Thread: Lebesgue integral and measure inequality

1. Lebesgue integral and measure inequality

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

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

2. Hello,
Originally Posted by Amanda1990
Let $\displaystyle f : \Re \rightarrow [-\infty,\infty]$ be an integrable function, and let a > 0. Prove

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

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

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

And you're done

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