2.1. Characterization of stationarity by a functional equation for F

Recall that any number $x\in[0, 1]$ has a base-q expansion $x=(0.x_1x_2\ldots)_q$ with $x_1,x_2,\ldots\in\{0,\ldots,q-1\}$ . This expansion is unique except when x is a base-q fraction in (0, 1), that is, when for some (necessarily unique) $n\in\mathbb N$ we have either $x_n<x_{n+1}=x_{n+2}=\cdots=q-1$ or $x_n>x_{n+1}=x_{n+2}=\cdots=0$ ; we refer to n as the order of x and denote the set of all base-q fractions in (0, 1) by $\mathbb{Q}_q$ .

Here is the first main result of our paper.

Remark 2.2. Clearly, when the $X_n$ are i.i.d., (2.1) is satisfied, and for the examples of i.i.d. $X_n$ discussed in Section 1.1, F was either the uniform CDF on [0, 1] or a singular continuous function.

Apart from these examples, the best-known example of a singular continuous CDF is probably Minkowski’s question-mark function (Fragefunktion ?(x)) restricted to [0, 1]; see, e.g., [Reference Denjoy4, Reference Denjoy5, Reference Kairies12, Reference Minkowski14]. Recall that for two reduced fractions $p/q>r/s$ such that $ ps-rq =1$ (i.e. two consecutive Farey fractions), Minkowski’s question-mark function is recursively defined by

\begin{equation*} ?(0/1)=0, \qquad ?(1/1)=1, \qquad ?\bigg(\frac{p+r}{q+s}\bigg)=( ?(p/q)+?(r/s))/2, \end{equation*}

and extended to any $x\in [0, 1]$ by continuity. As later shown in Corollary 2.1, the ?-function does not satisfy (2.1) for any $q\geq 2$ . Hence, if the ?-function is studied in the framework of (1.1) and (1.2), the process $\{X_n\}_{n\geq 1}$ would not be stationary.

In the case of a Bernoulli scheme it is well known that $\mathrm{d}F$ is self-similar in the sense of Hutchinson, see [Reference Hu9, Reference Hutchinson10, Reference Strichartz23]. For completeness we remind the reader that $\mathrm{d}F$ is self-similar if there exist $p_0,\dots, p_n>0$ with $\sum_{j=0}^n p_j=1$ and contractions $S_0,\dots,S_n$ on $\mathbb{R}$ such that $\mathrm{d}F(E)=\sum_{j=0}^n p_j \mathrm{d} F(S_j^{-1}(E))$ , for all Borel sets $E\subset \mathbb{R}$ . When $n=q-1$ and $S_j(x)=(x+j)/q$ for $j=0,\dots,q-1$ , we show in the next proposition that self-similarity occurs if and only if the $X_n$ are i.i.d.

Proposition 2.1. Under stationarity the $X_n$ are i.i.d. if and only if

(2.2) \begin{equation} F(x)=\sum_{j=0}^{q-1} \mathbb{P}(X_1=j) F(qx-j), \qquad x\in [0, 1]. \end{equation}

As any CDF is differentiable almost everywhere, we next consider the derivative of F when F satisfies (2.1). As usual, we let $L^1([0, 1])$ be the set of complex absolutely integrable Borel functions defined on [0, 1]. Note that F ^′(x) exists outside a set of Lebesgue measure zero, $M\subset [0, 1]$ , and, for all x not belonging to $M\cup \{ x\in [0, 1] \,:\, x\in qM-j$ for some $j\in \{0,\dots, q-1\} \}$ , it follows from (2.1) that $F^{\prime}(x)=q^{-1} \sum_{j=0}^{q-1} F^{\prime}((x+j)/q)$ , for almost all $x\in [0, 1]$ . Proposition 2.2 shows that F ^′ is almost everywhere on [0, 1] equal to a constant $c\in[0, 1]$ (with $c=1$ if and only if F is absolutely continuous), and hence that the only purely absolutely continuous $\mathrm dF$ is Lebesgue measure. Proposition 2.2 is a consequence of [Reference Parry15, Theorems 1 and 2], where the author considers absolutely continuous measures which are invariant under the transformation $T(x)=\beta x\pmod 1$ where $\beta>1$ . In Section 3.3 we give a proof for Proposition 2.2 which is simpler than the one in [Reference Parry15] due to the fact that q is integer.

