.. role:: premium

.. 1995 Nov 17
.. ============

Exam 3 (17--26)
=======================

1995 Nov
--------

**Instructions**. Each of the following 10 problems will be scored from 0 to 10. Please begin each problem on a separate sheet and write on only one side of each sheet.

The following notations are used in the problem statements:

* :math:`ℝ` is the set of real numbers; 
* :math:`ℕ` is the set :math:`\{0,1,2,\dots\}`; 
* :math:`m^∗` is Lebesgue outer measure, which is defined on the set of all subsets of :math:`ℝ`. 
* :math:`m` is Lebesgue measure, which is the restriction of :math:`m^∗` to the set of Lebesgue measurable susbsets of :math:`ℝ`.

.. _1995Nov_p1:

.. proof:prob::

   Define the following terms:

   a. Lebesgue outer measure (which is denoted by :math:`m^∗`).
   b. Lebesgue measurable subset of :math:`ℝ` (give Caratheodory's definition).
   c. Borel subsets of :math:`ℝ`.
   d. Lebesgue measurable function (defined on a subset of :math:`ℝ` and taking values in :math:`ℝ`).
   e. Borel function (defined on a subset of :math:`ℝ` and taking values in :math:`ℝ`).

.. _1995Nov_p2:

.. proof:prob::

   Suppose that :math:`A ⊆ ℝ` and :math:`m^∗ A = 0`. Prove that :math:`A` is a Lebesgue measurable subset of :math:`ℝ`.

.. _1995Nov_p3:

.. proof:prob::

   Suppose that :math:`f: [0,∞) → ℝ`, :math:`f` is :term:`uniformly continuous`, and :math:`∫_0^∞ |f(x)|\, dm(x) < ∞`. Prove that :math:`\lim\limits_{x→ ∞} f(x) = 0`.

.. _1995Nov_p4:

.. proof:prob::

   Suppose that :math:`f, g: ℝ → ℝ`, :math:`f` and :math:`g` are :term:`continuous <continuous function>`, :math:`∫_ℝ |f(x)|\, dm(x) < ∞` and :math:`∫_ℝ |g(x)|\, dm(x) < ∞`. For each :math:`x∈ ℝ`, define :math:`f ⋆ g (x) := ∫_ℝ f(x-t)g(t)\, dm(t)`. Prove that :math:`∫_ℝ |f ⋆ g (x)|\, dm(x) < ∞`.

.. _1995Nov_p5:

.. proof:prob::

   State a version of :term:`Fatou's lemma` (which concerns a relationship between integrals of limits and limits of integrals). Give an example to show that strict inequality can obtain in your version of Fatou's lemma.

.. _1995Nov_p6:

.. proof:prob::

   Suppose that :math:`f: [0,1]→ ℝ`, :math:`f` is a Lebesgue measurable function, and, whenever :math:`φ: [0,1]→ℝ` and :math:`φ` is continuous, we have :math:`∫_0^1 f(x) φ(x)\, dm(x) = 0`. Prove that :math:`f(x) = 0` for :math:`m`-almost every :math:`x∈ [0,1]`. (Hint: Prove that :math:`∫_0^xf(t) \, dm(t) = 0` for all :math:`x∈ [0,1]` and finish by citing a major theorem.)

.. _1995Nov_p7:

.. proof:prob::

   Suppose that :math:`(X, 𝔐, μ)` is a measure space, :math:`(A_n)_{n=1}^∞` is a sequence from :math:`𝔐`, and :math:`A_n ⊆ A_{n+1}` whenever :math:`n∈ ℕ`. Prove that 

   .. math:: \lim\limits_{n↑∞} μ A_n = μ \left(⋃_{n=0}^∞ A_n\right).

.. _1995Nov_p8:

.. proof:prob::

   Suppose that :math:`μ` is a measure on the :term:`Borel subsets <Borel set>` of :math:`[0,1]`, :math:`μ[0,1]<∞`, :math:`f:[0,1] → ℝ`, :math:`f` is a :term:`Borel function`, and :math:`f` is integrable with respect to :math:`μ` on :math:`[0,1]`. Define :math:`F: [0,1] → ℝ` by the rule :math:`F(x) = ∫_{[0,x]} f(t) \, dμ(t)`. Prove that :math:`F` is of :term:`bounded variation` on :math:`[0,1]`. State conditions on :math:`f` and :math:`μ` which imply that :math:`F` is :term:`absolutely continuous <absolutely continuous function>` on :math:`[0,1]`.

