$f:[0,1]rightarrow[0,1]$, measurable, and $int_{[0,1]}f(x)dx=yimplies m{x:f(x)frac{y}{2}}geqfrac{y}{2}$.

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Question: Suppose $f:[0,1]rightarrow[0,1]$ is a measurable function such that $int_0^1f(x)dx=y$.  Prove that $m{x:f(x)>frac{y}{2}}geqfrac{y}{2}$.
My thoughts: Since we have $int_0^1f(x)dx=yimplies frac{1}{2}int_0^1f(x)dx=frac{y}{2}$.  Now, I was hoping to be able to split the integral bounds and consider $int f$ over ${x:f(x)>frac{y}{2}}$ and over ${x:f(x)leqfrac{y}{2}}$, and somehow use Markov's inequality in there somewhere.  I think I have something off though and I am not sure if I can do what I am saying, or even if it is correct.  Any suggestions, ideas, etc. are greatly appreciated!  Thank you.

zkutch · Answer

Suppose $A = {x:f(x)>frac{y}{2}}$ and $m(A)<frac{y}{2}$. Also lets denote $B = {x:f(x)leqslantfrac{y}{2}}$. Now we have
$$intlimits_{0}^{1} = intlimits_{A}+ intlimits_{B}$$
For first we use that $f leqslant 1$ and $m(A)<frac{y}{2}$. For second we have estimation $leqslant frac{y}{2}$. So, sum cannot be $y$.

$f:[0,1]rightarrow[0,1]$, measurable, and $int_{[0,1]}f(x)dx=yimplies m{x:f(x)>frac{y}{2}}geqfrac{y}{2}$.

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