.. 2007 Nov 16
.. ===========

Exam 10 (71--76)
========================

2007 Nov
---------

**Notation**  :math:`ℝ` is the set of real numbers and :math:`ℝ^n` is :math:`n`-dimensional :term:`Euclidean space`. Denote by :math:`m` :term:`Lebesgue measure` on :math:`ℝ` and :math:`m_n` :math:`n`-dimensional Lebesgue measure. Be sure to give a *complete statement* of any theorems from analysis that you use in your proofs below.

.. _2007Nov_p1:

.. proof:prob::

   Let :math:`μ` be a :term:`positive measure` on a :term:`measure space` :math:`X`.  Assume that :math:`E_1, E_2, \dots` are :term:`measurable sets <measurable set>` in :math:`X` with the property that for :math:`n ≠ m, μ (E_n ∩ E_m) = 0`. Let :math:`E` be the union of these sets. Prove that

   .. math:: μ E = ∑_{n=1}^∞ μ E_n

   .. index:: Littlewood's principle, a.e. convergence, Egoroff's theorem, Fatou's lemma

.. _2007Nov_p2:

.. proof:prob::

   a. State a theorem that illustrates Littlewood's principle for pointwise a.e. convergence of a sequence of functions on :math:`ℝ`.

   b. Suppose that :math:`f_n ∈ L_1(m)` for :math:`n=1,2,\dots`. Assuming that :math:`\|f_n-f\|_1 → 0` and :math:`f_n → g` a.e. as :math:`n → ∞`, what relation exists between :math:`f` and :math:`g`? Make a conjecture and then prove it using the statement in Part a.

.. _2007Nov_p3:

.. proof:prob::

   Let :math:`K` be a :term:`compact set` in :math:`ℝ^3` and let :math:`f(x) = \mathrm{dist}(x,K)`.

   a. Prove that :math:`f` is a :term:`continuous function` and that :math:`f(x) = 0` if and only if :math:`x ∈ K`.

   b. Let :math:`g = \max(1-f,0)` and prove that :math:`\lim_{n → ∞} ∭ g^n` exists and is equal to :math:`m_3 K`.

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

.. _2007Nov_p4:

.. proof:prob::

   Let :math:`E` be a :term:`Borel subset <Borel set>` of :math:`ℝ^2`.

   a. Explain what this means.

   b. Suppose that for every real number :math:`t` the set :math:`E_t = \{(x,y) ∈ E ∣ x = t\}` is finite. Prove that :math:`E` is a :term:`Lebesgue null set`.

   .. index:: Radon-Nikodym derivative

.. _2007Nov_p5:

.. proof:prob::

   Let :math:`μ` and :math:`ν` be :term:`finite positive measures <finite measure>` on the :term:`measurable space` :math:`(X, 𝔐)` such that :math:`ν ≪ μ ≪ ν`, and let :math:`\frac{dν}{d(μ + ν)}` represent the :term:`Radon-Nikodym derivative` of :math:`ν` with respect to :math:`μ + ν`. Show that

   .. math:: 0 < \frac{dν}{d(μ + ν)} < 1 \quad μ\mathrm{-a.e.}

   .. index:: Banach space, separable space, second countable space, Riesz representation theorem

   .. index:: pair: Banach-Steinhauss; principle of uniform boundedness

.. _2007Nov_p6:

.. proof:prob::

   Suppose that :math:`1 < p < ∞` and that :math:`q = p/(p-1)`.

   a. Let :math:`a_1, a_2, \dots` be a sequence of real numbers for which the series :math:`∑_n a_n b_n` converges for all real sequences :math:`\{b_n\}` satisfying the condition :math:`∑ _n |b_n|^q < ∞`. Prove that :math:`∑_n |a_n|^p < ∞`.

   b. Discuss the cases of :math:`p = 1` and :math:`p = ∞`. Prove your assertions.

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     **Solution** to :numref:`Problem {number} <2007Nov_p1>`

  Define :math:`F_1 = E_1, \, F_2 = E_2 \setminus E_1,\, F_3 = E_3 \setminus (E_1 \cup E_2), \dots`, and, in general,

  .. math:: F_n = E_n \setminus \bigcup_{i=1}^{n-1} E_i \quad (n=2, 3, \dots).

