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Sequence of measurable functions and limit

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Given a [tex]\sigma[/tex]-algebra [tex](X,\mathcal{A})[/tex], let [tex]f_n : X \to
    [-\infty,\infty][/tex] be a sequence of measurable functions. Prove that
    the set [tex]\{ x \in X | \lim f_n (x) \text{ exists} \}[/tex] is in

    2. Relevant equations

    Let [tex](X,\mathcal{A})[/tex] be a [tex]\sigma[/tex]-algebra and [tex]M_{+}(X) := [/tex] set
    of all functions [tex]f : X \to [0,\infty][/tex] which are
    [tex]\mathcal{A}[/tex]-measurable. Let [tex]f_n \in M_+(X)[/tex], then if [tex]\displaystyle \lim_{n \to \infty} f_n(x) \exists[/tex] [tex]\forall
    x \in X[/tex], then [tex]\displaystyle f := \lim_{n \to \infty} f_n \in

    Any [tex]f : X \to [-\infty,\infty][/tex] can be written [tex]f = f_+ - f_-[/tex]
    where [tex]f_+(x) := \sup \{ f(x), 0 \}[/tex], [tex]f_-(x) := - \inf \{ f(x), 0
    \}[/tex]. So [tex]f[/tex] is [tex]\mathcal{A}[/tex]-measurable iff [tex]f_+, f_- \in M_+(X)[/tex].

    3. The attempt at a solution

    Should I use the above two facts to show that there exists a measurable [tex]\displaystyle f = \lim_{n \to \infty} f_n[/tex] and hence anything [tex]f^{-1}(E) \in \mathcal{A}, E \in \mathbb{R}[/tex]?
  2. jcsd
  3. Aug 10, 2010 #2
    Here's a sketch of a proof.
    1. Define f(x) =lim f_n(x) if f_n(x) converges ,
    = 0 otherwise.
    2. Using Lebesgue's dominated convergence theorem, show that f is measurable.
    3. If a measurable function exists on a set S, its characteristic function ,too, is measurable.
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