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Let $f$ be a nonnegative monotone function defined on the $[0,1]$. Prove that $$\mu_L(\{w\in I;f(w)>\alpha\})<\dfrac1\alpha\int_0^1fdx$$ where the measure is the Lebesgue measure, and the integral is the Riemann integral.

So let $f$ be monotone increasing (monotone decreasing should be similar). Suppose $f(w)=\alpha$ at $w=r$. So $\mu_L(\{w\in I;f(w)>\alpha\}) = 1-r$. Then partitioning $[0,1]$ into $[0,r],[r,1]$, we have $\int_0^1fdx$ is at least $\alpha(1-r)$. But still, it might be equal to $\alpha(1-r)$. How can we eliminate that possibility?

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  • $\begingroup$ $\int_0^1fdx>\int_r^1fdx\ge\alpha(1-r)$ $\endgroup$
    – Pocho la pantera
    Sep 5, 2013 at 2:00
  • $\begingroup$ @Pocholapantera That's not true. It is possible that $f(x)=0$ for $x\in[0,r)$. $\endgroup$
    – Paul S.
    Sep 5, 2013 at 2:09
  • $\begingroup$ You are right, sorry. I was thinking in a positive function. Well, then the proposition is false, unless some aditional hypothesis is added. $\endgroup$
    – Pocho la pantera
    Sep 5, 2013 at 2:42
  • $\begingroup$ @Pocholapantera Can you give a counterexample where it's false? $\endgroup$
    – Paul S.
    Sep 5, 2013 at 2:49
  • $\begingroup$ It is not my day. Now I was thinking in condition $f(w)\ge \alpha$. $\endgroup$
    – Pocho la pantera
    Sep 5, 2013 at 3:05

1 Answer 1

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I continue your reasoning.

Let $f$ be monotone increasing. Define $r=\sup \{w\,;\,f(w)\le \alpha\}$. Then

$\mu_L(\{w\in I;\,f(w)>\alpha\})=1−r$

For the other side,

$$ \int_0^1f\,dx\ge \int_r^1f\,dx>\alpha\,(1-r) %\int_r^{r+\epsilon}f+\int_{r+\epsilon}^1 f>\alpha \epsilon +\alpha (1-r-\epsilon) $$

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  • $\begingroup$ I believe that's finally correct :=) $\endgroup$
    – Paul S.
    Sep 5, 2013 at 4:08

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