  If :math:`\mathfrak{M}` is the :math:`\sigma`-algebra of :math:`\mu`-measurable subsets of :math:`X,` then :math:`F_n\in \mathfrak{M}` for each :math:`n\in \mathbb N,` since :math:`\mathfrak{M}` is a :math:`\sigma`-algebra. Also, :math:`F_i\cap F_j = \emptyset` for :math:`i\neq j`, and :math:`F_1\cup F_2 \cup \cdots \cup F_n = E_1\cup E_2 \cup \cdots \cup E_n` for all :math:`n\in \mathbb N`. Thus,

  .. math:: \bigcup_{n=1}^\infty F_n = \bigcup_{n=1}^\infty E_n \triangleq E,

  and, by :math:`\sigma`-additivity of :math:`\mu`,

  .. math:: \mu(E) = \mu(\bigcup_{n=1}^\infty F_n) = \sum_{n=1}^\infty \mu(F_n).

  Therefore, if we can show :math:`\mu(E_n) = \mu(F_n)` holds for all :math:`n\in \mathbb N`, the proof will be complete.

  Now, for each :math:`n = 2, 3, \dots`,

  .. math::  F_n = E_n \cap (\bigcup_{i=1}^{n-1}E_i)^c
     :label: 2007-11-1a

  and

  .. math:: \mu(E_n) = \mu(E_n \cap (\bigcup_{i=1}^{n-1}E_i)^c) + \mu(E_n \cap (\bigcup_{i=1}^{n-1}E_i)).
     :label: 2007-11-1b

  Equation :eq:`2007-11-1b` holds because :math:`\bigcup_{i=1}^{n-1}E_i` is a measurable set for each :math:`n = 2, 3, \dots`. Finally, note that

  .. math:: E_n \cap (\bigcup_{i=1}^{n-1}E_i) =    \bigcup_{i=1}^{n-1}(E_n\cap E_i),

  which implies

  .. math:: \mu(E_n \cap (\bigcup_{i=1}^{n-1}E_i)) \leq \sum_{i=1}^{n-1}\mu(E_n\cap E_i),

  by :math:`\sigma`-subadditivity. By assumption, each term in the last sum is zero, and therefore, by :eq:`2007-11-1a` and :eq:`2007-11-1b`,

  .. math:: \mu(E_n) = \mu(E_n \cap (\bigcup_{i=1}^{n-1}E_i)^c)  = \mu(F_n) \quad \text{holds for each $n=2, 3, \dots$}.

  For :math:`n=1`, we have :math:`F_1 = E_1`, by definition.
  
  ☐

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     **Solution** to :numref:`Problem {number} <2007Nov_p2>`

  a. One of Littlewood's principles involving a.e. convergence of a sequence of functions on :math:`ℝ` is :term:`Egoroff's theorem`.

     Another is this: if :math:`f_n → f` in :math:`L_1(ℝ)`, then there is a subsequence :math:`\{f_{n_j}\}` converging to :math:`f` a.e. in :math:`ℝ`.

  b. **Claim**. :math:`f = g` a.e.

     We give two alternative proofs.

     *Proof 1*. By the second of Littlewood's principles mentioned in Part a, :math:`f_{n_j} → f` a.e.

       If :math:`f_n → g` a.e., then every subsequence of :math:`\{f_n\}` converges to :math:`g` a.e.  In particular, :math:`f_{n_j} → g` a.e.

       Let :math:`A := \{x ∈ ℝ: f_{n_j}(x) ↛ f(x) \}` and :math:`B := \{x ∈ ℝ: f_{n_j}(x) ↛ g(x) \}`.

       Then :math:`m(A ∪ B) ≤ m A + m B = 0`, and off of :math:`A ∪ B` we have,

       .. math:: |f(x) - g(x)| ≤ |f(x) - f_{n_j}(x)| + | f_{n_j}(x)- g(x)| → 0 \; \text{ as } \; j → ∞.