Proposition 2.2. Let $f\in L^1([0, 1])$ be such that, for almost all $x\in [0, 1] $ (with respect to Lebesgue measure),

(2.3) \begin{equation} f(x)=q^{-1}\sum_{j=0}^{q-1}f((x+j)/q). \end{equation}

Then f is almost everywhere a (complex) constant equal to $\int_0^1 f(x)\,\mathrm{d} x$ .

In Section 3.3 we construct remarkable examples of functions $f\not\in L^1([0, 1])$ where (2.3) is satisfied, but in one case f is not absolutely integrable and in another case f is not measurable.

2.2. Characterization of stationarity by the characteristic function of X

Next, we characterize stationarity of $\{X_n\}_{n\geq 1}$ in terms of the characteristic function of X given by $f(t)\,:\!=\, \int \mathrm{e}^{itx} \, \mathrm{d} F(x)$ , $t\in \mathbb{R}$ . In particular, we discuss when $\mathrm{d} F$ is a Rajchman measure, meaning that $f(t)\rightarrow 0$ as $t\to \pm\infty$ , cf. Section 1.1.

By the Riemann–Lebesgue lemma, if $\mathrm{d} F$ is absolutely continuous relative to Lebesgue measure, then it is also a Rajchman measure. Thus, a corollary to Theorem 2.2(II) is that if F is a CDF satisfying (2.1), then $\mathrm{d} F$ is absolutely continuous with respect to Lebesgue measure if and only if F is the uniform CDF on [0, 1]. But a more direct argument for this is to apply Proposition 2.2; see the beginning of Section 3.5.

Salem in [Reference Salem22] asked whether the measure $\mathrm{d} ?$ corresponding to Minkowski’s question-mark function restricted to [0, 1] is a Rajchman measure. It has recently been shown that $\mathrm{d} ?$ is indeed a Rajchman measure [Reference Jordan and Sahlsten11, Reference Persson17]. Combining this with Theorem 2.2(II), we obtain the following corollary.

2.3. Characterization of stationarity by a decomposition result for F

The theorem below characterizes stationarity of $\{X_n\}_{n\geq 1}$ by properties of each part of the Lebesgue decomposition of F. It is a generalization of results obtained in [Reference Harris8] and [Reference Dym6]; our proof in Section 3.5 is based on Theorem 2.1, Proposition 2.2, Theorem 2.2, and a technical result (Lemma 3.3).

Table 1. All cycles up to equivalence when $q\in\{2,3\}$ and $n\in\{1,2,3\}$ .

We need the following notation and concepts. We call $s\in[0, 1]$ a purely repeating base-q number of order n if the base-q expansion of s is of the form $s=(0.\overline{t_1\dots t_n})_q\,:\!=\,(0.t_1 \ldots t_n t_1\ldots t_n\ldots)_q =\sum_{j=1}^n t_jq^{-j}/(1-q^{-n})$ , where n is the smallest possible positive integer and $t_1,\ldots,t_n\in\{0,\ldots,q-1\}$ . A purely repeating base-q number cannot be a base-q fraction; if $S_j(x)\,:\!=\,(x+j)/q$ then the purely repeating number $(0.\overline{t_1\dots t_n})_q$ is the unique fixed point of the function $S_{t_1}\circ\dots \circ S_{t_n}$ . For any $n\in\mathbb N$ we call $(s_1,\ldots,s_n)$ a cycle of order $n\in\mathbb N$ if, for some integers $t_1,\dots,t_n\in \{0,\ldots,q-1\}$ ,

(2.4) \begin{equation} s_1=(0.\overline{t_1t_2\dots t_n})_q, \quad s_2=(0.\overline{t_nt_1\dots t_{n-1}})_q, \quad \dots, \quad s_{n}=(0.\overline{t_2\dots t_nt_1})_q,\end{equation}

and $s_1,\dots,s_n$ are pairwise distinct. Note that for two cycles $(s_1,\dots,s_n)$ and $(s^{\prime}_1,\dots,s^{\prime}_m)$ , the sets $\{s_1,\dots, s_n\}$ and $\{s^{\prime}_1,\dots, s^{\prime}_m\}$ are either equal or disjoint. Table 1 shows the cycles up to the order of elements for $q=2,3$ and $n=1,2,3$ . Moreover, let H be the Heaviside function defined by $H(x)=0$ for $x<0$ and $H(x)=1$ for $x\ge0$ . Finally, we say that F is a mixture of an at most countable number of CDFs if there exist CDFs $F_1,F_2,\ldots$ and a discrete probability distribution $(\theta_1,\theta_2,\ldots)$ such that $\tilde F=\sum_{i}\theta_iF_i$ .

