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Math Help - Lebesgue integral and measure inequality

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    Lebesgue integral and measure inequality

    Let f : \Re \rightarrow [-\infty,\infty] be an integrable function, and let a > 0. Prove

    m({x:|f(x)|\geq a}) \leq 1/a \int |f|
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    Hello,
    Quote Originally Posted by Amanda1990 View Post
    Let f : \Re \rightarrow [-\infty,\infty] be an integrable function, and let a > 0. Prove

    m({x:|f(x)|\geq a}) \leq 1/a \int |f|
    |f| \geqslant a \bold{1}_{\{|f| \geqslant a\}}, where \bold{1} is the indicator function : \bold{1}_{\{|f| \geqslant a\}}=\left\{\begin{array}{ll} 1 \text{ if } |f| \geqslant a \\ 0 \text{ if } |f| < a\end{array} \right.

    This inequality is very easy to check if you consider |f| \geqslant a and |f| <a

    Since the integral is increasing ( f \leqslant g \Longrightarrow \int f ~dm \leqslant \int g ~dm), we can state :
    \int |f| ~dm \geqslant \int a \bold{1}_{\{|f| \geqslant a\}} ~dm
    By homogeneity of the integral, this equals :
    =a \int \bold{1}_{\{|f| \geqslant a\}} ~dm=a ~m(\{|f| \geqslant a\})=a ~m(\{x ~:~ |f(x)| \geqslant a\})


    And you're done


    As a sidenote : this is a form of Markov's inequality.
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