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Math Help - Proof: Every Positive Measureable Function There Exist Simple Measureable Function

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    Proof: Every Positive Measureable Function There Exist Simple Measureable Function

    Hello,
    This is my first post on this site, so please excuse me if the post is a bit rough. I'm currently reading Rudin's Real and Complex Book and ran across Theorem 1.17

    From Book:
    Theorem 1.17:
    Let  \small{f: X \rightarrow [0,\infty]} be measurable. There exist simple measurable functions s_n on X such that.
    (a) 0 \leq s_1 \leq s_2 \leq \ldots \leq f
    (b) s_n(x) \rightarrow f(x), n \rightarrow \infty, \forall x \in X

    Proof:
    For n=1,2,3\ldots and for 1\leq i \leq n2^n, define
    E_{n,i}=f^{-1}([\frac{i-1}{2^n},\frac{i}{2^n})) and F_n=f^{-1}([n,\infty]) and put s_n=\sum_{i=1}^{n2^n}\frac{i-1}{2^n}\chi_{E_{n,i}}+n\chi_{F_n} where \chi_{E_{n,i}},\chi_{F_n} are the characteristic functions of E and F, respectively.

    Questions:
    Part a) So we know that E and F are measurable sets because the pull back of a Borel set through a measurable function is a measurable set. And with a little algebra we can see that s_1 \leq s_2 \leq \ldots How do you prove that the simple functions are less than or equal to f for all n? I can do this with finite examples, but don't know how it to extend it to an arbitrary function.

    Part b) If f(x)=\infty then s_n(x)=n and we're good. If f(x)<\infty then we should be able to show that s_n(x)\geq f(x)-2^{-n} for large enough n. I am having problems seeing how to actually arrive at this result.

    Any help would be appreciated,
    Brad
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Fix n. Take any x \in X Then x belongs to exactly one of E_{n,i} or F_n. If x \in E_{n,i}, for example, this means that \frac{i-1}{2^n}\leq f(x) < \frac{i}{2^n}. But s_n(x) = \frac{i-1}{2^n}. Hence s_n(x) \leq f(x).
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