The CDF $F_{s_1,\ldots,s_n}$ given by (2.5) is just the empirical CDF at the points in the cycle $(s_1,\ldots,s_n)$ . Thus, the following corollary follows immediately from (2.5).

3.1. Proof of Theorem 2.1

Before proving Theorem 2.1, we need the following two lemmas.

Lemma 3.1. Suppose that $\{X_n\}_{n\geq 1}$ is stationary. Then the probability of all $X_n$ having the same value starting from some $n_0>1$ and at least one $X_m$ having a different value for some $m<n_0$ is zero:

(3.1) \begin{equation} \mathbb P\Bigg(\bigcup_{0<m<n_0<\infty}\{X_m\not= X_{n_0}=X_{n_0+1}=\cdots\}\Bigg)=0. \end{equation}

Proof. It suffices to verify that, for integers $0<m<n_0<\infty$ ,

(3.2) \begin{equation} \mathbb P\!\left(X_m\not= X_{n_0}=X_{n_0+1}=\cdots\right)=0, \end{equation}

where, without loss of generality, we may assume that $m=n_0-1$ . For any $k\in\{0,\ldots,q-1\}$ , we have, by the law of total probability for two events, that

\begin{align*} \mathbb P(X_{n_0}=X_{n_0+1}=\cdots=k) & = \mathbb P( X_{n_0-1}\neq k, X_{n_0}=X_{n_0+1}=\cdots=k) \\ & \quad + \mathbb P(X_{n_0-1}=X_{n_0}=\cdots=k). \end{align*}

Then (3.2) follows, since by stationarity of $\{X_n\}_{n\geq 1}$ we have $\mathbb P(X_{n_0}=X_{n_0+1}=\cdots=k)=\mathbb P(X_{n_0-1}=X_{n_0}=\ldots=k)$ . Thereby, (3.1) is verified.

Proof. Clearly, (2.1) is also true for $\tilde{F}(x)$ if $x=1$ , so using (2.1) we have, for any $\delta \in \mathbb{Q}_q$ and for any base-q fraction $x\in(\delta,1)$ or for $x=1$ , that

(3.3) \begin{equation} \tilde{F}(x)-\tilde{F}(x-\delta)=\sum_{j=0}^{q-1} [\tilde{F}((j+x)/q)-\tilde{F}((j+x)/q -\delta/q)]. \end{equation}

Now, we prove the lemma by induction, considering first base-q fractions of order one. Letting $x=1$ gives $\tilde{F}(1)-\tilde{F}(1-\delta) = \tilde{F}(1)-\tilde{F}(1-\delta/q)+ \sum_{j=0}^{q-2} [\tilde{F}((j+1)/q)-\tilde{F}((j+1)/q -\delta/q)]$ , and letting $\delta\downarrow 0$ we see that all the jumps of $\tilde{F}$ at $1/q, \ldots, (q-1)/q$ must be zero, so $\tilde{F}$ is continuous at these points, which are first-order base-q fractions. Next, let us assume that, for a given order $n \geq 1$ , $\tilde{F}$ is continuous at all base-q fractions of order n. Let $x=(0.k_1k_2\ldots k_n)_q$ with $k_n \neq 0$ . By using (3.3) and taking $\delta \downarrow 0$ , we obtain that $\tilde{F}$ is continuous at all numbers of the form $(x+j)/q=(0.jk_1,\ldots k_n)_q$ for all $j \in \{0,\ldots,q-1\}$ . This shows that $\tilde{F}$ is continuous at all base-q fractions of order $n+1$ , which completes the proof.

Now, to prove Theorem 2.1 we need the following notation and observations. Let $(x_1,\ldots,x_n)$ , $(t_1,\ldots,t_n)\in\{0,\ldots,q-1\}^n$ . We write $(x_1,\ldots,x_n)\le(t_1,\ldots,t_n)$ if $\sum_{k=1}^n x_kq^{-k}\le\sum_{k=1}^n t_kq^{-k}$ . Define $t\,:\!=\,\sum_{j=1}^nt_jq^{-j}$ . Note that $x=\sum_{i=1}^\infty x_i q^{-i}\in [0, 1]$ satisfies $x\leq t+q^{-n}$ if and only if one of the following two statements holds true:

• The first n digits of x obey $(x_1,\ldots,x_n)\le(t_1,\ldots,t_n)$ (regardless of the values of the next digits $x_{n+1}, x_{n+2},\ldots$ ).