       Therefore, :math:`f = g` on :math:`ℝ - (A ∪ B)` and :math:`m(A ∪ B) = 0`, so :math:`f = g` a.e.
       
     ☐

     *Proof 2*. First, we claim that if :math:`f = g` a.e. on :math:`[-n,n]` for every :math:`n ∈ ℕ`, then :math:`f=g` a.e. in :math:`ℝ`.

       To see this, let :math:`B_n = \{x ∈ [-n,n]: f(x) ≠ g(x)\}`. Then :math:`m B_n = 0` for all :math:`n ∈ ℕ`, so that if :math:`B = \{x ∈ ℝ: f(x) ≠ g(x)\}`, then :math:`B = ⋃_n B_n` and :math:`m B ≤ ∑ m B_n = 0`, as claimed.

       Thus, to prove the conjecture it is enough to show that :math:`f = g` for almost every :math:`x ∈ [-n, n]`, for an arbitrary fixed :math:`n ∈ ℕ`.

       Fix :math:`n ∈ ℕ`, and suppose we know that :math:`f - g ∈ L_1([-n,n], m)`. (This will follow from the fact that :math:`f, g ∈ L_1([-n,n],m)`, which we prove below.)

       Then, for all :math:`ε > 0` there is a :math:`δ > 0` such that :math:`∫_E |f-g|\, dm < ε` for all measurable :math:`E ⊆ [-n,n]` with :math:`m E < δ`.

       Now apply :term:`Egoroff's theorem` to find a set :math:`A ⊆ [-n,n]` such that :math:`m([-n,n] - A) < δ` and :math:`f_n → g` uniformly on :math:`A`. Then

       .. math::

           ∫_{-n}^n|f-g|\,dm &= ∫_{[-n,n]- A} |f-g|\, dm + ∫_{A} |f-g|\,dm \\
                             &≤ ε + ∫_{A} |f-f_n|\, dm + ∫_{A} |f_n-g|\,dm \\
                             &≤ ε + \|f-f_n\|_1 + mA\,\sup_{x∈A}|f_n(x)-g(x)|,

       where :math:`\|f-f_n\|_1 → 0` (by assumption) and :math:`\sup_{x ∈ A}|f_n(x)-g(x)| → 0` since :math:`f_n → g` uniformly on :math:`A`.
        
       Since :math:`ε > 0` was arbitrary, it follows that :math:`∫_{-n}^n |f - g| \, dm = 0` and, for functions :math:`f, g ∈ L_1([-n,n],m)`, this implies that :math:`f = g` a.e. on :math:`[-n,n]`.

       It remains to show that :math:`f, g ∈ L_1([-n,n],m)`.

       It's clear that :math:`f ∈ L_1` since :math:`\|f\|_1 ≤ \|f - f_n\|_1 + \|f_n\|_1 < ∞`.
        
       To prove :math:`g ∈ L_1([-n,n],m)` note that :math:`f_n \xrightarrow{a.e.} g` implies :math:`\lim_n |f_n(x)| =|g(x)|` for almost all :math:`x`, so by :term:`Fatou's lemma`,

       .. math:: \|g\|_1 = ∫ |g|\, dm = ∫ \lim \inf |f_n| \, dm ≤ \lim \inf ∫ |f_n| \, dm = \lim \|f_n\|_1 = \|f\|_1 < \infty.

       The last equality holds because :math:`\bigl| \|f_n\|_1 - \|f\|_1 \bigr| ≤ \|f_n - f\|_1 → 0`, by the triangle inequality.
       
     ☐

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      **Solution** to :numref:`Problem {number} <2007Nov_p3>`

   a. Define :math:`\mathrm{dist}(x,K) = f(x) = \inf_{k∈K} |x-k|`.
   
