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 $t\gt 0$.
An Application of Chebyshev's inequality?
1
$\begingroup$
real-analysis
-
0Are you allowed to use Chebyshev's inequality to prove that? What do you mean by application? – 2012-04-05
1 Answers
2
Hint: if $I(x) = \cases{1 & if $|f(x)| \ge t$\cr 0 & otherwise\cr}$, what can you say about $f(x)^2/t^2 - I(x)$?
-
1OK, now integrate that with $d\mu$, and what do you get? – 2012-04-05