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Math Help - Evalute an integral with a weird measure.

  1. #1
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    Evalute an integral with a weird measure.

    Let  \mu be a measure on the set of natural numbers.

    And define  \mu ( \{ n, n+1, ... \} ) = \frac {n}{2^n} for  n = \{ 1, 2, 3, ... \}

    Find  \int x d \mu (x)

    My solution so far:

    To be honest I'm a bit lost in this one. So should I have something like  \sum ^ \infty _{x=1}\frac {x}{2^x } ?
    Last edited by tttcomrader; August 26th 2011 at 08:09 AM.
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    Re: Evalute an integral with a weird measure.

    Quote Originally Posted by tttcomrader View Post
    Let  \mu be a measure on the set of natural numbers.

    And define  \mu ( \{ n, n+1, ... \} ) = \frac {n}{2^n} for  n \in \mathbb {N}
    Taking n=0 we get that \mu(\mathbb N)=0:maybe there is a typo.
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    Re: Evalute an integral with a weird measure.

    Sorry, n should be equal to 1, 2, 3, ... I changed it in the original post.
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    Super Member girdav's Avatar
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    Re: Evalute an integral with a weird measure.

    Ok. Hint: write the integral as \sum_{n=1}^{+\infty}\int_{\{n\}}xd\mu(x) (use monotone convergence theorem) and you are almost done. Compute the measure of \{n\} for n\geq 1.
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    Re: Evalute an integral with a weird measure.

    So would I have  \sum ^ \infty _{n=1} x \frac {n}{2^n} ?
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    Re: Evalute an integral with a weird measure.

    Quote Originally Posted by tttcomrader View Post
    So would I have  \sum ^ \infty _{n=1} x \frac {n}{2^n} ?
    There shouldn't be any x in the sum, since you sum over n. We have \mu(\{n\})=\mu(\{n,n+1,\ldots\})-\mu(\{n+1,\ldots\})=\frac{n}{2^n}-\frac{n+1}{2^{n+1}} = \frac{2n-n-1}{2^{n+1}}.
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  7. #7
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    Re: Evalute an integral with a weird measure.

    Thanks! I understand how the measure will sum up now. But what would x become? Does it become n, n+1, ...?
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    Super Member girdav's Avatar
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    Re: Evalute an integral with a weird measure.

    We have \int_{\{n\}}xd\mu(x)=n\mu(\{n\}), since on \{n\} we must have x=n.
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