      Clearly, for all :math:`k ∈ K`, :math:`f(x) ≤ |x-k|`. Therefore, by the triangle inequality, for all :math:`x, y ∈ ℝ^3`,

       .. math::  f(x) ≤ |x-y|+|y-k|, \quad ∀ k ∈ K,

       and so, taking the infimum over :math:`k ∈ K` on the right,

      .. math:: f(x) ≤ |x-y|+f(y).
         :label: 2007-3-1 

      Similarly,

      .. math:: f(y) ≤ |x-y|+f(x).
         :label: 2007-3-2

      Obviously, for each :math:`x ∈ ℝ^3`, :math:`f(x)` is finite. Therefore, :eq:`2007-3-1` and :eq:`2007-3-2` together imply

      .. math:: |f(x)-f(y)| ≤ |x-y|, \quad ∀ x, y ∈ ℝ^3.

      Whence :math:`f` is :term:`Lipschitz continuous`.

      Now, if :math:`x ∈ K`, then it's clear that :math:`f(x) = 0`.
      
      Suppose :math:`x ∉ K`; that is, :math:`x ∈ K^c`. Since :math:`K` is :term:`closed <closed set>`, :math:`K^c` is :term:`open <open set>` and we can find an :math:`ε`-:term:`neighborhood` about :math:`x` fully contained in :math:`K^c`, in which case :math:`f(x) > ε`.
      
      We have thus proved that :math:`f(x) = 0` if and only if :math:`x ∈ K`.
      
      ☐

   b. First, observe that :math:`f(x)=0` for all :math:`x ∈ K`, and :math:`f(x)>0` for all :math:`x ∉ K`.

      Define :math:`K_1` to be a :term:`closed <closed set>` and :term:`bounded set` containing :math:`K` on which :math:`f(x) ≤ 1`. That is, :math:`K_1` is the set of points that are a distance of not more than 1 unit from the set :math:`K`. In particular, :math:`K ⊂ K_1`.

      Notice that :math:`g = \max(1-f,0) = (1-f)χ_{K_1}`. Also, if :math:`x ∈ K_1 - K`, then :math:`0 ≤ 1-f(x) < 1`, so :math:`g^n → 0` on the set :math:`K_1 - K`, while on the set :math:`K`, :math:`g^n = 1` for all :math:`n ∈ ℕ`. Therefore, :math:`g^n → χ_K`.

      Finally, note that :math:`g^n ≤ χ_{K_1} ∈ L_1(ℝ^3)` so the :term:`dominated convergence theorem` applies and yields :math:`\lim_{n → ∞} ∭ g^n = ∭ χ_K = m_3 K.`
      
      ☐

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      **Solution** to :numref:`Problem {number} <2007Nov_p4>`

  a. Recall, the :term:`Borel σ-algebra` of :math:`ℝ^2`, denoted :math:`ℬ(ℝ^2)`, is the smallest :term:`σ-algebra` that contains the :term:`open <open set>` subsets of :math:`ℝ^2`. The sets belonging to :math:`ℬ(ℝ^2)` are called :term:`Borel sets <Borel set>` in :math:`ℝ^2`.

     ☐

  b. Observe that if :math:`G` is a :term:`finite set` in :math:`ℝ`, then :math:`G` is a :term:`Lebesgue null set`. That is, :math:`m G = 0`.

     In problems involving 2-dimensional :term:`Lebesgue measure`, distinguishing :math:`x` and :math:`y` coordinates sometimes clarifies things. To wit, let :math:`(X, ℬ(X), μ)` and :math:`(Y, ℬ(Y), ν)` be two copies of the :term:`measure space` :math:`(ℝ, ℬ(ℝ), m)`, and represent Lebesgue measure on :math:`ℝ^2` by :math:`(X × Y, ℬ(X) ⊗ ℬ(Y), μ × ν) =(ℝ^2, ℬ(ℝ^2), m_2)`. [1]_

     Our goal is to prove that :math:`m_2 E = 0`.
      
     Observe that

     .. math:: m_2 E = (μ × ν)(E) = ∫_{X × Y} χ_E \, d(μ × ν).

     The integrand :math:`χ_E` is non-negative and :term:`measurable <measurable set>` (since :math:`E` is :term:`Borel <Borel set>`). Therefore, by :term:`Tonelli's theorem`),

     .. math:: m_2 E = ∫_Y ∫_X χ_E(x,y) \, dμ(x)\, dν(y)  = ∫_X ∫_Y χ_E(x,y) \, dν(y) \, dμ(x).
        :label: 200711-4-1

     Now, let

     .. math:: G_t = \{y ∈ ℝ: (x,y) ∈ E \text{ and } x=t\}.

