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Math Help - Limit of the Measure of a Set

  1. #1
    Senior Member roninpro's Avatar
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    Limit of the Measure of a Set

    Hello. I am working on the following problem:

    Let (X,\mathcal{A},\mu) be a measure space and let f be an extended real-valued \mathcal{A}-measurable function on X such that \int_X |f|^p\text{ d}\mu<\infty$ for some $p\in (0,\infty). Show that \lim_{\lambda\to \infty} \lambda^p \mu\{X:|f|\geq \lambda\}=0.

    Now, I let X_\lambda=\{X:|f|\geq\lambda\} and tried to consider the inequalities

    \displaystyle \int_X |f|^p\text{ d}\mu\geq \int_{X_\lambda} |f|^p\text{ d}\mu\geq \int_{X_\lambda} \lambda^p\text{ d}\mu=\lambda^p \mu(X_\lambda)

    The expression on the left side is finite, but as far as I can tell, this doesn't give much information about the expression on the right. At this point, I'm at a loss on this problem, so I would appreciate if anybody had some suggestions.

    Thanks.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by roninpro View Post
    Hello. I am working on the following problem:

    Let (X,\mathcal{A},\mu) be a measure space and let f be an extended real-valued \mathcal{A}-measurable function on X such that \int_X |f|^p\text{ d}\mu<\infty$ for some $p\in (0,\infty). Show that \lim_{\lambda\to \infty} \lambda^p \mu\{X:|f|\geq \lambda\}=0.

    Now, I let X_\lambda=\{X:|f|\geq\lambda\} and tried to consider the inequalities

    \displaystyle \int_X |f|^p\text{ d}\mu\geq \int_{X_\lambda} |f|^p\text{ d}\mu\geq \int_{X_\lambda} \lambda^p\text{ d}\mu=\lambda^p \mu(X_\lambda)

    The expression on the left side is finite, but as far as I can tell, this doesn't give much information about the expression on the right. At this point, I'm at a loss on this problem, so I would appreciate if anybody had some suggestions.

    Thanks.
    Surely, the second integral from the left goes to 0 as \lambda \to\infty, for if |f|^p is integrable, it must be <\infty almost everywhere.
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  3. #3
    Senior Member roninpro's Avatar
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    Thanks! That was really dumb of me.
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