Exercise 6, § 5

For every integer $n \geqslant 1$, let $f_n(x) = [2^n x] - 2[2^{n-1} x]$. In the compact space of mappings of $\mathbf{R}$ into $\{0, 1\}$ (equipped with the topology of pointwise convergence), let $f$ be a cluster point of the sequence $(f_n)$. Show that, for every dyadic number $r$, $f(r + x) = f(x)$ for all $x \in \mathbf{R}$ and $f(r - x) = 1 - f(x)$ for every $x \in \mathbf{R}$ different from a dyadic number. From this, deduce that for the Lebesgue measure $\mu$, $f$ is not measurable. (Argue by contradiction; let $A$ be the set of $x \in [0, 1]$ such that $f(x) = 1$, and suppose that $A$ is measurable and that $\mu(A) = \alpha > 0$; show that there exists a set $I$, the finite union of open intervals contained in $]0, 1[$, such that $\mu(I \cap \complement A) \leqslant \alpha/4$ and $\mu(A \cap \complement I) \leqslant \alpha/4$; on considering the intervals contained in $I$ and of the form $]k/2^n, (k+1)/2^n[$, show that one obtains a contradiction with the relation $f(r - x) = 1 - f(x)$ for $r$ dyadic and $x$ non-dyadic.)