# Prove that there is a set $E$ such that $f(x)\ge\frac{1}{2\mu(E)}\int_Ef(y)\,\text{d}\mu(y)$

Here is another task from the preparatory set I do not know how to solve. I hope you could help me.

Let $\mu$ be a Radon measure on $\mathbb{R}$ and let $f$ be a $\mu$–measurable, nonnegative function on $\mathbb{R}$. Prove that there is a $\mu$–measurable subset $E\subset\mathbb{R}$ with $0\lt\mu(E)\lt\infty$ such that for all $x\in E$ $$f(x)\ge\frac{1}{2\mu(E)}\int_Ef(y)\,\text{d}\mu(y).$$

The solution inspired by did's hints:

Consider two cases: either $1)$ $f$ is $0$ $\mu$–a.e. or $2)$ there exists $\delta\gt0$ such that $\mu([f\ge \delta])\gt0$.

Ad $1)$
Then $\int_A f=0$ for all $A\subset \mathbb{R}$ so all we want is a set E satisfying $0\lt\mu(E)\lt\infty$. As such we may take one of the balls of positive measure $B(0,n_0)$. Note that at least one such a ball exist since $\mathbb{R}=\bigcup_{j=1}^\infty B(0,j)$. Also $\mu(B(0,n_0))\lt\infty$ for $\mu$ is Radon. Thus $$f(x)\ge \int_{B(0,n_0)} f(y)\,\text{d}\mu(y)=0$$ and we are done.

Ad $2)$
Since $[f\ge\delta]=\bigcup_{j=0}^\infty \,[2^j\delta\le f\lt 2^{j+1}\delta]$ there exists $k\in\{0,1,\ldots\}$ such that the measure of $E_k=[2^k\delta\le f\lt 2^{k+1}\delta]$ is positive. In order to ensure the finiteness of the desired set's measure we can define $E_{k}^n=E_{k}\cap B(0,n)$ for $n\in\mathbb{N}$. Again, we know that there is a radius $n_0$ such that $0 \lt \mu(E_k^{n_0})\lt\infty$ since $\bigcup_{n=1}^\infty E_k^n =E_k$ and $\mu$ is Radon.
Then $\int_{E_k^{n_0}}f(y)\,\text{d}\mu(y)\lt2^{k+1}\delta\mu(E_k^{n_0})$ and, hence, $$\frac{1}{2\mu(E_k^{n_0})}\int_{E_k^{n_0}}f(y)\,\text{d}\mu(y)\lt2^k\delta\le f(x)$$ holds for all $x\in E_k^{n_0}$.

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Hint: Consider the sets $E_{s,t}=[s\lt f\leqslant t]$ and show that some suitable $0\lt s\lt t$ exist.

Sub-hint: for every $x$ in $E_{s,t}$, $f(x)\geqslant s$ and $\int\limits_{E_{s,t}}f\leqslant t\mu(E_{s,t})$ since $f\leqslant t$ uniformly on $E_{s,t}$. This shows that the inequality one is interested in holds as soon as:

• $s\geqslant\frac12t$,
• $\mu(E_{s,t})\ne0$.

It remains to show that some $(s,t)$ exists such that both these properties hold at the same time.

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So I should assume that for all $f^{-1}((s,t])$ the opposite inequality $f(x)\lt \frac{1}{2\mu((s,t])}\int_E f(y)\,\text{d}\mu(y)$ holds and derive a contradiction? – Kuba Helsztyński Nov 19 '12 at 0:05
No, you should try to find some s and t such that the corresponding set E solves your problem. – Did Nov 19 '12 at 7:29
Still, I have no idea how to make use of your hint. Do you mind elaborating? – Kuba Helsztyński Nov 19 '12 at 9:25
See "sub-hint". – Did Nov 19 '12 at 9:40