     This is the :math:`x`-:term:`section` of :math:`E` at the point :math:`x=t`, which is a subset of :math:`ℝ`, though we can view it as a subset of :math:`ℝ^2` by simply identifying each point :math:`y ∈ G_t` with the point :math:`(t,y) ∈ E_t = \{(x,y) ∈ E: x = t\}`.
      
     Then, for each :math:`t ∈ ℝ`, :math:`G_t` is a finite subset of :math:`ℝ`, so :math:`m G_t = 0`. [2]_
     
     Finally, by :eq:`200711-4-1`,

     .. math:: m_2 E = ∫_X ∫_Y χ_{G_t}(y) \, dν(y) \, dμ(t) = ∫_X ν (G_t) \, dμ(t) = 0.

     since :math:`ν G_t := m G_t = 0`.
     
     ☐

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      **Solution** to :numref:`Problem {number} <2007Nov_p5>`

   First note that :math:`ν ≪ μ` implies :math:`ν ≪ μ + ν`, so, by the :term:`Radon-Nikodym theorem`, there is a unique :math:`f ∈ L_1(μ + ν)` such that

   .. math:: ν E = ∫_E f \, d(μ+ν) \quad ∀ E ∈ 𝒜.

   Indeed, :math:`f` is the :term:`Radon-Nikodym derivative`; i.e., :math:`f = \frac{dν}{d(μ + ν)}`. We want to show :math:`0 < f(x) < 1` holds for :math:`\mu`-almost every :math:`x \in X`. Since we're dealing with :term:`positive measures <positive measure>`, we can assume :math:`f(x) ≥ 0` for all :math:`x ∈ X`.

   If :math:`B_0 = \{x ∈ X: f(x) = 0\}`, then

   .. math:: ν B_0 = ∫_{B_0} f \, d(μ + ν) = 0.

   Therefore, :math:`μ B_0 = 0`, since :math:`μ ≪ ν`, which proves that :math:`f(x) > 0`, :math:`μ`-a.e.

   If :math:`B_1 = \{x ∈ X: f(x) ≥ 1\}`, then

   .. math:: ν B_1 = ∫_{B_1} f \, d(μ + ν) ≥ (μ + ν)(B_1) = μ B_1 + ν B_1.

   Since :math:`ν` is finite by assumption, we can subtract :math:`ν B_0` from both sides to obtain :math:`μ B_1 = 0`. This proves :math:`f(x) < 1`, :math:`μ`-a.e.
   
   ☐

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      **Solution** to :numref:`Problem {number} <2007Nov_p6>`

  a. For each :math:`k ∈ ℕ`, define :math:`T_k: ℓ_q(ℕ) → ℝ` by :math:`T_k b = ∑_{n=1}^k a_n b_n`, for :math:`b ∈ ℓ_q(ℕ)`.
  
     Then :math:`\{T_k\}` is a family of pointwise :term:`bounded linear functionals <bounded linear functional>`. That is, each :math:`T_k` is a :term:`linear functional`, and for each :math:`b ∈ ℓ_q(ℕ)` there is an :math:`M_b ≥ 0` such that :math:`|T_k b| ≤ M_b` holds for all :math:`k ∈ ℕ`.
  
     To see this, simply note that a convergent sequence of real numbers is bounded, and, in the present case, we have

     .. math:: S_k := ∑_{n=1}^k a_n b_n → ∑_{n=1}^∞ a_n b_n = x ∈ ℝ.

     Thus, :math:`\{T_k b\} = \{S_k\}` is a convergent sequence of real numbers, so, if :math:`N ∈ ℕ` is such that :math:`k ≥ N` implies :math:`|S_k - x| < 1`, and if :math:`M_b'` is defined to be :math:`\max\{|S_k|: 1 ≤ k ≤ N\}`, then for each :math:`k ∈ ℕ`,

     .. math:: |T_k b| ≤ M_b := \max\{ M_b', x+1\}.