.. _1995Nov_p9:

.. proof:prob::

   Let :math:`(Ω_j, τ_j)` be a :term:`topological space` for each :math:`j∈ J`. Define the corresponding :term:`product topology`, :math:`τ`, on the Cartesian product :math:`∏_{j∈ J} Ω_j`. State :term:`Tychonoff's theorem`.

.. _1995Nov_p10:

.. proof:prob::

   State the :term:`Baire category theorem` for a general :term:`complete metric space` and cite an application of the theorem; no details are required in the citation.

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

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     **Solution** to :numref:`Problem {number} <1995Nov_p1>`

  a. See :term:`Lebesgue measure`.
  b. See :term:`Lebesgue measurable set` and :term:`measurable set`.
  c. See :term:`Borel set`.
  d. See :term:`Lebesgue measurable function`.
  e. See :term:`Borel measurable function`.

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     **Solution** to :numref:`Problem {number} <1995Nov_p2>`

  Let :math:`A ⊆ ℝ` and assume :math:`m^∗ A = 0`. We must show that :math:`A` is :term:`Lebesgue measurable <Lebesgue measurable set>` subset of :math:`ℝ`.
  
  Equivalently, we show that :math:`m^∗(B) = m^∗(B ∩ A) + m^∗(B ∩ A^c)` for every :math:`B ⊆ ℝ`.

  Indeed, by :term:`subadditivity <subadditive>` of :math:`m^∗` we have :math:`m^∗(B) ≤ m^∗(B ∩ A) + m^∗(B ∩ A^c)`, so we need only prove :math:`≥`.
  
  Recall, a :term:`measure` is called :term:`complete <complete measure>` if every subset of a :term:`negligible` set is :term:`measurable <measurable set>` and negligible.
  
  Since :math:`B∩ A ⊆ A` and since Lebesgue outer measure is complete, we have

  .. math:: m^∗(B ∩ A) = 0.
     :label: 2.1

  By monotonicity of :math:`m^∗` under the partial ordering :math:`⊆`, we have

  .. math:: m^∗(B ∩ A^c) ≤ m^∗(B) \quad (∵ B∩ A^c ⊆ B).
     :label: 2.2
  
  By :eq:`2.1` and :eq:`2.2`, :math:`m^∗(B) ≥ m^∗(B ∩ A) + m^∗(B ∩ A^c)`.

  ☐

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     **Solution** to :numref:`Problem {number} <1995Nov_p3>`

  Suppose that :math:`f: [0,∞) → ℝ`, :math:`f` is :term:`uniformly continuous`, and :math:`∫_0^∞ |f(x)|\, dm(x) < ∞`. Prove that :math:`\lim\limits_{x→ ∞} f(x) = 0`.

  Suppose on the contrary that :math:`\lim\limits_{x→ ∞} f(x) ≠ 0`.  Then there exists :math:`ε>0` such that for all :math:`M>0` :math:`∃ x > M`

  .. math:: (∃ε>0)\, (∀ M>0) \, (∃ x > M), \, |f(x)| > ε.

  Fix such an :math:`ε>0`. Since :math:`f` is uniformly continuous, :math:`∃ δ>0` such that 

  .. math:: |f(x) - f(y)| < ε/2 \quad \text{ for all }\; |x - y| < δ.

  Taking :math:`δ' = \min \{δ, 1/2\}` if necessary, we can assume without loss of generality that our original :math:`δ` is :math:`< 1`.

  Now

  .. math:: ∑_{n=0}^∞ ∫_{[n,n+1]} |f(x)| \, dm(x) ≤ ∫_0^∞ |f(x)|\, dm(x) < ∞,
  
  so there exists :math:`N>0` such that 
  
  .. math:: ∑_{n=N+1}^∞ ∫_{[n,n+1]} |f(x)| \, dm(x) < ε ⋅ δ.
     :label: 3.1

  Next, find :math:`x > N+2` such that :math:`|f(x)| >ε` and 
  
  :math:`|f(x) - f(y)| < ε/2` for all :math:`|x - y| < δ`.

  By the :term:`triangle inequality`, :math:`|f(x)| ≤ |f(x) - f(y)| + |f(y)|`, so 

  .. math:: |f(y)| ≥ |f(x)|- |f(x) - f(y)| > ε/2 \quad \text{ for all } \; |x-y|< δ.
  
  Therefore,
  
  .. math:: ∫_{[x-δ , x+δ]} |f(x)| \, dm(x) ≥ \frac{ε}{2} m[x-δ, x+δ] = ε ⋅ δ.
     :label: 3.2

  On the other hand, since :math:`x>N+2`, we have :math:`[x-δ , x+δ] ⊆ [N,∞)`.
  
