.. role:: premium


.. 1998 Apr 03
.. ===========

Exam 4 (27--35)
=========================

1998 Apr
---------

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

Part A.
-------

.. _1998Apr_p1:

.. proof:prob::

   Let :math:`\{x_n\}_{n=1}^∞` be a bounded sequence of real numbers, and for each positive :math:`n` define

   .. math:: \hat{x}_n = \sup_{k\geq n} x_k

   a. Explain why the limit :math:`ℓ = \lim_{n→∞}x̂_n` exists.

   b. Prove that, for any :math:`ε > 0` and positive integer :math:`N`, there exists an integer :math:`k` such that :math:`k ≥ N` and :math:`|x_k - ℓ| < ε`.

.. _1998Apr_p2:

.. proof:prob::

   Let :math:`C` be a collection of subsets of the real line :math:`ℝ`, and define

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

   a. Prove that :math:`A_σ (C)` is a :math:`σ`-algebra, that :math:`C ⊂ A_σ (C)`, and that :math:`A_σ (C) ⊂ A` for any other :math:`σ`-algebra :math:`A` containing all the sets of :math:`C`.

   b. Let :math:`O` be the collection of all finite open intervals in :math:`ℝ`, and :math:`F` the collection of all finite closed intervals in :math:`ℝ`. Show that

       .. math:: A_\sigma (O) = A_\sigma (F).

   .. index:: monotone convergence theorem
   .. _1998Apr3:

.. _1998Apr_p3:

.. proof:prob::

   Let :math:`(X, 𝔐, μ)` be a measure space, and suppose :math:`X = ⋃_n X_n`, where :math:`\{X_n\}_{n=1}^∞` is a pairwise disjoint collection of measurable subsets of :math:`X`. Use the :term:`monotone convergence theorem` and linearity of the integral to prove that, if :math:`f` is a non-negative measurable real-valued function on :math:`X`,

   .. math:: ∫_X f\, dμ = ∑_n ∫_{X_n} f\, dμ.

   .. index:: pair: Fubini's theorem; Tonelli's theorem

.. _1998Apr_p4:

.. proof:prob::

   Using the Fubini/Tonelli theorems to justify all steps, evaluate the integral

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

.. _1998Apr_p5:

.. proof:prob::

   Let :math:`I` be the interval :math:`[0,1]`, and let :math:`C(I)` (resp., :math:`C(I × I)`) denote the space of real valued continuous functions on :math:`I` (resp., :math:`I\times I`), with the usual supremum norm. Show that the collection of *finite* sums of the form

   .. math:: f(x,y) = ∑_i φ_i(x) ψ_i(y),

   where :math:`φ_i, ψ_i ∈ C(I)` for each :math:`i`, is dense in :math:`C(I × I)`.

   .. index:: Radon-Nikodym derivative, absolutely continuous

.. _1998Apr_p6:

.. proof:prob::

   Let :math:`m` be Lebesgue measure on the real line :math:`ℝ` and for each Lebesgue measurable subset :math:`E` of :math:`ℝ` define

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

   Show that :math:`m` is :term:`absolutely continuous <absolutely continuous measure>` with respect to :math:`μ` and compute the Radon-Nikodym derivative :math:`dm/dμ`.

-------------------------------

Part B.
-------

.. _1998Apr_p7:

.. proof:prob::

   Let :math:`φ(x,y) = x^2 y` be defined on the square :math:`S = [0,1] × [0,1]` in the plane, and let :math:`m` be two-dimensional Lebesgue measure on :math:`S`.

   Given a Borel subset :math:`E` of the real line :math:`ℝ`, define

   .. math:: μ \,E = m\, φ^{-1}(E).

   (a) Show that :math:`μ` is a :term:`Borel measure` on :math:`ℝ`.

   (b) Let :math:`χ_E` denote the :term:`characteristic function` of the set :math:`E`. Show that

       .. math:: ∫_ℝ χ_E \, dμ = \int_S  χ_E ∘ φ \, dm.

   (c) Evaluate the integral

       .. math:: ∫_{-∞}^∞ t^2  \, dμ(t).

   .. index:: absolutely continuous

.. _1998Apr_p8:

.. proof:prob::

   Let :math:`f` be a real-valued increasing function on the real line :math:`ℝ`, such that :math:`f(-∞)=0` and :math:`f(∞)=1`.

