Let $f$ be integrable on $\Bbb{R}$. Let a sequence $a_n>0$ have $\sum a_n=1$. Prove or disprove that there exists a partition of $\Bbb{R}$ into measurable sets $D_n$ such that $\mu(f\Bbb{1}_{D_n})=a_n\mu(f)$.
I think this is true, and my approach was to create a sequence of points that converges to a limit to create the required partition. It's a little messy, and I'm not sure about some parts. Does anyone else have a hint for this?
EDIT: I think I can do this for a bounded domain, such that we have a finite number of $a_n$'s such that $\sum\limits_{i=1}^n a_i=1$. But maybe the proof doesn't work for an unbounded domain like $\Bbb{R}$?
Without loss of generality assume that $\mu (f) >0$. The function $x \to \int_{-\infty} ^{x} f(t)dt$ is continuous, approaches $0$ as $ x\to -\infty$ and $\mu (f)$ as $x \to \infty$. Hence there exists $x_1$ such that $\int_{-\infty} ^{x} f(t)dt=a_1\mu (f)$. Now consider $\int_{x_1} ^{x} f(t)dt$. Note that as $x \to \infty$ this approaches $\mu (f)-a_1 \mu (f)>a_2 \mu (f)$ since $a_1+a_2 <1$. Hence there exits $x_2$ such that $\int_{x_1} ^{x_2} f(t)dt=a_2\mu (f)$. Proceed by induction to get disjoint intervals $(-\infty,x_1),(x_1,x_2),...$. Since $\sum a_n =1$ it follows that if $x_n$'s increase to a finite limit $y$ then $\int_y ^{\infty} f =0$ so we can include $(y,\infty)$ in one of the other intervals to make sure that we get a partition of the entire line.