• We have $x_1=t_1,\ldots,x_{n-1}=t_{n-1},x_n=t_n+1,x_{n+1}=x_{n+2}=\cdots=0$ .

Define

\begin{align*} \mathcal{F}_1(t_1,\dots,t_n) & \,:\!=\, \mathbb P((X_1,\ldots, X_n)\leq {(t_1,\dots,t_n)}) \\ & = \sum_{(x_1,\ldots,x_n)\leq (t_1,\ldots,t_n)}\mathbb P(X_1=x_1,\dots,X_n=x_n).\end{align*}

Stationarity of $\{X_n\}_{n\geq 1}$ is equivalent to that for every $(t_1,\ldots,t_n)\in \{0,\ldots,q-1\}^n$ , so $\mathcal{F}_1(t_1,\ldots,t_n)$ equals

\begin{equation*}\mathcal{F}_2(t_1,\dots,t_n)\,:\!=\,\sum_{(x_2,\ldots,x_{n+1})\leq (t_1,\ldots,t_n)}\mathbb P(X_2=x_2,\dots,X_{n+1}=x_{n+1}),\end{equation*}

cf. Kolmogorov’s extension theorem. Furthermore, by (1.1) and (1.2) we have

(3.4) \begin{multline} F(t+q^{-n}) = \mathcal{F}_1(t_1,\dots,t_n)\\ + \mathbb P(X_1=t_1,\ldots,X_{n-1}=t_{n-1}, X_n=t_n+1,X_{n+1}=X_{n+2}=\cdots=0). \end{multline}

Proof of Theorem 2.1(I).Assume that $\{X_n\}_{n\geq 1}$ is stationary. Using (3.4) together with Lemma 3.1 we obtain

(3.5) \begin{equation} F(t+q^{-n})=\mathcal{F}_1(t_1,\dots,t_n). \end{equation}

Furthermore, stationarity of $\{X_n\}_{n\geq 1}$ implies that X and the ‘left-shifted’ stochastic variable $\sum_{n=1}^\infty X_{n+1}q^{-n}=qX-X_1$ are identically distributed. Thus, $F(x)=\mathbb P(qX-X_1\leq x)=\sum_{j=0}^{q-1}\mathbb P(X_1=j,\,X\le (x+j)/q)$ . We see that $\mathbb P(X_1=0,\,X\le x/q) = \mathbb P(X=0)+\mathbb P(0<\,X\le x/q) = F(0)+(F(x/q)-F(0))$ . Further, for $j\in\{1,\ldots, q-1\}$ ,

\begin{align*} \mathbb P(X_1=j,\,X\le (x+j)/q) & = \mathbb P(X_1=j, X_2=X_3=\cdots=0) + \mathbb P(j/q<\,X\le (x+j)/q) \\ & = F((x+j)/q)-F(j/q), \end{align*}

where we used (3.1) in order to get the second identity. This leads to (2.1).

Conversely, assume that F satisfies (2.1). Then, since F is right continuous and $\mathbb{Q}_q$ constitutes a dense subset of [0, 1], (2.1) holds for all $x\in[0, 1]$ . Further, by Kolmogorov’s extension theorem we can define the distribution of $\{X_n\}_{n\geq 1}$ on $\{0,\ldots,q-1\}$ by specifying the finite-dimensional distribution $\mathcal{F}_1$ o f $( X_1,\dots, X_n)$ for every integer $n\ge1$ in a consistent way, setting

(3.6) \begin{equation} \mathcal{F}_1(t_1,\dots,t_n)=\mathbb{P}((X_1,\dots,X_n)\leq (t_1,\dots,t_n))\,:\!=\,F(t+q^{-n}). \end{equation}