     Next note that :math:`ℓ_q` is a :term:`Banach space`, so the (:term:`Banach-Steinhauss <Banach-Steinhaus theorem>`) principle of uniform boundedness implies that there is a single :math:`M > 0` such that :math:`\|T_k\| ≤ M` for all :math:`k ∈ ℕ`.
      
     In other terms,

     .. math:: (∃ M>0) \, (∀ b ∈ ℓ_q) \, (∀ k ∈ ℕ) \; |T_k b| ≤ M \|b\|.

     Define
      
     .. math:: T b := ∑_{n=1}^∞ a_n b_n = \lim_{k → ∞} T_k b,
      
     which exists by assumption. Since :math:`| \; |` is :term:`continuous <continuous function>`, we conclude that :math:`\lim_{k → ∞}|T_k b| = |T b|`.
      
     Finally, since :math:`|T_k b| ≤ M \|b\|` for all :math:`k ∈ ℕ`, we have :math:`|T b| ≤ M \|b\|`. That is :math:`T` is a :term:`bounded linear functional` on :math:`ℓ_q(ℕ)`.

     Now, by the :term:`Riesz representation theorem <Riesz representation theorem (version 2)>`, if :math:`1 ≤ q < ∞`, then every bounded linear functional :math:`T ∈ ℓ_q^∗` is uniquely representable by some :math:`α = (α_1, α_2, \dots) ∈ ℓ_p` as

     .. math:: T b = ∑_{n=1}^∞ α_n b_n.
        :label: RRT-20070601

     On the other hand, by definition :math:`T b = ∑_{n=1}^∞ a_n b_n` for all :math:`b ∈ ℓ_q`.
      
     Since the representation in :eq:`RRT-20070601` is unique, :math:`a = α ∈ ℓ_p`. That is, :math:`∑_n |a_n|^p < ∞`.
     
     ☐

   b. Consider the case :math:`p = 1` and :math:`q = ∞`.
   
      Recall that the :term:`Riesz representation theorem` says that every :math:`T ∈ ℓ_q^∗ \, (1 ≤ q < ∞)` is uniquely representable by some :math:`α ∈ ℓ_p` (where :math:`p = q/(q-1)`, so :math:`1 < p ≤ ∞`). That is, :math:`ℓ_p` is the dual of :math:`ℓ_q`, when :math:`1 ≤ q < ∞` and :math:`p = q/(q-1)`.

      In the present case we have :math:`q = ∞` and :math:`p = 1` and :math:`ℓ_1` is *not* the :term:`dual` of :math:`ℓ_∞`. [3]_ So we can't use the same method of proof for this case.
      
      Nonetheless, the result still holds by the following simple argument: Define :math:`b = (b_1, b_2, \dots)` by

      .. math:: b_n = \mathrm{sgn}(a_n)
                    = \begin{cases} ā_n/|a_n|, & a_n ≠ 0,\\
                                    0,         & a_n= 0,
                       \end{cases} \quad (n ∈ ℕ).

      Then :math:`∑_n |a_n| = ∑_n a_n b_n` converges by the hypothesis, since :math:`|b_n| ∈ \{0, 1\}` implies :math:`b ∈ ℓ_∞`. Therefore, :math:`a ∈ ℓ_1`.

      Finally, in case :math:`p = ∞` and :math:`q=1`, the :term:`Riesz representation theorem <Riesz representation theorem (version 2)>` can be applied as in Part a.
      
      ☐

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     .. rubric:: Footnotes

  .. [1]
     The expression :math:`ℬ(X) ⊗ ℬ(Y)` denotes the :term:`σ-algebra` generated by all sets :math:`A × B ⊆ X × Y` with :math:`A ∈ ℬ(X)` and :math:`B ∈ ℬ(Y)`. (See :term:`Fubini's theorem`. In this case, :math:`ℬ(X) ⊗ ℬ(Y)` is the same as :math:`ℬ(ℝ^2)`.

