Kolmogorov’s Block Construction and Almost-Everywhere Divergence: Examples

Multiplication gives $Q(x)=e^{10\pi ix}+e^{12\pi ix}=e_5(x)+e_6(x)$. Both coefficients are one and all others zero by the character-integral calculation in F1, so its support is exactly $\{5,6\}$. The cutoff through four is the empty supported sum, through five is $e_5$, and through six is $e_5+e_6=Q$.

Since $\mathrm{\mid }{e}_5(x)\mathrm{\mid }=1$, $\mathrm{\mid }Q(x)\mathrm{\mid }=\mathrm{\mid }1+e^{2\pi ix}\mathrm{\mid }$. Factoring gives $1+e^{2\pi ix}=2e^{\pi ix}\cos (\pi x)$, hence both magnitudes are $2\mathrm{\mid }\cos (\pi x)\mathrm{\mid }$. At x=0 they equal two, and at x=1/2 they vanish. This concrete shift changes the frequencies while preserving every pointwise magnitude.

Finite geometric cancellation gives ${\sum }_{j=1}^J{2}^{-j}=1-2^{-J}$, hence ${\sum }_{j\ge 1}m(E_j)\le 1<\mathrm{\infty }$. More precisely, the tail satisfies ${\sum }_{j\ge J}m(E_j)\le 2^{1-J}$.

F1 applies to these measurable sets and the finite sum proved in step 1.1; it gives $m({\bigcap }_J{\bigcup }_{j\ge J}{E}_j)=0$. Equivalently, outside this null intersection, some J has the point outside every $E_j$ for $j\ge J$, which is the asserted eventual good-set property. The proof of F1 bounds the limsup measure by each tail, here explicitly $2^{1-J}\to 0$. No independence or choice assumption on the supplied sequence is needed.