Furthermore, for any $\epsilon>0$ and $x_1,\dots, x_n\in \{0,\dots,q-1\}$ , the inequality $\mathbb{P}(X_1=x_1,\ldots, $ $X_n=x_n,X_{n+1}=\cdots=0) + \mathbb P(X_1=x_1,\ldots,X_{n-1}=x_{n-1}, X_n=x_n-1,X_{n+1}=\cdots=q-1) \leq$ $\mathbb P(x-\epsilon<X\leq x)=F(x)-F(x-\epsilon)$ holds, and hence by Lemma 3.2 the probability of realizing a base-q fraction is 0. Hence, the second term on the right-hand side of (3.4) is 0, and so (3.5) again holds true. Consequently, for any $n\in\mathbb N$ and $(t_1,\ldots,t_n)\in\{0,\ldots,q-1\}^n$ ,

\begin{align*} \mathcal{F}_2(t_1,\dots,t_n)&=\sum_{j=0}^{q-1}\sum_{(x_2,\ldots,x_{n+1})\leq (t_1,\ldots,t_n)}\mathbb P(X_1=j,X_2=x_2,\dots,X_{n+1}=x_{n+1}) \\ &=\mathbb P(X=0)+\sum_{j=0}^{q-1} \mathbb P(j/q<X\leq j/q + t/q +q^{-n-1})\\ &=F(0)+\sum_{j=0}^{q-1} (F((t+q^{-n}+j)/q)-F(j/q)) \\ &=F(t+q^{-n}), \end{align*}

using in the first identity the law of total probability, in the second that the probability of realizing a base-q point is zero, in the third (1.2), and in the last (2.1). Thereby, (3.6) gives that $\mathcal{F}_1(t_1,\dots,t_n)=\mathcal{F}_2(t_1,\dots,t_n)$ for every $n\in\mathbb N$ and every $(t_1,\ldots,t_n)\in\{0,\ldots,q-1\}^n$ , so $\{X_n\}_{n\geq 1}$ is stationary.

3.2. Proof of Proposition 2.1

Suppose that the $X_n$ are i.i.d. Since (2.2) holds for $x=1$ , F is right continuous, and $\mathbb{Q}_q$ is dense on (0, 1), it suffices to show that (2.2) holds for all $x \in \mathbb{Q}_q$ . Let $x=(0.x_1\dots x_n)_q$ with $x_1,\dots,x_n\in \{0,\dots,q-1\}$ where $x_n \neq 0$ . By Lemma 3.1, $\mathbb{P}(X_1=x_1,\dots, X_n=x_n, X_{n+1}=X_{n+2}=\cdots=0)=0$ . Hence, setting $\sum_{j=0}^{-1} \dots =0$ ,

\begin{align*} F(x)&=\sum_{j=0}^{x_1-1}\mathbb{P}(X_1=j)+\mathbb{P}(X_1=x_1)\sum_{j=0}^{x_2-1}\mathbb{P}(X_2=j)\\ &\quad+\dots+\mathbb{P}(X_1=x_1)\sum_{j=0}^{x_n-1} \mathbb{P}(X_2=x_2,\dots, X_{n-1}=x_{n-1},X_n=j) \\ &=\sum_{j=0}^{x_1-1} \mathbb{P}(X_1=j)+\mathbb{P}(X_1=x_1)F(qx-x_1),\end{align*}

using that the $X_n$ are independent in the first identity and identically distributed in the second. Furthermore, for $j\in \{0,\dots,q-1\}$ ,

(3.7) \begin{equation} F(qx-j)= \begin{cases} 0 & \textrm{if } x_1<j, \\[4pt] 1 & \textrm{if } x_1>j, \\[4pt] F((0.x_2\ldots,x_n)_q) & \textrm{if } x_1=j, \end{cases}\end{equation}

and so we conclude that (2.2) holds at x.

Conversely, suppose that (2.2) holds. Then $F(0)=\mathbb{P}(X_1=0)F(0)$ , so either $\mathbb{P}(X_1=0)=1$ or $F(0)=0$ . In the first case, since by stationarity the $X_n$ are identically distributed, we obtain immediately that the $X_n$ are i.i.d. So, assume that $F(0)=0$ and let $x=(0.x_1\dots x_n)_q$ with $x_1,\dots,x_n\in \{0,\dots,q-1\}$ . Then $\mathbb{P}(X_1=x_1,\dots, X_n=x_n)=\mathbb{P}(x\leq X\leq x+q^{-n})=F(x+q^{-n})-F(x)$ , using in the last identity that $\mathbb{P}(X=x)=0$ , cf. Lemma 3.1. Thus,

