Kolmogorov block maxima diverge off a null limsup set

For any integer k and integrable functions u,v, the definition gives $\mathrm{\mid }\widehat{u}(k)-\widehat{v}(k)\mathrm{\mid }\le \int \mathrm{\mid }u-v\mathrm{\mid }$, since $\mathrm{\mid }{e}_{-k}\mathrm{\mid }=1$. Hence the coefficients of F1's partial block sums converge to those of f. For each fixed k, the polynomial coefficients stabilize once the block containing k has been included; if no block contains k, all are zero. Thus f has exactly the prescribed block coefficients and no negative coefficients. In particular every fixed symmetric partial sum is determined by finitely many of these coefficients, with no pointwise infinite-series interchange.

By F4 and F1, $m({\mathrm{lim\hspace{0.17em}sup}}_j{E}_j)=0$. Remove this set and F1's null set on which $G(x)={\sum }_j\mathrm{\mid }{Q}_j(x)\mathrm{\mid }$ may be infinite. Fix any remaining x. There is $j_0(x)$ such that for all $j\ge j_0(x)$ the point lies in the strict good set of F2, so some $0\le r_j\le d_j$ has $\mathrm{\mid }{A}_{r_j}{P}_j(x)\mathrm{\mid }>j{2}^j$. These witnesses may be taken as least indices; no choice of a measurable family of cutoffs is needed for a pointwise supremum.

Set $N_j=m_j+r_j$. At this cutoff every earlier block is completed and every later block is absent. Steps 1.1 and F3 give the exact identity $S_{N_j}f(x)={\sum }_{i<j}{Q}_i(x)+a_j{e}_{m_j}(x)A_{r_j}{P}_j(x)$. Consequently $\mathrm{\mid }{S}_{N_j}f(x)\mathrm{\mid }>j-{\sum }_{i<j}\mathrm{\mid }{Q}_i(x)\mathrm{\mid }\ge j-G(x)\to \mathrm{\infty }$. Also $N_j\ge m_j\ge j$, so these are unbounded cutoffs. This proves the assertion. AC is inherited from F1's countable block choice; Borel–Cantelli requires no independence.