There's a "construction" of a non-measurable set that seems very keen to me; wondering whether anyone's seen it before.
Tedious part:
For $n,j\in\mathbb Z$ define the dyadic interval $I_{n,j}$ by $$I_{n,j}=[j2^{-n},(j+1)2^{-n}).$$
If $I$ is a half-open interval say $I_L$ and $I_R$ are the left half and right half of $I$, and say $|I|$ is the length of $I$. Note that $t\mapsto t+|I|/2$ maps $I_L$ onto $I_R$.
For $n=1,2,\dots$ the Rademacher function $r_n:[0,1)\to\{1,-1\}$ is defined by $$r_n=\sum_{j=0}^{2^{n}-1}(-1)^j\chi_{I_{n,j}}.$$(Draw a picture of $r_1$ and $r_2$ if this is your first time.)
Define $$f_n=\prod_{j=1}^nr_j.$$And note this: If $I=I_{n,j}$, $t\in I_L$ and $m>n$ then $$f_m(t+|I|/2)=-f_m(t).$$
Fun part:
Note that $$f_n\in K=\{1,-1\}^{[0,1)}.$$Here $K=\{1,-1\}^{[0,1)}$ is the set of all functions mapping $[0,1)$ to $\{1,-1\}$, which is compact in the product topology. Let $f$ be an accumulation point in $K$ of the sequence $(f_n)$.
It follows that if $I$ is a dyadic interval and $t\in I_L$ then $$f(t+|I|/2)=-f(t).$$And hence $f$ cannot be measurable:
If $f$ were measurable then we would have $\int_If=0$ for every dyadic interval $I$, so the Lebesgue Differentiation Theorem would imply $f=0$ almost everywhere, which isn't so. (We've shown that $f$ measurable implies $0\in\{1,-1\}$; we leave it as an exercise to derive the canonical contradiction $0=1$ from this.)
Seems neat to me. Like $f$ can't be measurable because it has too much symmetry or something. "Constructing" a non-measurable function as the limit of a net of step functions...
So. We seen this before or can I call it the Ullrich function or what?
