# How can I find a subset of a set with “half the size” of the original?

let $E\subset\Bbb R$ Lebesgue-measurable with $m(E)\lt \infty$, prove that exist $A\subset E$ Lebesgue-measurable such that $m(A)=m(E)/2$.

I've tried this

For every $k\ge 1$ exist $\{I_n^k\}$ with $E\subset \cup_1^\infty I_n^k$ such that $m(E)+ \frac{1}{k} \ge \sum_{n\in\Bbb N} m(I_n^k)$

cuttin every $I_n^k$ in half (for k fixed), we got $\frac{m(E)+ \frac{1}{k}}{2} \ge \sum_{n\in\Bbb N} m(\dot I_n^k)$ (where $\dot I_n^k$ is the "cutted in half $I_n^k$)

I don't know what to do next.

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– Giuseppe Negro Mar 14 '14 at 23:49

## 3 Answers

Consider the function $f : \Bbb R \to [0, \infty)$, $f(x) = m(E \cap [-x, x])$. This function is continuous as can be easily shown. Since $f(0) = 0$, $\lim_{x\to\infty} f(x) = m(E)$, the intermediate value theorem gives the desired result. In fact, it gives the stronger result that for any $\lambda \in [0, m(E)]$, one can find a subset of $E$ with measure $\lambda$.

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Consider your set intersected with the collection of rational intervals of the form $(k/2^m,(k+1)/2^m)$ for integer $m \geq 0$ and all integers $k$. If you choose $m$ small enough, you will be able to find a subset of intervals that gets you to within $1/2^m$ of your desired measure $\leq m(E)$ when you intersect your set $E$ with the collection of intervals. Then, depending on how close you are to your desired target measure, you choose a new $m$ and repeat the process, choosing intervals that are not subsets of the intervals you chose in previous iterations. And so forth. When you take the union over all $m$ that you used, and intersect the collection of intervals you used with your set $E$, you will get a set that has the desired target measure $\leq m(E)$. In particular you can get a set with measure $m(E) / 2$.

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A natural approach that I like and works in any non-atomic measure space is as follows:

(Non-atomic means here that whenever $E$ is measurable and $0<\mu(E)<\infty$, there is a measurable $A\subset E$ with $0<\mu(A)<\mu(E)$.)

Note first that for any $\epsilon>0$ we may assume that $A$ as above has measure smaller than $\epsilon$: Divide $E$ into two pieces of positive measure. One of them has measure at most $\mu(E)/2$. Repeat.

Now, given $E$, the above gives us that $E$ has a subset $A$ of positive measure at most $\mu(E)/2$, say $t$. It $t<\mu(E)/2$, the remainder $E\setminus A$ has a subset of positive measure at most $\mu(E)/2-t$. Repeat. We may have to repeat transfinitely, but eventually we will get a disjoint collection of measurable subsets of $E$ whose measures add up to precisely $\mu(E)/2$. Two points need to be mentioned: The process stops after countably many steps. Otherwise, $E$ contains uncountably many disjoint subsets of positive measure, and therefore has infinite measure, against the assumption. Since the process stops after countably many steps, the unions of these disjoint sets is measurable and has measure precisely $\mu(E)/2$.

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