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Thread: Evalute an integral with a weird measure.

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

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

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

    Find $\displaystyle \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 $\displaystyle \sum ^ \infty _{x=1}\frac {x}{2^x } $?
    Last edited by tttcomrader; Aug 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 $\displaystyle \mu $ be a measure on the set of natural numbers.

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

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

    So would I have $\displaystyle \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 $\displaystyle \sum ^ \infty _{n=1} x \frac {n}{2^n} $?
    There shouldn't be any $\displaystyle x $ in the sum, since you sum over $\displaystyle n$. We have $\displaystyle \mu(\{n\})=\mu(\{n,n+1,\ldots\})-\mu(\{n+1,\ldots\})=\frac{n}{2^n}-\frac{n+1}{2^{n+1}} = \frac{2n-n-1}{2^{n+1}}$.
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    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 $\displaystyle \int_{\{n\}}xd\mu(x)=n\mu(\{n\})$, since on $\displaystyle \{n\}$ we must have $\displaystyle x=n$.
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