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Math Help - Integration of a piecewise function using measures

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    Integration of a piecewise function using measures

    Problem: Define a function f by f(x)= \begin{cases} x, \quad& 0 \le x \le 1 \\0, \quad& otherwise \end{cases}.

    Given that  \mu is the Lebesgue measure on \mathbb {R} i.e. \mu ((a,b)) = b-a

    Show that  \int f d \mu = \frac {1}{2}

    My proof so far:

    Using the hints in the text, I'm defining a series of function f_n that converges to f, with  \int f_n d \mu \rightarrow \int f d \mu

    For  x \in (0,1] , let  f_n(x) = \frac { [2^nx]}{2^n} where  [x] a rounded down to the nearest integer.

     f_n(x)= \begin{cases} \frac { [2^nx]}{2^n}, \quad& x \in (0,1] \\ 0, \quad& otherwise \end{cases}.

    Then I have  \int f_n (x) d \mu =  \sum ^n _{k=1} ( \frac {1}{2^k} ) \mu \{ ( \frac {1}{2^k} , \frac {1}{2 ^{k-1}} )
     = \sum ^n _{k=1} \frac {1}{4^k}
    \frac {1}{3}

    But the answer is suppose to be  \frac {1}{2} ! I'm guessing my integration of the series of function is wrong, but why?
    Last edited by tttcomrader; July 18th 2011 at 02:59 PM.
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    MHF Contributor Siron's Avatar
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    Re: Integration of a piecewise function using measures

    For entering latex you have to use: [ tex ]... [ / tex ] not [ math ] ... [/ math ]
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