  Therefore, by :eq:`3.1` and :eq:`3.2`,

  .. math:: ε ⋅ δ ≤ ∫_{[x-δ , x+δ]} |f(x)| \, dm(x) ≤  ∫_{[N+1, ∞)} |f(x)| \, dm(x) = ∑_{n=N+1}^∞ ∫_{[n,n+1]} |f(x)| \, dm(x) < ε ⋅ δ,

  which is a contradiction.  Therefore, :math:`f(x) → 0` as :math:`x → ∞`.

  ☐  

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     **Solution** to :numref:`Problem {number} <1995Nov_p4>`

  Since the function :math:`ψ(x,y) := x-y` is a continuous mapping from :math:`ℝ^2` to :math:`ℝ`, the function
  
  .. math:: F(x,y) := (f ∘ ψ)(x,y) = f(x-y)`

  is also continuous and therefore measurable.

  Since the second projection function :math:`π_2: ℝ^2 → ℝ`---defined for all :math:`(x,y) ∈ ℝ^2` by :math:`π_2(x,y) := y`---is continuous, so is the function :math:`G(x,y) := (g ∘ π_2)(x,y) = g(y)`; the latter is thus measurable.

  Therefore, the function
  
  .. math:: H(x,y) = F(x,y) G(x,y) = f(x-y) g(y)

  is a measurable function on :math:`ℝ^2`.

  Next note the analogy with the functions defined in :term:`Fubini's theorem`:

  .. math:: φ(x) = ∫_Y H(x,y)\, dν(y) = ∫_ℝ f(x-y)g(y) \, dm(y) = (f ∗ g)(x).
     :label: phi

  In the present context, :math:`(X, 𝔐, μ) = (Y, 𝔑, ν) = (ℝ, ℬ(ℝ), m)`.

  Now applying Part 2 of :term:`Fubini's theorem`, we consider the integral with respect to :math:`x` first:

  .. math:: ∬_{ℝ^2} |f(x-y)||g(y)| \, dm(x) \, dm(y) &=∫_ℝ |g(y)| ∫_ℝ |f(x-y)| \, dm(x) \, dm(y)\\
                               &=∫_ℝ |g(y)| \, dm(y) ∫_ℝ |f(x)| \, dm(x) < ∞.

  The second equality holds by :term:`translation invariance` of Lebesgue measure.

  Therefore, Fubini's theorem implies :math:`H ∈ L_1(m × m)`.

  Finally, Fubini's theorem Part 3 (iii) says that :math:`H ∈ L_1(m × m)` implies :math:`φ ∈ L_1(m)`.

  By :eq:`phi`, this is equivalent to 

  .. math:: ∫ |φ| \, dm = ∫_ℝ | (f ∗ g)(x)| \, dm(x) < ∞.

  ☐  

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     **Solution** to :numref:`Problem {number} <1995Nov_p5>`

  :term:`Fatou's lemma` (relating integrals of limits and limits of integrals) asserts the following: if :math:`\{f_n\}` is a sequence of nonnegative, extended, real-valued, measurable functions then :math:`∫\varliminf f_n ≤ \varliminf ∫ f_n`. 
   
  We now describe a witness to strict inequality. Define :math:`f_n = \frac{1}{n}χ_{[0,n]}` for :math:`n =1,2,\dots`.  Then :math:`\varliminf f_n(x)  = 0` for all :math:`x` and :math:`∫ f_n = 1` for all :math:`n`, so  

  .. math:: ∫\varliminf f_n  = 0 < 1 = \varliminf ∫ f_n. 

  ☐  

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     **Solution** to :numref:`Problem {number} <1995Nov_p6>`

  Let :math:`f: [0,1]→ ℝ` be a :term:`Lebesgue measurable function`, and assume that for all :math:`φ ∈ C([0,1],ℝ)` we have :math:`∫_0^1 f(x) φ(x)\, dm(x) = 0`.
  
  We wish to prove that :math:`f(x) = 0` for almost every :math:`x∈ [0,1]`.
  
  First observe that :math:`f∈ L_1[0,1]`. Indeed, take :math:`φ` to be the constant function :math:`φ ≡ 1`, so that
  
  .. math:: ∫_0^1f(t) \, dm(t) = ∫_0^1f(t) φ(x)\, dm(t) = 0, 
  
  so the :term:`integral` :math:`∫_0^1f(t) \, dm(t)` exists, which by :term:`definition <Lebesgue integral>` means that :math:`∫_0^1 f^+` and :math:`∫_0^1 f^-` both exist, which is true if and only if :math:`∫_0^1 |f|` exists. (See the theorem on :term:`integrability of a measurable function`.)
  
