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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 $t\gt 0$.

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Are you allowed to use Chebyshev's inequality to prove that? What do you mean by application? –  user1736 Apr 5 '12 at 4:05

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

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I can say that $f(x)^2/t^2 - I(x) \geq 0$. –  Joel Apr 5 '12 at 3:20
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OK, now integrate that with $d\mu$, and what do you get? –  Robert Israel Apr 5 '12 at 7:12

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