\begin{align*} \mathbb{P}(X_1=x_1,\dots, &\,X_n=x_n)=F(x+q^{-n}) - F(x) \\ &=\sum_{j=0}^{q-1}\mathbb{P}(X_1=j) \big (F(q(x+q^{-n})-j)-F(qx-j)\big )\\ &=\mathbb{P}(X_1=x_1) \big (F((0.x_2\dots x_n)_q+q^{-n+1})-F((0.x_2\dots x_n)_q)\big ),\end{align*}

where in the second equality we use (2.2) and in the third we use (3.7). The dependence on $x_1$ has now been isolated in the factor $\mathbb{P}(X_1=x_1)$ , and using the same method for the other variables $x_2,\ldots,x_n$ we obtain

\begin{align*} \mathbb{P}(&X_1=x_1,\dots,X_n=x_n)\\ &=\mathbb{P}(X_1=x_1)\sum_{j=0}^{q-1}\mathbb{P}(X_1=j) \big (F(x_2-j+(0.x_3\dots x_n)_q+q^{-n+2})\\ &\phantom{=\mathbb{P}(X_1=x_1)\sum\mathbb{P}(X_1=j) \big (}-F(x_2-j+(0.x_3\dots x_n)_q)\big ) \end{align*}

\begin{align*} &=\mathbb{P}(X_1=x_1)\mathbb{P}(X_1=x_2) \big (F((0.x_3\dots x_n)_q+q^{-n+2})-F((0.x_3\dots x_n)_q)\big )\\ &\;\,\vdots\\ &=\mathbb{P}(X_1=x_1)\cdots\mathbb{P}(X_1=x_{n-1})\sum_{j=0}^{q-1}\mathbb{P}(X_1=j) \big (F(x_n-j+1)-F(x_n-j)\big )\\ &=\mathbb{P}(X_1=x_1)\cdots \mathbb{P}(X_{1}=x_{n-1})\big (\mathbb{P}(X_1=x_n)(1-F(0))+\mathbb{P}(X_1=x_{n}+1)F(0)\big ).\end{align*}

Consequently, since $F(0)=0$ and by stationarity the $X_n$ are identically distributed, $\mathbb{P}(X_1=x_1,\dots, X_n=x_n)=\mathbb{P}(X_1=x_1)\cdots \mathbb{P}(X_{1}=x_{n})=\mathbb{P}(X_n=x_1)\cdots \mathbb{P}(X_{n}=x_{n})$ . Therefore, the $X_n$ are i.i.d.

3.3. Proof of Proposition 2.2 and related examples

Proof of Proposition 2.2.We first claim that for every $n\in\mathbb N$ , there exists a Borel set $A_n\subseteq [0, 1]$ of Lebesgue measure 1 such that, for all $x\in A_n$ ,

(3.8) \begin{equation} f(x)=q^{-n}\sum_{j=0}^{q^n-1} f((x+j)/q^n). \end{equation}

The case $n=1$ follows by the assumption in Proposition 2.2. Define

\begin{equation*} \tilde{A}_2\,:\!=\, \bigcup_{j=0}^{q-1} \frac{A_1+j}{q},\qquad A_2\,:\!=\, A_1 \cap \tilde{A}_2. \end{equation*}

Since the Lebesgue measure of $({A_1+j})/{q} \cap ({A_1+k})/{q}$ is 0 for $j \neq k$ , $\tilde{A}_2$ and hence $A_2$ have Lebesgue measure 1. For every $x\in A_2$ we have $(x+j)/q\in A_1$ and hence, using (2.3) for $f((x+j)/q)$ , we obtain

\begin{align*} f(x)=q^{-2} \sum_{j=0}^{q-1} \sum_{k=0}^{q-1} f\bigg( \frac{x+j}{q^2} + \frac{k}{q} \bigg) = q^{-2} \sum_{j=0}^{q^2-1} f\bigg( \frac{x+j}{q^2}\bigg). \end{align*}