  .. [2]
     In fact, it is easy to prove that if :math:`G` is any countable subset of :math:`ℝ`, then :math:`m G = 0`. Just fix :math:`ε > 0` and cover each point :math:`x_n ∈ G` with a set :math:`E_n` of measure less than :math:`ε 2^{-n}`. Then :math:`m G ≤ ∑_n m E_n < ε`.

  .. [3]
     One way to see that :math:`ℓ_1` is not the dual of :math:`ℓ_∞` this is to note that :math:`ℓ_1` is :term:`separable` but :math:`ℓ_∞` is not. For the collection of :math:`a\in \ell_\infty` such that :math:`a_n\in \{0, 1\}, \,n\in \mathbb N`, is uncountable and, for any two distinct such sequences :math:`a, b\in \{0, 1\}^{\mathbb N}`, we have :math:`\|a - b\|_\infty = 1`, so there cannot be a countable base, so :math:`\ell_\infty` is not second countable, and a metric space is separable iff it is second countable.)

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

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.. *Proof 2.* [1]_ First recall that :math:`L_1` convergence implies convergence in measure. That is, if :math:`\{f_n\} \subset L_1(m)` and :math:`\|f_n-f\|_1 \rightarrow 0`, then :math:`f_n \rightarrow f` in measure.

..   (*Proof.* :math:`m(\{x:|f_n(x) - f(x)| >\epsilon\}) \leq \frac{1}{\epsilon}\|f_n-f\|_1 \rightarrow 0.`)

..   Next recall that if :math:`f_n\rightarrow f` in measure then there is a subsequence :math:`\{f_{n_j}\} \subseteq \{f_n\}` that converges a.e. to :math:`f` as :math:`j\rightarrow \infty`. [2]_

..   Combining these two results in the present context (Lebesgue measure on the real line), we have the following: [3]_

..   If :math:`\{f_n\} \subset L_1(m)` and :math:`\|f_n-f\|_1\rightarrow 0` then there is a subsequence :math:`\{f_{n_j}\} \subseteq \{f_n\}` with the property :math:`f_{n_j}(x) \rightarrow f(x)` for almost all :math:`x\in \mathbb R`.

..   Now, if :math:`f_n(x)\rightarrow g(x)` for almost all :math:`x\in \mathbb R`, and if :math:`B_1` be the set of measure zero where :math:`f_n(x) \nrightarrow g(x)`, then off of :math:`B_1` the sequence :math:`f_n`, as well as every subsequence of :math:`f_n`, converges to :math:`g`.

..   Let :math:`\{f_{n_j}\}` be the subsequence mentioned above which converges to :math:`f` almost everywhere. Then

..   .. math:: |f(x) - g(x)| \leq |f(x) - f_{n_j}(x)|+|f_{n_j}(x) - g(x)|.
..    :label: 2007-2-1

..   Define :math:`B_2 = \{x\in \mathbb R : f_{n_j}(x) \nrightarrow f(x)\}`. Then the set :math:`B = B_1 \cup B_2` has measure zero and, for all :math:`x\in \mathbb R\setminus B`, :math:`f_{n_j}(x)\rightarrow f(x)` and :math:`f_{n_j}(x)\rightarrow g(x)` . Therefore, by :eq:`2007-2-1`, :math:`|f(x) -g(x)| = 0` for all :math:`x\in \mathbb R \setminus B`.

..   It follows that the set :math:`\{x\in \mathbb R : f_(x) \neq g(x)\} \subset B`, as a subset of a null set, must itself be a null set (since :math:`m` is complete). That is, :math:`f = g` a.e. and the conjecture is proved.  
.. (1) Note that *Proof 1*, which seems to me the more natural one, doesn't use :ref:`Egoroff's theorem <egoroff>`, so either the examiners were looking for a different proof, or a different claim, or perhaps Egoroff's theorem was not the Littlewood principle they had in mind. In any event, there is a way to prove the conjecture which does make use of Egoroff's theorem---namely, *Proof 2*.
.. (2) Folland :cite:`Folland:1999`, Theorem 2.30 and its corollary.
.. (3) Perhaps this statement is the version of the Littlewood principle dealing with a.e. convergence that we were meant to cite in part (a).