Finite Rademacher blocks are equidistributed

Binary counting. Set $J:=j_m+1\ge 1$ and $d_i:=j_m-j_i$, so $0=d_m<d_{m-1}<\cdots <d_1<J$. Successive division by $2$ gives a unique remainder in $\{0,1\}$ and a quotient less than $2^{J-1}$; induction on $J$, starting with $J=0$ and $k=0$, therefore gives every integer $k$ with $0\le k<2^J$ a unique binary expansion $k={\sum }_{r=0}^{J-1}{\beta }_r(k)2^r$ with digits ${\beta }_r(k)\in \{0,1\}$. For every integer $d$ with $0\le d<J$ one has $\lfloor k\mathrm{/}{2}^d\rfloor ={\sum }_{r=d}^{J-1}{\beta }_r(k)2^{r-d}$, whose parity is ${\beta }_d(k)$ because all terms with $r>d$ are even multiples of $2$; hence $(-1{)}^{\lfloor k\mathrm{/}{2}^d\rfloor }=(-1{)}^{{\beta }_d(k)}$. The map $k\mapsto ({\beta }_{d_1}(k),\dots ,{\beta }_{d_m}(k))$ from $\{0,\dots ,2^J-1\}$ onto $\{0,1{\}}^m$ has every fiber of cardinality exactly $2^{J-m}$: for prescribed digits $b_1,\dots ,b_m$ the solutions are precisely $k={\sum }_{i=1}^m{b}_i{2}^{d_i}+{\sum }_{r\mathrm{\notin }\{d_1,\dots ,d_m\},r<J}{c}_r{2}^r$ with $c_r\in \{0,1\}$, and distinct choices of the free digits $c_r$ give distinct $k$ by uniqueness of the binary expansion, while there are $J-m$ free digits.

Constant values on the dyadic intervals. Let $I_k:=[k{2}^{-J},(k+1)2^{-J})$ for $k=0,\dots ,2^J-1$; these intervals partition $I$ exactly. If $t\in I_k$ then $2^{j_i+1}t\in [k\mathrm{/}{2}^{d_i},(k+1)\mathrm{/}{2}^{d_i})\subset [\lfloor k\mathrm{/}{2}^{d_i}\rfloor ,\lfloor k\mathrm{/}{2}^{d_i}\rfloor +1)$, because $k=2^{d_i}\lfloor k\mathrm{/}{2}^{d_i}\rfloor +r$ with $0\le r<2^{d_i}$ gives $(k+1)\mathrm{/}{2}^{d_i}=\lfloor k\mathrm{/}{2}^{d_i}\rfloor +(r+1)\mathrm{/}{2}^{d_i}\le \lfloor k\mathrm{/}{2}^{d_i}\rfloor +1$; hence $\lfloor 2^{j_i+1}t\rfloor =\lfloor k\mathrm{/}{2}^{d_i}\rfloor$ and, by [F1], ${\epsilon }_{j_i}(t)=(-1{)}^{\lfloor k\mathrm{/}{2}^{d_i}\rfloor }=(-1{)}^{{\beta }_{d_i}(k)}$. Therefore the sign vector $({\epsilon }_{j_1}(t),\dots ,{\epsilon }_{j_m}(t))$ equals $s(k):=((-1{)}^{{\beta }_{d_1}(k)},\dots ,(-1{)}^{{\beta }_{d_m}(k)})$ on all of $I_k$.

The level sets have measure $2^{-m}$. For a sign pattern $s\in \{\pm 1{\}}^m$ put $E_s:=\{t\in I:({\epsilon }_{j_1}(t),\dots ,{\epsilon }_{j_m}(t))=s\}$ and let $N_s:=\mathrm{\#}\{k<2^J:s(k)=s\}$. By step 2.1 the set $E_s$ equals the union of the intervals $I_k$ over those $k$; the intervals are pairwise disjoint, each has measure $2^{-J}$ by [F2], and the defining map $s(\cdot )$ is a bijection between sign patterns and digit vectors, so by step 1.1 $N_s=2^{J-m}$; hence $\lambda (E_s)=N_s{2}^{-J}=2^{-m}$.

The identity for $F\ge 0$. Suppose first that $F\ge 0$. The composition $F\circ \mathrm{\Phi }$, $\mathrm{\Phi }:=({\epsilon }_{j_1},\dots ,{\epsilon }_{j_m})$, is a nonnegative simple measurable function constant on the finitely many measurable sets $E_s$: it equals $F(s)$ on $E_s$, the union of the $E_s$ is all of $I$, and measurability follows from [F5]. Hence, by [F3], ${\int }_0^{1}F\circ \mathrm{\Phi }d\lambda$ equals its simple integral ${\sum }_{y\ge 0}y\lambda (\{F\circ \mathrm{\Phi }=y\})={\sum }_{s}F(s)\lambda (E_s)=2^{-m}{\sum }_{s}F(s)$ by step 3.1, where the last sum groups the $s$ with equal value $F(s)$. For a real-valued $F$ write $F=F^{+}-F^{-}$; then $(F\circ \mathrm{\Phi }{)}^{\pm }=F^{\pm }\circ \mathrm{\Phi }$ and the real integral is the difference of the two nonnegative integrals, so the identity holds; for complex-valued $F$ apply this to $\mathrm{Re}F$ and $\mathrm{Im}F$ and combine with the definition of the complex integral in [F3].

The monomial and orthonormality formulas. If $N=0$, the empty product is $1$ and its integral is $1$. For $N\ge 1$, let $l_1,\dots ,l_N\ge 0$ have distinct values $j_1<\cdots <j_m$ with multiplicities $e_1,\dots ,e_m\ge 1$. Apply step 4.1 to the function $F(s_1,\dots ,s_m):={\prod }_{i=1}^m{s}_i^{e_i}$: since $s^e=1$ for even $e$ and $s^e=s$ for odd $e$, the average factorises as $2^{-m}{\sum }_s{\prod }_i{s}_i^{e_i}={\prod }_{i=1}^m\left(\frac{1}{2}{\sum }_{u=\pm 1}{u}^{e_i}\right)$, and each factor is $1$ for even $e_i$ and $\frac{1}{2}(1-1)=0$ for odd $e_i$. Hence the integral is $1$ when every index occurs an even number of times and $0$ otherwise. Taking $m=1$ gives ${\int }_0^1{\epsilon }_j=0$; taking $N=2$ gives ${\int }_0^1{\epsilon }_j{\epsilon }_k=1$ for $j=k$ and $=0$ for $j\mathrm{\ne }k$, that is ${\delta }_{jk}$.