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

Question

Question: Suppose $f:[0,1]\rightarrow[0,1]$ is a measurable function such that $\int_0^1f(x)dx=y$. Prove that $m\{x:f(x)>\frac{y}{2}\}\geq\frac{y}{2}$.

My thoughts: Since we have $\int_0^1f(x)dx=y\implies \frac{1}{2}\int_0^1f(x)dx=\frac{y}{2}$. 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 $\{x:f(x)\leq\frac{y}{2}\}$, 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.

Answer 1

Suppose $A = \{x:f(x)>\frac{y}{2}\}$ and $m(A)<\frac{y}{2}$. Also lets denote $B = \{x:f(x)\leqslant\frac{y}{2}\}$. Now we have $$\int\limits_{0}^{1} = \int\limits_{A}+ \int\limits_{B}$$ 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 $y$.