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1998 Apr

1998 April 03

Instructions. Do at least four problems in Part A, and at least two problems in Part B.

PART A

  1. Let \(\{xₙ\}_{n=1}^∞\) be a bounded sequence of real numbers, and for each positive \(n\) define \(\hat{x}_n = \sup_{k ≥ n} x_k\).

    a. Explain why the limit \(ℓ = \lim_n \hat{x}_n\) exists.
    b. Prove that for every \(ϵ > 0\) and every positive integer \(N\) there exists an integer \(k\) such that \(k ≥ N\) and \(| xₖ - ℓ | < ϵ\).

  2. Let \(C\) be a collection of subsets of the real line ℝ and define

    \(A_σ(C) = ⋂ \{A : C ⊆ A \text{ and } A \text{ is a σ-algebra of subsets of } ℝ\}\).

    a. Prove that \(A_σ (C)\) is a σ-algebra, that \(C ⊆ A_σ(C)\), and that \(A_σ(C) ⊆ 𝔄\) for every σ-algebra 𝔄 containing all the sets in \(C\).
    b. Let \(O\) be the collection of all finite open intervals in ℝ and \(F\) the collection of all finite closed intervals in ℝ. Show that \(A_σ(O) = A_σ(F)\).

  3. Let \((X, A, μ)\) be a measure space, and suppose \(X = ⋃_n X_n\), where \(\{X_n\}_{n=1}^∞\) is a pairwise disjoint collection of measurable subsets of \(X\). Use the Monotone Convergence Theorem and linearity of the integral to prove that, if \(f\) is a nonnegative measurable real-valued function on \(X\), then \(∫_X f \, dμ = ∑_n ∫_{X_n} f\, dμ\).

  4. Using the Fubini/Tonelli Theorems to justify all steps, evaluate the integral \(∫_0^1 ∫_y^1 x^{-3/2} \cos \left(\frac{π y}{2x}\right) \, dx\, dy\).

  5. Let \(I\) be the interval \([0, 1]\). Let \(C(I)\), \(C(I × I)\) denote the spaces of real valued continuous functions on \(I\) and \(I × I\), respectively, with the usual supremum norm on these spaces. Show that the collection of finite sums of the form \(f(x, y) = ∑_i ϕ_i(x) ψ_i(y)\) (where \(ϕ_i, ψ_i ∈ C(I)\) for each \(i\)) is dense in \(C(I × I)\).

  6. Let \(m\) be Lebesgue measure on the real line ℝ, and for each Lebesgue measurable subset \(E\) of ℝ define \(μ(E) = ∫_E \frac{1}{1+x^2}\, dm(x)\). Show that \(m\) is absolutely continuous with respect to \(μ\), and compute the Radon-Nikodym derivative \(dm/dμ\).

PART B

  1. Let \(ϕ(x,y) = x^2 y\) be defined on the square \(S = [0, 1] × [0, 1]\) in the plane, and let \(m\) be two-dimensional Lebesgue measure on \(S\). Given a Borel subset \(E\) of the real line ℝ, define \(μ(E) = m(ϕ^{-1}(E))\).

    a. Show that \(μ\) is a Borel measure on ℝ.
    b. Let \(χ_E\) denote the characteristic function of the set \(E\). Show that \(∫_ℝ χ_E \, dμ = ∫_S χ_E ∘ ϕ \, dm\).
    c. Evaluate the integral \(∫_{-∞}^∞ t^2 \, dμ(t)\).

  2. Let \(f\) be a real valued and increasing function on the real line ℝ, such that \(f(-∞) = 0\) and \(f(∞) = 1\). Prove that \(f\) is absolutely continuous on every closed finite interval if and only if \(∫_ℝ f'\, dm = 1\).

  3. Let \(F\) be a continuous linear functional on the space \(L^1[-1, 1]\), with the property that \(F(f) = 0\) for all odd functions \(f\) in \(L^1[-1, 1]\). Show that there exists an even function \(ϕ\) such that \(F(f) = ∫_{-1}^1 f(x) ϕ(x)\, dx\), for all \(f ∈ L^1[-1, 1]\).
    (Hint. One possible approach is to use the fact that any function in \(L^p[-1, 1]\) is the sum of an odd function and an even function.)

