Here's something that's pretty neat: find a measurable subset $A$ of $[0,1]$ such that for any subinterval $I$ of $[0,1]$, the Lebesgue measure $\mu(A\cap I)$ has $0 < \mu(A\cap I) < \mu(I)$. There's an explicit construction of such a set in Rudin, who describes such sets as "well-distributed". Balint Virag (and maybe others) found a very slick probabilistic construction.
Let $X_1, X_2, \ldots$ be i.i.d. coin flips, i.e. $X_1$ is $1$ with probability $1/2$ and $-1$ with probability $1/2$. Consider the (random) series
$$S:=\sum_{n=1}^\infty X_n/n.$$
By the Kolmogorov three-series theorem, it converges almost surely. However, it's a simple exercise to see that for any $x$, a$, the event ${S > x}$ a}$ has non-trivial measure: for $a>0$, there's a positive chance of the first $e^a$ terms of the series being positive, so the $e^a$-th partial sum is positive, and the tail is independent and positive or negative with equal probability, due to symmetry. For $a\leq 0$, it's trivial, again because of symmetry.
A common way of realizing i.i.d. coin flips on the unit interval is as Rademacher functions: for $x\in[0,1]$, let ${b_n}$ be its binary expansion, and $X_n(x) = (-1)^{b_n}$. Realized this way, the random sum $S$ becomes an almost everywhere finite measurable function from $[0,1]$ to $\R$, and \R$. It only takes a bit more work to see that the set ${S>x}$ {S>a}$ is exactly a well-distributed set.
Alex Bloemendal has written this up in a short note, but I'm not sure if he's published it anywhere.