  Next, we prove the assertion given in the hint.  That is, :math:`∫_0^xf(t) \, dm(t) = 0` for all :math:`x∈ [0,1]`.
  
  Indeed, if :math:`x=1`, take :math:`φ ≡ 1` so that :math:`∫_0^xf(t) \, dm(t) = ∫_0^1f(t) φ(x)\, dm(t) = 0`.
  
  If :math:`0≤ x<1`, then for each :math:`n =1,2,\dots` define
  
  .. math:: φ_n(t) = \begin{cases} 1, & 0≤ t ≤ x-1/n\\ -n t,& x-1/n < t≤ x\\0, & x<t≤ 1.\end{cases}
  
  Then :math:`φ_n ∈ C([0,1],ℝ)` and :math:`\lim_{n→∞}f(x)φ_n(x) = f(x)χ_{[0,x]}(x)`.  Also,
  
  .. math:: -|f| ≤ -|f| φ_n ≤ f φ_n ≤ |f| φ_n ≤ |f|,

  holds for all :math:`n`, so the :term:`bounded convergence theorem` gives the penultimate eequality in the following calculation.

  .. math:: ∫_0^x f(t)\, dm(t) &=  ∫_0^1f(x)χ_{[0,x]}(x)\, dm(t)\\  &= ∫_0^1 \lim_{n→∞}f(x)φ_n(x)\, dm(x)\\ &= \lim_{n→∞}∫_0^1f(x)φ_n(x) = 0.

  It follows for all :math:`0≤ a ≤ b ≤ 1` that

  .. math:: ∫_0^a f(t)\, dm(t) = ∫_0^bf(x)\, dm(t) = 0,
  
  and therefore,
  
  .. math:: ∫_a^b f(t)\, dm(t) = ∫_0^bf(x)\, dm(t) - ∫_0^a f(t)\, dm(t) = 0.
     :label: zero integral

  Now let :math:`B := \{x ∈ [0,1] : f(x) >0\} = f^{-1}((0,∞])`, which is measurable since :math:`f` is measurable. We will prove :math:`m B = 0`.
  
  Recall the (:term:`Borel <Borel set>`) :term:`measurable sets <measurable set>` in :math:`[0,1]` are generated by the half-open intervals of the form :math:`(a,b]`.
  
  Since :math:`B` is measurable, either we can find an interval :math:`(a,b] ⊆ B`, or else :math:`B` is a set of measure 0.

  If :math:`m B > 0`, then there exists :math:`0≤ a < b ≤ 1` such that :math:`(a,b] ⊆ B`. Since :math:`f>0` on :math:`(a,b]`, we then have :math:`∫_a^b f(x) \, dm(x) >0`, which contradicts :eq:`zero integral`.  Therefore, we must have :math:`m B = 0`.

  Similarly, the set :math:`C := \{x∈ [0,1] : f(x) <0\}` is :term:`negligible`, so
  
  .. math:: B ∪ C = \{x ∈ [0,1] : f(x) ≠ 0\}
  
  is negligible.  This proves that :math:`f(x) = 0` for almost every :math:`x∈ [0,1]`.

  ☐

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

  Let :math:`(X, 𝔐, μ)` be a :term:`measure space` and :math:`\{A_n\}_{n=1}^∞ ⊆ 𝔐` a sequence of :term:`measurable sets <measurable set>` such that :math:`A_n ⊆ A_{n+1}` for all :math:`n∈ ℕ`. We must show,

  .. math:: \lim\limits_{n → ∞} μ A_n = μ \left(⋃_{n=1}^∞ A_n\right).
     :label: goal p7

  Define :math:`A:= ⋃_{n=1}^∞ A_n` and consider the sequence :math:`\{χ_{A_n}\}` of characteristic functions of the :math:`A_n`.
  
  Clearly :math:`A_n ⊆ A_{n+1}` (:math:`∀ n`) implies :math:`0≤ χ_{A_1} ≤ \cdots ≤ χ_{A_n} ≤ χ_{A_{n+1}} ≤ \cdots ≤ χ_A ≤ χ_X = 1`, and :math:`\lim_{n→ ∞} χ_{A_n} = χ_A`, so the :term:`monotone convergence theorem` yields the second equality in the following calculation:

  .. math:: \lim_{n→ ∞} μ A_n = \lim_{n→ ∞} ∫χ_{A_n}\, dμ = ∫\lim_{n→ ∞} χ_{A_n}\, dμ = ∫χ_A\, dμ = μA = μ\left(⋃_{n=1}^∞A_n\right).

