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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.For each n,define E_n= \{ x \in X : f(x)>n \}.
    Show that lim_{n \rightarrow \infty} m(E_n) =0.
    ( m(A) is the measure of A.)

    I was trying to prove that since f is integrable,then f < \infty   a.e..Since E_n is a decreasing sequence of sets,so lim_{n \rightarrow \infty} m(E_n) = m( \bigcap_{n \in N} E_n)
    Since f< \infty   a.e so \bigcap_{n \in N} E_n = \phi

    so m( \phi ) = 0 and hence get the result.
    but the problem is I have to assume m(E_1) < \infty to claim the result.I was stuck here.
    Can anyone help?Or if this is not the correct approach,can anyone show me the right way?
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  2. #2
    Moo
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    Hello,

    What do you think of that ?

    I first started by looking at the integral of f over E_n :

    \int_{E_n} f ~d\mu=\int_{\mathbb{R}} f \cdot \bold{1}_{\{f>n\}} ~d\mu

    But in this case, we have f>n. Otherwise, it's 0.
    So \int_{E_n} f~d\mu>\int_{\mathbb{R}} n \cdot \bold{1}_{\{f>n\}} ~ d\mu=n\int_{\mathbb{R}} \bold{1}_{\{f>n\}} ~d\mu=n\mu(E_n)

    Now, we know that f is integrable. This means that \int_{\mathbb{R}}f ~d\mu<\infty

    And since E_n\subset \mathbb{R}, we have \int_{\mathbb{R}}f ~d\mu\geq \int_{E_n}f ~d\mu


    Finally, we get \infty > \int_{\mathbb{R}} f ~d\mu> n\mu(E_n)

    Now what happens if n goes to infinity ? Unless \mu(E_n)\to 0, n\mu(E_n) \to \infty...
    A simple proof by contradiction can finish it
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