Solution A.3.

Note that the hypotheses imply \(ν(E) = ∫_E f \, dμ\) is a measure, from which the desired conclusion immediately follows.

Of course, this does not answer the question as stated, since the examiners specifically require the use of the MCT and linearity of the integral.

Define \(fₙ = ∑_{k=1}^n f χ_{X_k} = f χ_{∪_1^n X_k}\).

Then it's clear that the hypotheses of the MCT are satisfied. That is, \(∀ x ∈ X\),

  1. \(0 ≤ f₁(x) ≤ f₂(x) ≤ ⋯ ≤ f(x)\), and

  2. \(\lim_n fₙ(x) = f(x) χ_X(x) = f(x)\).

Therefore, by linearity of the integral,

\(∑_{k=1}^∞ ∫_{Xₖ} f\, dμ = \lim_n ∑_{k=1}^n ∫_X f χ_{Xₖ}\, dμ\\ = \lim_n ∫_X ∑_{k=1}^n f χ_{Xₖ}\, dμ.\)

which, by definition of \(fₙ\), is equal to \(\lim_n ∫_X fₙ \, dμ\).

By the MCT, \(∫_X \lim_n fₙ \, dμ = ∫_X f\, dμ\).

∎

Solution A.4.

By Tonelli's Theorem, if 1. \(f(x,y) ≥ 0\) is measurable and 2. one of the iterated integrals \(∬ f(x,y)\,dx\,dy\) or \(∬ f(x,y)\,dy\,dx\) exists, then they both exist and are equal.

Moreover, if one of the iterated integrals is finite, then \(f(x,y) ∈ L^1(dx,dy)\).

Let \(g(x, y) = x^{-3/2}\cos(\pi y/2x)\), and apply Tonelli's theorem to the non-negative measurable function \(|g(x, y)|\) as follows:

\(∫_0^1 ∫_0^x |g(x, y)| \, dy \, dx = ∫_0^1 ∫_0^x |x|^{-3/2} \left|\cos\left(\frac{π y}{2x}\right)\right|\,dy\,dx ≤ ∫_0^1 ∫_0^x x^{-3/2}\cdot 1 \,dy\,dx = ∫_0^1 x^{-1/2} \,dx = 2\).

Thus one of the iterated integrals is finite, so \(g(x, y) ∈ L^1(dx, dy)\) by Tonelli's theorem.

Therefore, Fubini's Theorem applies to \(g(x, y)\), and yields

\(∫_0^1 ∫_y^1 x^{-3/2} \cos \left(\frac{π y}{2x}\right) \, dx \, dy = ∫_0^1 ∫_0^x x^{-3/2}\cos\left(\frac{\pi y}{2x}\right)\,dy\,dx\).

The latter is clearly equal to the first (hence the last) of the equivalent expressions below.

\(∫_0^1 x^{-3/2} \cdot \frac{2x}{\pi} \left[\sin\left(\frac{\pi y}{2x}\right)\right]_{y=0}^{y=x} \, dx = ∫_0^1 \frac{2}{\pi}x^{-1/2}\, dx = \frac{2}{π} \left[2 x^{1/2}\right]_{x=0}^{x=1} = \frac{4}{π}\).

∎

Solution A.6

Obviously both measures are non-negative. We must first prove \(m ≪ μ\). To this end, suppose \(m(E) > 0\), where \(E ∈ σ(m)\), the σ-algebra of Lebesgue measurable sets. If we can show \(μ(E) > 0\), this will establish that the implication \(μ(E) = 0 \, ⇒ \, m(E) = 0\) holds for all \(E ∈ σ(m)\); i.e., \(m ≪ μ\).

For \(n = 1, 2, …\), define \(A_n = \{x ∈ ℝ : \frac{1}{n+1} < \frac{1}{1+x^2} ≤ \frac{1}{n}\}\).

Then \(A_i ∩ A_j = ∅\) for all \(i ≠ j\) in ℕ, and, for all \(n = 1, 2, …\), \(μ(A_n) ≥ \frac{1}{n+1} m(A_n)\).

Also, \(ℝ = ⋃_n A_n\), since \(0 < \frac{1}{1+x^2} \leq 1\) holds for all \(x ∈ ℝ\).

