Lebesgue integral and measure inequality

**Amanda1990** · February 8th 2009, 01:37 AM

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

$m({x:|f(x)|\geq a}) \leq 1/a \int |f|$

**Moo** · February 8th 2009, 02:15 AM

Hello,

Originally Posted by Amanda1990

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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