   Prove that :math:`f` is :term:`absolutely continuous <absolutely continuous function>` on every closed finite interval if and only if

   .. math:: ∫_ℝ f' \, dm = 1.

   .. index:: linear functional, odd function, even function

.. _1998Apr_p9:

.. proof:prob::

   Let :math:`F` be a continuous :term:`linear functional` on the space :math:`L_1[-1,1]`, with the property that :math:`F(f) = 0` for all *odd* functions :math:`f` in :math:`L_1[-1,1]`. Show that there exists an *even* function :math:`φ` such that

   .. math:: F(f) = ∫_{-1}^1 f(x) φ(x) \, dx, \quad \text{ for all } f∈ L_1[-1,1].

   [Hint: One possible approach is to use the fact that every function in :math:`L_p[-1,1]` is the sum of an odd function and an even function.]

-------------------------------

Part A Solutions
----------------

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  .. container:: header

     **Solution** to :numref:`Problem {number} <1998Apr_p1>`

  a. By assumption
  
     .. math:: x̂_n = \sup_{k≥n} x_k = \sup \{x_n, x_{n+1}, \dots \} ≥ \sup \{x_{n+1}, x_{n+2}, \dots \} = x̂_{n+1},
     
     and there exists :math:`M>0` such that :math:`\{x_n\} ⊆ [-M, M]` for all :math:`n`.
     
     Therefore, :math:`x̂_n ∈ [-M,M]` for all :math:`n`, so :math:`\{x̂_n\}` is a bounded monotone decreasing sequence of real numbers.
     
     By the :term:`Bolzano-Weierstrass theorem`, :math:`x̂_n` converges to some :math:`ℓ := \lim_{n→ ∞} x̂_n ∈ [-M,M]`.
     
     ☐

  b. Fix :math:`ε > 0` and :math:`N>0`.  By definition of :math:`ℓ`, there exists :math:`N_0 ≥ N` such that :math:`|x̂_k - ℓ| < ε/2` for all :math:`k ≥ N_0`.

     By definition of supremum, there exists :math:`j>0` such that

     .. math:: |x_{N_0+j} - \hat{x}_{N_0}| = |x_{N_0+j} - \sup \{x_{N_0}, x_{N_0+1}, \dots\}| < ε/2.

     Therefore,

     .. math:: |x_{N_0+j} - ℓ| ≤ |x_{N_0+j} - x̂_{N_0}| + |x̂_{N_0} - ℓ| ≤ ε/2 + ε/2 = ε.

     Thus, for arbitrary :math:`ε > 0` and :math:`N>0`, we have found an integer :math:`k = N_0 + j ≥ N` such that :math:`|x_k - ℓ| < ε`, as desired.
     
     ☐

.. container:: toggle

  .. container:: header

     :premium:`Solution to` :numref:`Problem {number} <1998Apr_p2>`

  (coming soon)

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    .. container:: header

     **Solution** to :numref:`Problem {number} <1998Apr_p3>`

    Define :math:`f_n = ∑_{k=1}^n f χ_{X_k} = f χ_{⋃_1^n X_k}`.
    
    Then it is clear that the hypotheses of the :term:`monotone convergence theorem` are satisfied. That is, for all :math:`x ∈ X`,

    #. :math:`0≤ f_1(x) ≤ f_2(x) ≤ \cdots ≤ f(x)`, and

    #. :math:`\lim_{n → ∞} f_n(x) = f(x)χ_X(x) = f(x)`.

    Therefore,

    .. math:: ∑_{k=1}^∞ ∫_{X_k} f\, dμ &=  \lim_{n → ∞} ∑_{k=1}^n ∫_X f χ_{X_k}\, dμ &\\
              &= \lim_{n→∞} ∫_X ∑_{k=1}^n f χ_{X_k}\, dμ & \text{ (by linearity of the integral)}\\
              &=  \lim_{n→∞} ∫_X f_n \, dμ & (\text{by definition of} f_n)\\
              &= ∫_X \lim_{n → ∞} f_n \, dμ & (\text{by the monotone convergence theorem})\\
              &= ∫_X f\, dμ. &

    ☐

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    .. container:: header

       **Solution** to :numref:`Problem {number} <1998Apr_p4>`

    By :term:`Tonelli's theorem`, if :math:`f(x,y)≥ 0` is :term:`measurable <measurable function>` and one of the iterated integrals :math:`∬ f(x,y)\,dx\,dy` or :math:`∬ f(x,y)\,dy\,dx` exists, then they both exist and are equal.
    
