k{A'k}, which implies (a).
(b) The relation sup{ak + bk, ak+1 + bk+1, ...} ≤ sup{ak, ak+1, ...} + sup{bk, bk+1, ...} is clear, so this implies (b). To see that the inequality in (b) can be strict, consider a1 = 1, ai = 0 for i > 1, and b1 = −1, bi = 0 for i > 1. Then lim sup(an + bn) = 0, but lim sup an + lim sup bn = 1.
(c) Now suppose that an ≤ bn for all n. Then inf{ak, ak+1, ...} ≤ inf{bk, bk+1, ...} for all k, so (c) follows.
Exercise 5
- (a) Suppose f : X → [-1, 1] and g : X → [-1, 1] are measurable. Prove that the sets {x | f(x) < g(x)}, {x | f(x) = g(x)} are measurable.
- (b) Prove the set of points at which a sequence of measurable real-valued functions converges (to a finite limit) is measurable.
Solution
(a) First note that f(x) - g(x) is a measurable function. The first set of (a) is the preimage of the open set [-1, 0], so is measurable. Also, the set where f and g agree is the complement of where f(x) - g(x) ≠ 0, which is measurable.
(b) As for (b), let {fn} be a sequence of measurable real functions, and let E be the set of x such that fn(x) converges as n → ∞. Define f = lim sup fn. Then f is measurable (Theorem 1.14), and f agrees with lim fn on E. For each n, the function f - fn is measurable (1.22), so the set En,r which is defined to be the preimage of f - fn of (−r, r) is measurable. Then E = ∩_r=1 ∪_n=1 En,r, so is measurable.
Exercise 6
Let X be an uncountable set, let M be the collection of all sets E ⊂ X such that either E or Ec is at most countable, and define μ(E) = 0 in the first case, μ(E) = 1 in the second. Prove that M is a σ-algebra in X and that μ is a measure on M. Describe the corresponding measurable functions and their integrals.
Solution
Since Xc = ∅ is at most countable, X ∈ M. Also, if E ∈ M, then either E or Ec is at most countable, so the same is true for Ec since (Ec)c = E, and so Ec ∈ M. Now suppose En ∈ M for all n, and put E = ∪_n En. Let I be the set of n for which En is at most countable, and let J be the set of n for which En is uncountable, but Enc is at most countable, so that E = ∪_n∈I En ∪ ∪_n∈J En. If J = ∅, then E is a countable union of countable sets, and hence is countable. Otherwise, Ec = ∩_n∈I Enc ∩ ∩_n∈J Enc, so Ec ⊆ ∩_n∈J Enc, which is countable since J ≠ ∅, so E ∈ M. Thus, M is a σ-algebra.
Now write a measurable set A as a disjoint union of measurable sets An. If A is at most countable, then so is each An, so μ(A) = ∑ μ(An) = 0. In case Ac is at most countable, then A is uncountable, so at least one Ai is uncountable. Suppose that Ai and Aj are both uncountable for i ≠ j. Then Ac_i ∪ Ac_j is countable and equal to X since Ai and Aj are disjoint. But this contradicts that X is uncountable, so exactly one Ai is uncountable, which means that μ(A) = ∑ μ(An) = 1. Hence μ is a measure on M.
The measurable functions on M consist of those functions f : X → R such that for each r ∈ R, f−1(r) is either at most countable, or f−1(R \ {r}) is at most countable. If we let A ⊂ R denote the set of points such that f−1(r) is not countable, then the integral of f is ∑_r∈A r.
Exercise 7
Suppose {fn : X → [0, 1]} is measurable for n = 1, 2, 3, ..., f1 ≥ f2 ≥ f3 ≥ ... ≥ 0, fn(x) → f(x) as n → ∞, for every x ∈ X, and f1 ∈ L1(μ). Prove that
lim_{n→∞} ∫_X fn dμ = ∫_X f dμ
and show that this conclusion does not follow if the condition "f1 ∈ L1(μ)" is omitted.
Solution
If we first assume that f1(x) < 1 for all x, then the conclusion is a consequence of Lebesgue’s dominated convergence theorem (Theorem 1.34) using g(x) = f1(x) since f1(x) ≥ fn(x) ≥ 0 implies that f1(x) ≥ |fn(x)|. Otherwise, let E = {x ∈ X | f1(x) = 1}. If μ(E) > 0, then ∫_X |f1| dμ = ∞, which contradicts f1 ∈ L1(μ). So we conclude that μ(E) = 0, in which case, we can ignore E when integrating over X, and we are back to the above discussion.
Now suppose that f1 ∈ L1(μ) no longer holds. Take X = R, and μ(E) is the length of E. Then define fn(x) = 1 for x ∈ [0, 1/n], and 0 elsewhere, so that fn → 0. Then ∫_X fn dμ = 1 for all n, but ∫_X 0 dμ = 0.
2. Positive Borel Measures
Content for Positive Borel Measures will go here...
3. Lp-Spaces
Content for Lp-Spaces will go here...
4. Elementary Hilbert Space Theory
Content for Elementary Hilbert Space Theory will go here...
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