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Thread: measure theory intergration

  1. #1
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    measure theory intergration

    The question is:
    Show that if $\displaystyle g:\mathbb{R} \longrightarrow [0,\infty) $ equals 0 outside a bounded interval and $\displaystyle \int g^2 < \infty $ , then $\displaystyle \int g < \infty $

    It doesn't appear so difficult, but i dont know how to progress in it. I say let I be the bounded interval, then $\displaystyle \int g^2 = \int_{\mathbb{R}} g^2 \chi_I < \infty $ how do i progress from here? i dont know how to get $\displaystyle \int g $ from this! thanks
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  2. #2
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    Divide $\displaystyle I=A\cup B$, $\displaystyle A=\{x\in I: |f(x)|\leq 1\}$ and $\displaystyle B:=I\setminus A$. Both sets are measurable. Show that $\displaystyle f$ is integrable both in $\displaystyle A$ and $\displaystyle B$ because it is dominated by the identically 1 function in $\displaystyle A$ (which is of finite measure) and by $\displaystyle f^2$ in $\displaystyle B$, integrable by hypothesis.
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  3. #3
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    Quote Originally Posted by Enrique2 View Post
    Divide $\displaystyle I=A\cup B$, $\displaystyle A=\{x\in I: |f(x)|\leq 1\}$ and $\displaystyle B:=I\setminus A$. Both sets are measurable. Show that $\displaystyle f$ is integrable both in $\displaystyle A$ and $\displaystyle B$ because it is dominated by the identically 1 function in $\displaystyle A$ (which is of finite measure) and by $\displaystyle f^2$ in $\displaystyle B$, integrable by hypothesis.
    Thanks a lot!!!!!!! I wouldn't have ever thought of dividing I!!!
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