Therefore, \(μ(E) = μ(E ∩ ⋃_n A_n) = μ(⋃_n (A_n ∩ E)) = ∑_n μ(A_n ∩ E)\).

The last equality might need a bit of justification. Since \(f(x) = \frac{1}{1+x^2}\) is continuous, hence measurable, the sets \(\{A_n\}\) are measurable. Therefore, the last equality holds by countable additivity of disjoint measurable sets.

Note that \(m(E) = ∑_n m(A_n ∩ E) > 0\) implies the existence of an \(n ∈ ℕ\) such that \(m(A_n ∩ E) > 0\). Therefore,

\(μ(E) ≥ μ(A_n ∩ E) ≥ \frac{1}{n+1} m(A_n ∩ E) > 0\),

which proves that \(m ≪ μ\).

By the Radon-Nikodym Theorem there is a unique \(h ∈ L^1(μ)\) such that

\(m(E) = ∫ h \, dμ\), and \(∫ f\, dm = ∫ f \, h \, dμ\), for all \(f ∈ L^1(m)\).

In particular, if \(E ∈ σ(m)\) and \(f(x) = \frac{1}{1+x^2} χ_E\), then

\(μ(E) = ∫_E \frac{1}{1+x^2} \, dm(x) = ∫_E \frac{h(x)}{1+x^2}\, dμ(x)\).

That is, \(∫_E \, dμ = ∫_E \frac{h(x)}{1+x^2}\, dμ(x)\) holds for all measurable sets \(E\), which implies (see note blow) that, \(\frac{h(x)}{1+x^2} = 1\) holds for \(μ\)-almost every \(x ∈ ℝ\). Therefore, \(\frac{dm}{dμ}(x) = h(x) = 1+x^2\).

∎

Remarks.

  • The penultimate implication comes from the following standard result: if \(f\) and \(g\) are integrable functions such that \(∫_E f = ∫_E g\) holds for all measurable sets \(E\), then \(f = g\), \(μ\)-a.e.

  • \(h\) is uniquely defined only up to an equivalence class of functions that are equal to \(1+x^2\), \(μ\)-a.e.

Solution B.2.

Preliminary remark. I have worked this problem a number of times, and what follows is the clearest and most instructive proof I have come up with. It is by no means the shortest, most elegant solution, and probably not the type of detailed answer one ought to provide on an actual exam, when time is a factor. However, some of the facts that I prove in detail here have appeared as separate questions on other exams, so the proofs are worth knowing.

Note that \(f\) is increasing, so \(f'\) exists for a.e. \(x ∈ ℝ\) and \(f'(x) ≥ 0\) wherever \(f'\) exists. Also, \(f'\) is measurable. To see this, define

\(g(x) = \limsup_{n → ∞} \, [f(x+ \frac{1}{n})-f(x)] \, n\).

As a lim sup of a sequence of measurable functions, \(g\) is measurable (Theorem 1.14 of Rudin (1987)).

Let \(E\) be the set on which \(f'\) exists. Then \(m(ℝ - E) = 0\), and \(f' = g\) on \(E\) (by definition of derivative), so \(f'\) is measurable.

  • (⇐) Suppose \(∫_ℝ f' \, dm = 1\). We must show \(f ∈ AC[a, b]\) for all \(-∞ < a < b < ∞\). First, check that \(f' ∈ L^1(ℝ)\), since

    \(1 = ∫_ℝ f' \, dm = ∫_{ℝ - E} f' \, dm + ∫_E f' \, dm = ∫_E f' \, dm\),

    (The last equality holds because \(m(ℝ - E) = 0\).)

    Since \(f\) is increasing, \(f' ≥ 0\) on \(E\), so

    \(∫_ℝ |f'| \, dm = ∫_{ℝ - E} |f'| \, dm + ∫_E |f'| \, dm = ∫_E |f'| \, dm = ∫_E f' dm = 1\).

    Thus, \(f' ∈ L^1(ℝ)\) as claimed.

    A couple of lemmas will be needed to complete the ⇐ direction of the proof. The first is proved in the appendix while the second can be found on page 100 of Royden (1988).

