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 $x\in E_k^{n_0}$.