  Thus :eq:`goal p7` is proved.

  ☐
 
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     :premium:`Solution to` :numref:`Problem {number} <1995Nov_p8>`

  Let :math:`μ` be a finite measure on the :term:`Borel subsets <Borel set>` of :math:`[0,1]`. Let :math:`f:[0,1] → ℝ` be a :term:`Borel function`, and assume :math:`f ∈ L_1([0,1],μ)`. Define :math:`F: [0,1] → ℝ` by the rule
  
  .. math:: F(x) = ∫_{[0,x]} f(t) \, dμ(t).
     :label: 8.1
  
  **Claim**. :math:`F` is of :term:`bounded variation` on :math:`[0,1]`.

  *Proof*. By :eq:`8.1` we have :math:`F(x_{i})-F(x_{i-1}) = ∫_{x_{i-1}}^{x_{i}} f(t) \, dμ(t)`,

    so 

    .. math:: |F(x_{i})-F(x_{i-1})| ≤ ∫_{x_{i-1}}^{x_{i}} |f(t)| \, dμ(t).

    Since :math:`f∈ L_1` we have :math:`∫_0^1 |f(t)| \, dμ(t) < ∞`, and for every partition :math:`0=x_0 < \cdots < x_n = 1`,
  
    .. math:: ∑_{i=1}^n|F(x_{i})-F(x_{i-1})| &≤ ∑_{i=1}^n\left|∫_{x_{i-1}}^{x_{i}} f(t) \, dμ(t)\right|\\
                                             &≤ ∑_{i=1}^n∫_{x_{i-1}}^{x_{i}} |f(t)| \, dμ(t)\\
                                             &= ∫_0^1 |f(t)| \, dμ(t) < ∞,
    
    which proves that :math:`F ∈ BV[0,1]`.

    ☐

   .. todo:: To complete the solution we must give conditions on :math:`f` and :math:`μ` under which :math:`F` is :term:`absolutely continuous <absolutely continuous function>` on :math:`[0,1]`.
  

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     **Solution** to :numref:`Problem {number} <1995Nov_p9>`

  For the first part, see the :term:`definition of product topology <product topology>` in the :ref:`appendix of definitions <appendix-definitions>`.

  For the second part, see the :term:`statement of Tychonoff's theorem <Tychonoff's theorem>`  in the :ref:`appendix of theorems <appendix-theorems>`.
  
  ☐  

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     **Solution** to :numref:`Problem {number} <1995Nov_p10>`

  Here's a version of the :term:`Baire category theorem`:
  
    Let :math:`X` be a :term:`complete metric space` and let :math:`G⊆ X` be an :term:`open set`.  If :math:`G = ⋃_{n=1}^∞ G_n` then :math:`(Ḡ_n)^° ≠ ∅` for at least one :math:`n∈ ℕ`.

  Here's another version:

    If :math:`\{A_n: n ∈ ℕ\}` is a (countable) collection of :term:`dense <dense set>` subsets of :math:`X`, then :math:`⋂_{n=0}^∞ A_n` is dense.

  Here's a corollary:

    A :term:`complete metric space` is not a countable union of :term:`nowhere dense` sets.

  Here's an application:

    Often we want to prove that a particular set has a nonempty interior. For example, the Baire category theorem can be used to prove the following:

    **Proposition**. If :math:`T: X → Y` is a linear map from one :term:`Banach space` to another, then :math:`T` is bounded iff the set :math:`T^{-1}\{y∈ Y ∣ \|y\|_Y ≤ 1\}` has a nonempty interior.

  A related application due to `Stefan Banach`_ and `Hugo Steinhaus`_ is the `uniform boundedness principle`_ which asserts the following:

    Let :math:`ℱ` be a family of mappings from one :term:`Banach space` :math:`X` to another :math:`Y`.  Suppose that for each :math:`x ∈ X` the set :math:`\{f(x) ∣ f ∈ ℱ\}` is bounded; i.e., :math:`∃ M_x` such that :math:`\|f(x)\|_Y ≤ M_x` for all :math:`f ∈ ℱ`.  Then :math:`ℱ` is uniformly bounded; i.e., :math:`∃ M` such that  that :math:`\|f(x)\|_Y ≤ M \|x\|_X` for all :math:`x∈ X` and all :math:`f ∈ ℱ`. Equivalently, :math:`\|f\| = \sup\{\|f(x)\|_Y : \|x\|_X = 1\} ≤ M`.

  ☐  

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.. .. rubric:: Footnotes

.. .. [1]

.. include:: hyperlink_references.rst