    Lemma 8.1. Let \(f: ℝ → ℝ\) be a function. If \(f\) is differentiable on \([a, b]\), \(f' ∈ L^1([a, b])\), and \(∫_a^x f'(t) \, dt = f(x) - f(a)\) for \(a ≤ x ≤ b\), then \(f ∈ AC[a,b]\).

    The converse of this lemma is also true.

    See Folland (1999) for a nice treatment of the Fundamental Theorem of Calculus (FTC) for Lebesgue integration.

    Lemma 8.2. If \(f: ℝ → ℝ\) is increasing and \(f' ∈ L^1([a,b])\), then \(∫_a^x f'(t) \, dt ≤ f(x) - f(a)\).

    To finish the ⇐ direction of the proof, by Lemma 8.1, it suffices to show that \(∫_ℝ f' \, dm = 1\) implies \(∫_a^b f'(t) \, dt = f(b) - f(a)\) holds for all \(-∞ < a < b < ∞\).

    By Lemma 8.2, we have \(∫_a^b f'(t) \, dt ≤ f(b) - f(a)\), so we need only show that strict inequality cannot hold.

    Suppose, by way of contradiction, that \(∫_a^b f'(t) \, dt < f(b) - f(a)\) holds for some \(-∞ < a < b < ∞\). Then,

    \(1 = ∫_ℝ f' \, dm = ∫_{-∞}^a f' \, dm + ∫_a^b f' \, dm + ∫_b^∞ f' \, dm\)

    \(< [f(a) - f(-∞)] + [f(b) - f(a)] + [f(∞) - f(b)] = f(∞) - f(-∞) = 1\).

    Therefore, \(∫_ℝ f' \, dm = 1\) implies that

    \(∫_a^b f'(t) \, dt = f(b) - f(a)\) holds for all \(-∞ < a < b < ∞\), as desired.

  • (⇒) Assume \(f ∈ AC[a, b]\) for all \(-∞ < a < b < ∞\). We must show \(∫_ℝ f' \, dm = 1\). By assumption \(f(∞) - f(-∞) = 1\), so this is equivalent to showing

    \(\lim_{x → ∞} ∫_{-x}^x f'(t) dm(t) = \lim_{x → ∞}[f(x) - f(-x)]\).

    Let \(x∈ ℝ\), \(x > 0\), and \(f ∈ AC[-x,x]\). We claim \(f(x)-f(-x) = ∫_{-x}^x f' dm\). Assuming the claim is true---see p. 110 of Royden (1988) for the proof---we have

    \(1 = \lim_{x → ∞} [f(x) - f(-x)] = \lim_{x → ∞}∫_{-x}^x f'(t) dm(t) = ∫_ℝ f' dm\).

    ∎

Solution B.3.

Since \(F ∈ L^1[-1, 1]^*\), then by the Riesz Representation Theorem there is a unique \(h ∈ L^∞[-1,1]\) such that

\(F(f) = ∫^1_{-1} f(x) h(x) \, dx\) for all \(f ∈ L^1[-1,1]\).

Let \(h = ϕ + ψ\), where \(ϕ\) and \(ψ\) are the even and odd functions,

\(ϕ(x) = \frac{h(x) + h(-x)}{2}\) and \(ψ(x) = \frac{h(x) - h(-x)}{2}\).

Let \(f = f_e + f_o\) be the decomposition of \(f\) into a sum of even and odd functions. Then, by linearity of \(F\), and since \(F(f_o) = 0\) by hypothesis,

\(F(f) = F(f_e) + F(f_o) = F(f_e) = ∫_{-1}^1 f_e h = ∫_{-1}^1 f_e ϕ +∫_{-1}^1 f_e ψ\).

Now note that \(f_e ψ\) is an odd function (even times odd), so \(∫_{-1}^1 f_e ψ = 0\), since \([-1, 1]\) is symmetric. Similarly, \(∫_{-1}^1 f_o ϕ = 0\). Therefore,

\(F(f) = F(f_e) = ∫_{-1}^1 f_e ϕ = ∫_{-1}^1 f_e ϕ + ∫_{-1}^1 f_o ϕ = ∫_{-1}^1 (f_e+f_o) ϕ = ∫_{-1}^1 f ϕ\).

∎