How can I find a subset of a set with "half the size" of the original?

Solution 1:

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 $\lambda$.

Solution 2:

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 $\mu(E)/2$.