    Moreover, if one of the iterated integrals is finite, then :math:`f(x,y) ∈ L_1(dx,dy)`.
    
    :term:`Fubini's theorem` states that :math:`f(x,y)∈ L_1(dx,dy)` implies the iterated integrals exist and are equal.

    Now let :math:`g(x,y) = x^{-3/2}\cos(π y/2x)`, and apply :term:`Tonelli's theorem` to the (nonnegative measurable) function :math:`|g(x,y)|` as follows:

    .. math:: ∫_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} ⋅ 1 \,dy\,dx\\
                                           &= ∫_0^1 x^{-1/2} \,dx = 2.

    Thus one of the iterated integrals is finite which, by :term:`Tonelli's theorem`, implies :math:`g(x,y)∈ L_1(dx,dy)`.
    
    Therefore, :term:`Fubini's theorem` applies to :math:`g(x,y)` and gives the first of the following equalities:

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

    ☐

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    .. container:: header

       :premium:`Solution to` :numref:`Problem {number} <1998Apr_p5>`

    (coming soon)

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    .. container:: header

       **Solution** to :numref:`Problem {number} <1998Apr_p6>`

    Obviously both measures are nonnegative.
    
    We must first prove :math:`m ≪ μ`. To this end, let :math:`𝔐` denote the :term:`σ-algebra` of (Lebesgue) measurable sets and suppose :math:`E ∈ 𝔐` and :math:`m E >0`.
    
    Then, if we can show :math:`μ E > 0`, this will establish that the following implication holds for all :math:`E ∈ 𝔐`: :math:`μ E = 0 \, ⇒ \, m E=0`; that is, :math:`m ≪ μ`.

    For :math:`n = 1, 2, \ldots`, define

    .. math:: A_n = \left\{x∈ ℝ: \frac{1}{n+1} < \frac{1}{1+x^2} ≤ \frac{1}{n}\right\}.

    Then :math:`A_i ∩ A_j = ∅` for all :math:`i≠ j` in :math:`ℕ`, and, for all :math:`n = 1, 2, \ldots`,

    .. math:: \mu(A_n) \geq \frac{1}{n+1} m(A_n).

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

    .. math:: \mu(E) = \mu(E\cap (\cup_n A_n)) = \mu(\cup_n (A_n \cap E)) = \sum_n \mu(A_n \cap E).

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

    Now note that :math:`m E = ∑_n m(A_n ∩ E) >0` implies the existence of an :math:`n ∈ ℕ` such that :math:`m (A_n ∩ E) >0`. Therefore,

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

    which proves that :math:`m ≪ μ`.
    
    By the :term:`Radon-Nikodym theorem`, there is a unique :math:`h ∈ L_1(μ)` such that

    .. math:: m(E) = ∫ h \, dμ, \quad \text{ and } \quad ∫ f\, \, dm  = ∫ f h \, dμ \quad  ∀ f ∈ L_1(m).

    In particular, if :math:`E ∈ 𝔐` and :math:`f(x) = \frac{1}{1+x^2} χ_E`, then

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

    That is, :math:`∫_E \, dμ = ∫_E \frac{h(x)}{1+x^2}\, dμ(x)` holds for all measurable sets :math:`E`. It follows by :term:`integral extensionality` that :math:`\frac{h(x)}{1+x^2} = 1` for :math:`μ`-almost every :math:`x ∈ ℝ`. Therefore,

    .. math:: \frac{\, dm}{dμ}(x) = h(x) = 1+x^2.

    One final note: :math:`h` is uniquely defined only up to an :term:`equivalence class` of functions that equal :math:`1+x^2` for :math:`μ`-almost every :math:`x ∈ ℝ`.
    
    ☐

----------------------

Part B Solutions
----------------

.. container:: toggle

  .. container:: header

     :premium:`Solution to` :numref:`Problem {number} <1995Nov_p7>`

  (coming soon)

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   .. container:: header

     **Solution** to :numref:`Problem {number} <1998Apr_p8>`

   First note that :math:`f` is increasing, so by :term:`differentiability of increasing functions`, :math:`f'` exists for a.e. :math:`x ∈ ℝ`, and :math:`f'(x) ≥ 0` wherever :math:`f'` exists.

