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Math Help - Lebesgue integral

  1. #1
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    Lebesgue integral

    Suppose f is a non-negative integrable function on X and F_n=\{ x \in X : f(x)<1/n \}.
    Show that lim_{n \rightarrow \infty} \int_{F_n} f = 0.
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    Hello,

    Let G_n=X\backslash F_n=\{x\in X ~:~ f(x)\geq 1/n\}

    We have \int_X f ~d\mu=\int_{F_n} f ~d\mu+\int_{G_n} f ~ d\mu (*)

    Let's study \int_{G_n} f ~ d\mu=\int_X f \cdot\bold{1}_{G_n} ~d\mu

    Since \left(\tfrac 1n\right)_{n\geq 1} is a decreasing sequence, G_n \subseteq G_{n+1} and hence (f \cdot\bold{1}_{G_n})_{n\geq 1} is a non-negative integrable and increasing sequence.

    So we can apply Lebesgue monotone convergence theorem, and we have f \cdot\bold{1}_{G_n} \xrightarrow[]{n\to\infty} f.

    Hence by making the limit as n tends to infinity in (*), we have \int_X f ~d\mu=\lim_{n\to\infty} \int_{F_n} f ~d\mu+\int_X f ~d\mu

    And hence the result...



    Another direct way would've been to use Lebesgue's dominated convergence theorem... :
    \int_{F_n} f~d\mu=\int_X f \cdot\bold{1}_{F_n} ~d\mu

    Note that f \cdot\bold{1}_{F_n} \leq f, which is integrable.

    And since \lim_{n\to\infty}F_n=\emptyset, f\cdot\bold{1}_{F_n}\xrightarrow[] {n\to\infty} 0

    And hence the result.
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