Continuing in this way the claim is verified. Define $\mathcal{I}\,:\!=\,\int_0^1 f(t) \, \mathrm{d} t\in \mathbb{C}$ and $I_{j,n}\,:\!=\,(jq^{-n},jq^{-n}+q^{-n})$ . Then (3.8) gives, for all $x\in A_n$ , $f(x)-\mathcal{I}=\sum_{j=0}^{q^n-1} \int_{I_{j,n}} \{ f((x+j)/q^n) -f(y) \}\,\mathrm{d} y$ , so $|f(x)-\mathcal{I}|\leq \sum_{j=0}^{q^n-1} \int_{I_{j,n}} \vert f((x+j)/q^n) -f(y) \vert \,\mathrm{d} y$ . Integrating with respect to $x\in[0, 1]$ and making the change of variable $t =(x+j)/q^n\in I_{j,n}$ , we obtain

(3.9) \begin{equation} \int_0^1|f(x)-\mathcal{I}|\,\mathrm{d} x\leq q^n\sum_{j=0}^{q^n-1} \int_{I_{j,n}}\int_{I_{j,n}}|f(t) -f(y)|\,\mathrm{d} y\,\mathrm{d} t. \end{equation}

Now, for any $\varepsilon>0$ , there exists a uniformly continuous function $g_\varepsilon \,:\, [0, 1]\to\mathbb R$ such that $\Vert f-g_\varepsilon\Vert_{L^1}\leq \varepsilon /3$ , where we consider the usual $L^1$ -norm. Writing $|f(t)-f(y)|\leq |f(t)-g_\varepsilon(t)|+| f(y)-g_\varepsilon(y)|+|g_\varepsilon (t)-g_\varepsilon(y)|$ , we obtain from (3.9) that $\int_0^1|f(x)-\mathcal{I}|\,\mathrm{d} x\leq 2\varepsilon/3 +\sup_{|t-y|\leq q^{-n}} |g_\varepsilon (t)-g_\varepsilon(y)|$ . Since $g_\varepsilon$ is uniformly continuous, for any sufficiently large $n=n(\varepsilon)\in\mathbb N$ we have $\sup_{|t-y|\leq q^{-n(\varepsilon)}} |g_\varepsilon (t)-g_\varepsilon(y)|\leq \varepsilon/3$ . Thus, $\int_0^1|f(x)-\mathcal{I}|\,\mathrm{d} x\leq \varepsilon$ . Consequently, $f=\mathcal{I}$ almost everywhere on [0, 1].

We end this section by presenting two examples of functions $f\not\in L^1([0, 1])$ where (2.3) is satisfied but in one case f is not absolutely integrable and in the other case f is not measurable.

Example 3.1. (A solution to (2.3) which is not in $L^1([0, 1])$ .) Let $q=2$ . Below we construct a solution to (2.3) which is piecewise smooth on [0, 1], has finite jumps at all dyadic fractions $1-2^{-n}$ with $n\in\mathbb N$ , but whose integral diverges.

For this, we notice the following. For any function $f\,:\,[0, 1]\rightarrow\mathbb R$ , define $g(x)\,:\!=\,f(x)-1/(1-x)$ for $x\in[0, 1)$ , and let g(1) be any number. Then f satisfies (2.3) if and only if g satisfies the equation

(3.10) \begin{equation} g(x)=g(x/2)/2 +g((x+1)/2)/2 +1/(2-x)\qquad\mbox{for almost all $x\in[0, 1]$}. \end{equation}

To construct a particular solution g to (3.10), we start by setting $g(x)\,:\!=\,0$ for all $x\in \big[0,\frac12\big)$ . Then $g(x/2)=0$ for all $x\in [0, 1)$ , and in accordance with (3.10) we should have

(3.11) \begin{equation} g((x+1)/2)=2g(x)-2/(2-x)\qquad\mbox{for all $x\in[0, 1)$}, \end{equation}

which is possible because of the following observations. For each $n\in\mathbb N\cup\{0\}$ , the map $[1-2^{-n},1-2^{-n-1})\ni {\color{black}x\mapsto (x+1)/2}\in [1-2^{-n-1},1-2^{-n-2})$ is a bijection. Thus, for $n=1,2,\ldots$ , we can inductively use (3.11) to compute g(x) for all $x\in [1-2^{-n},1-2^{-n-1})$ . We see that g becomes more and more negative near 1. Now, the function $f(x)=g(x)+1/(1-x)$ is not constant on [0, 1), since $f(x)=1/(1-x)$ on $[0,\frac12)$ . Although f satisfies (2.3) and is smooth on each interval $[1-2^{-n},1-2^{-n-1})$ with $n\in\mathbb N$ , f cannot have a finite integral due to Proposition 2.2. Figure 1 shows a plot of f.