   Also, :math:`f'` is measurable. To see this, define

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

   As a limsup of a sequence of :term:`measurable functions <measurable function>`, :math:`g` is measurable (:cite:`Rudin:1987`, Theorem 1.14).

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

   (⇒) Assume :math:`f∈ AC[a,b]` for all :math:`-∞<a<b<∞`. We must show :math:`∫_ℝ f' \, dm =1`, which is equivalent to showing

     .. math:: \lim_{x\rightarrow \infty}\int_{-x}^x f'(t) \, dm(t) = \lim_{x\rightarrow \infty}[f(x)-f(-x)],

     since :math:`f(\infty)-f(-\infty) = 1`, by assumption.

     Let :math:`x∈ ℝ`, :math:`x>0`, and :math:`f\in AC[-x,x]`. Then  we claim :math:`f(x)-f(-x) = \int_{-x}^x f' \, dm`. Assuming the claim is true (see :cite:`Royden:1988`, page 110 for the proof), we have

     .. math:: 1 = \lim_{x\rightarrow \infty}[f(x)-f(-x)] = \lim_{x\rightarrow \infty}\int_{-x}^x f'(t) \, dm(t) = \int_\BbbRf' \, dm.

   (⇐) Assume :math:`∫_{ℝ} f' \, dm =1`. We must show :math:`f∈ AC[a,b]` for all :math:`-∞ < a < b < ∞`.

     Observe that :math:`f' ∈ L_1(ℝ)`, since

     .. math:: 1=∫_ℝ f' \, dm  = ∫_{ℝ-E} f' \, dm + ∫_{E} f' \, dm = ∫_{E} f' \, dm,

     and, since :math:`f` is increasing, :math:`f'≥ 0` on :math:`E`, so

     .. math:: ∫_ℝ|f'| \, dm  = ∫_{ℝ- E} |f'| \, dm + ∫_{E} |f'| \, dm = ∫_{E} |f'| \, dm =  ∫_{E} f' \, dm = 1.

     Thus, :math:`f' ∈ L_1(ℝ)` as claimed. 

     Since :term:`AC is equivalent to being an indefinite integral`, it suffices to show that :math:`∫_ℝ f' \, dm =1` implies :math:`\int_a^x f'(t)\, dt = f(x) - f(a)` holds for all :math:`-∞ < a < x < ∞`.

     By the :term:`differentiability of increasing functions`, we have :math:`∫_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 :math:`∫_a^b f'(t) dt < f(b) - f(a)` holds for some :math:`-∞ < a < b < ∞`. Then,

     .. math::

         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.

     This contradiction proves that :math:`∫_ℝ f' \, dm =1` implies :math:`∫_a^b f'(t) dt = f(b) - f(a)` holds for all :math:`-∞ < a < b < ∞`, as desired. ☐

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     **Solution** to :numref:`Problem {number} <1998Apr_p9>`

    Since :math:`F ∈ L_1^∗[-1,1]`, then by the :term:`Riesz representation theorem <Riesz representation theorem (version 2)>` there is a unique :math:`h∈ L_∞[-1,1]` such that

    .. math:: F(f) = ∫^1_{-1} f(x) h(x) \, dx \qquad (∀ f ∈ L_1[-1,1])

    Now (using the hint) write :math:`h = φ + ψ`, where :math:`φ` and :math:`ψ` are the even and odd functions,

    .. math::

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

    Similarly, let :math:`f = f_e + f_o` be the decomposition of :math:`f` into a sum of even and odd functions.
    
    Then, by linearity of :math:`F` and since :math:`F(f_o) = 0` by hypothesis,

    .. math::  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.

    As the product of even and odd, :math:`f_e ψ` is odd; so :math:`∫_{-1}^1 f_e ψ = 0` (since :math:`[-1,1]` is symmetric).
    
    Similarly, :math:`∫_{-1}^1 f_o φ = 0`. Therefore,

    .. math::

       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.

    (See also :numref:`Problem {number} <2001Nov_p3>`.)

-----------------------------