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Levi Convergence Theorem for step functions

  1. Jun 24, 2009 #1
    Hi Everyone,

    I am currently reading Apostol's "Mathematical Analysis" to introduce myself to some advanced calculus techniques I will need in probability. Usually the proofs in this book are fairly straight forward, without too many missed steps, which is why I think this book is great for self study. I was reading about the Levi's convergence theorem for step functions and didn't like the proof that Apostol gives. To me it seems unnecessarily long and counter-intuitive. I think I have an alternative proof which is simpler to understand and shorter. But I started doubting myself, why would he present a long winded proof, when there is a simpler available - maybe something in my proof is wrong and I just can't spot it? That is why I want to share it here, maybe someone would be able to quickly see an error in my reasoning. Thank you very much in advance for anyone caring to taking a look.

    Theorem:
    Let [tex]\{s_n\}[/tex] be a sequence of step functions such that:

    1). for every [tex]x \in I[/tex], [tex]s_{n+1}(x) \ge s_n(x)[/tex]

    2). [tex]$\lim_{n \to \infty} \int_I s_n$[/tex] exists.

    Then [tex]\{s_n\}[/tex] converges almost everywhere on [tex]I[/tex] to a limit function [tex]f[/tex].

    Proof:
    Since integrals of [tex]\{s_n\}[/tex] converge, they must be bounded. Suppose that
    [tex]$\lim_{n \to \infty} \int_I s_n < M$[/tex]. Let [tex]D[/tex] be a set of points x of I such that [tex]s_n(x)[/tex] diverges. Given any [tex]$\epsilon$[/tex], define:
    [tex]B_n = \{x | s_n(x) > \frac{2M}{\epsilon}\}[/tex]
    Then we have:

    [tex]B_n \subseteq B_{n+1} [/tex]

    [tex]D \subseteq \bigcup_{n} B_n [/tex]

    Also, each [tex]B_n[/tex] is a finite collection of intervals. Define [tex]|B_n|[/tex] to be the sum of the lengths of these intervals.

    Now for any n:

    [tex]M > \int_I s_n \ge \int_{B_n} s_n > \int_{B_n}\frac{2M}{\epsilon} = \frac{2M}{\epsilon}\int_{B_n} 1 = \frac{2M}{\epsilon}|B_n|[/tex]

    So that [tex] |B_n| < \epsilon/2 [/tex]. From this we get that [tex]$\lim_{n\to\infty} |B_n| \le \epsilon/2$[/tex].
    Since [tex]|\bigcup_{n=1}^m B_n| = |B_m|[/tex], [tex]D[/tex] is covered by a countable collection of intervals the sum of whose lengths is less than any [tex]\epsilon[/tex], therefore [tex]D[/tex] has measure zero.
    From this it follows that [tex]s_n[/tex] is bounded almost everywhere, and so converges almost everywhere on I to some function [tex]f[/tex].
     
    Last edited: Jun 25, 2009
  2. jcsd
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