Rademacher functions on the unit interval

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) for the Lebesgue-measure facts below. Let $I:=[0,1)$ and let $\lambda$ be Lebesgue measure restricted to the Borel subsets of $I$ (Lebesgue measurable sets, the family $\mathcal{L}({\mathbb{R}}^n)$, and the restricted set function ${\lambda }_n$). For integers $k\ge 1$ let $b_k(t):=\lfloor 2^{k}t\rfloor -2\lfloor 2^{k-1}t\rfloor \in \{0,1\}$ be the $k$-th binary digit, with $\lfloor \cdot \rfloor$ the integer part (Integer part: for every real $x$ there is exactly one integer $m$ with $m\le x<m+1$); the inclusion $b_k(t)\in \{0,1\}$ follows from $\lfloor 2u\rfloor \in \{2\lfloor u\rfloor ,2\lfloor u\rfloor +1\}$ for $u=2^{k-1}t$. For integers $j\ge 0$ define the $j$-th Rademacher function ${\epsilon }_j:I\to \{\pm 1\}$ by ${\epsilon }_j(t):=1-2b_{j+1}(t)\mathrm{.}$

Then:

1. Each ${\epsilon }_j$ is Borel measurable (Borel measurable and Lebesgue measurable functions on ${\mathbb{R}}^n$) and $\mathrm{\mid }{\epsilon }_j\mathrm{\mid }\equiv 1$; indeed $b_k$ is the parity of $\lfloor 2^{k}t\rfloor$, so ${\epsilon }_j$ is constant on the $2^{j+1}$ half-open dyadic intervals of generation $j+1$, which are measurable (A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included), and the arithmetic of Arithmetic and lattice operations preserve measurability whenever they are defined preserves measurability.
2. ${\epsilon }_0=1$ on $[0,1\mathrm{/}2)$ and $-1$ on $[1\mathrm{/}2,1)$; more generally, for every $j\ge 0$ the function ${\epsilon }_j$ is constant, with alternating signs, on each half-open dyadic interval $[k{2}^{-(j+1)},(k+1)2^{-(j+1)})$, $k=0,\dots ,2^{j+1}-1$, where it equals $(-1{)}^k$: on such an interval $2^{j+1}t\in [k,k+1)$, so $\lfloor 2^{j+1}t\rfloor =k$ and $2^{j}t\in [k\mathrm{/}2,(k+1)\mathrm{/}2)$ gives $\lfloor 2^{j}t\rfloor =\lfloor k\mathrm{/}2\rfloor$, hence ${\epsilon }_j(t)=1-2(k-2\lfloor k\mathrm{/}2\rfloor )=(-1{)}^k$.
3. Consequently, since each half-open dyadic interval of generation $j+1$ has Lebesgue measure $2^{-(j+1)}$ (A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included) and the intervals partition $I$, ${\int }_0^1{\epsilon }_j(t)dt={\sum }_{k=0}^{2^{j+1}-1}(-1{)}^k{2}^{-(j+1)}=0(j\ge 0),$ with the finite sum evaluated by pairing consecutive terms. Indeed, ${\epsilon }_j^{+}$ and ${\epsilon }_j^{-}$ are the indicators of the unions of the even and odd indexed intervals respectively, each of measure $1\mathrm{/}2$. Their nonnegative integrals equal $1\mathrm{/}2$ (The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions), so ${\epsilon }_j$ is integrable and its signed integral is their difference (Integrable real and complex functions, and their integrals). The displayed identity is the $m=1$ case of the equidistribution proved by the finite-block lemma below.

All integrals of functions of finitely many Rademacher functions on this page are integrals over $(I,\lambda )$ and are written ${\int }_0^1$. The exponent $2^{j+1}$ is an integer power in the sense of Integer powers $a^m$. No choice principle is used to construct the binary digits or sign functions; the measure and integral assertions inherit Countable Choice from the cited Lebesgue-measure suppliers.