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Thread: Lebesgue integral

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    Lebesgue integral

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

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

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

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

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

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

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

    And hence the result...



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

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

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

    And hence the result.
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