1
$\begingroup$

Let $(X,\Omega,\mu)$ be a measure space and let $f$ be an extended real valued measurable function defined on $X$. I want help in showing that $ \mu\left(\{x\in X : |f(x)|\geq t\}\right) \leq \frac{1}{t^2}\int_X f^2~d\mu$ for any real number $t\gt 0$.

  • 0
    Are you allowed to use Chebyshev's inequality to prove that? What do you mean by application?2012-04-05

1 Answers 1

2

Hint: if $I(x) = \cases{1 & if $|f(x)| \ge t$\cr 0 & otherwise\cr}$, what can you say about $f(x)^2/t^2 - I(x)$?

  • 1
    OK, now integrate that with $d\mu$, and what do you get?2012-04-05