Figure 1. The function $f(x)=g(x)+1/(1-x)$ for $x\in[0,1-2^{-4}]$ , where g is given in Example 3.1.

3.4. Proof of Theorem 2.2

First, we prove Theorem 2.2(I) with $\tilde X$ , $\tilde{F}$ , and $\tilde{f}$ as in the theorem and $S_j(x)=(x+j)/q$ as in Section 2.3. Note that, for any $k\in\mathbb{Z}$ ,

(3.12) \begin{align} \tilde{f}(2\pi k q)&=\int_0^1 \exp\!(2\pi \mathrm{i} x kq)\, \mathrm{d} \tilde F(x)\nonumber =\sum_{j=0}^{q-1}\int_{j/q}^{(j+1)/q} \exp\!(2\pi \mathrm{i} x k q)\, \mathrm{d} \tilde F(x)\\\nonumber&=\sum_{j=0}^{q-1}\int_{0}^{1} \exp\!(2\pi \mathrm{i} (x+j) k)\, \mathrm{d} (\tilde F\circ S_j)(x)\\&=\int_{0}^{1} \exp\!(2\pi \mathrm{i} x k) \,\mathrm{d} \Bigg(\sum_{j=0}^{q-1}\tilde F\circ S_j\Bigg)(x).\end{align}

If $\tilde F$ satisfies (2.1) then $\mathrm{d} \tilde F=\sum_{j=0}^{q-1} \mathrm{d} (\tilde F\circ S_j)$ , which, combined with (3.12), shows that

(3.13) \begin{equation}\tilde f(2\pi kq)= \tilde f(2\pi k)\end{equation}

for all $k\in \mathbb Z$ .

Now, suppose that (3.13) holds for all $k\in \mathbb{Z}$ . Define $\mathrm{d} \tilde G\,:\!=\,\mathrm{d} \big(\sum_{j=0}^{q-1} \tilde F\circ S_j\big)$ and $\tilde g(t)\,:\!=\,\int \mathrm{e}^{\mathrm{i}xt} \, \mathrm{d} \tilde G(x)$ . From (3.12) we have that $\tilde{f}(2\pi kq)=\tilde{g}(2\pi k)$ which, together with (3.13), implies $\tilde f(2\pi k)=\tilde g(2\pi k)$ for all $k\in \mathbb{Z}$ . Recall that any continuous $\mathbb{Z}$ -periodic function $\varphi\,:\, \mathbb{R}\to \mathbb{C}$ is a uniform limit of trigonometric polynomials $\sum_{k=- N}^N c_k^N\mathrm{e}^{2\pi \mathrm{i} kx}$ , where each $c_k^N\in \mathbb{C}$ and $N\in \mathbb{N}$ [Reference Rudin21]. Taking the limit $N\to \infty$ and using Lebesgue’s dominated convergence theorem, we get

(3.14) \begin{equation} \int_{[0, 1]} \varphi(x)\,\mathrm{d} \tilde F(x)=\int_{[0, 1]} \varphi(x) \,\mathrm{d} \tilde G(x).\end{equation}

The remaining part of this proof consists of verifying (3.14) when $\varphi$ is merely continuous and then applying the Riesz–Markov theorem, which implies that a positive linear functional on $C_0([0, 1])$ can be represented by a unique measure (see [Reference Riesz19]). First, we establish equality of $\mathrm{d} \tilde F$ and $\mathrm{d} \tilde G$ at the endpoints of the interval [0, 1]. Since the indicator function of $\mathbb{Z}$ can be pointwise approximated by uniformly bounded continuous $\mathbb{Z}$ -periodic functions, another application of Lebesgue’s dominated convergence theorem in (3.14) gives $\mathrm{d}\tilde F(\{0\})+\mathrm{d}\tilde F(\{1\})=\mathrm{d}\tilde G(\{0\})+\mathrm{d}\tilde G(\{1\})$ , where, by the definition of $\mathrm d\tilde G$ , $\mathrm{d}\tilde G(\{0\})=\mathrm{d} \tilde F(\{0\}+\mathrm{d} \tilde F(\{1/q\})+\dots+\mathrm{d} \tilde F(\{(q-1)/q\})$ and $\mathrm{d}\tilde G(\{1\})=\mathrm{d} \tilde F(\{1/q\})+\dots+\mathrm{d} \tilde F(\{(q-1)/q\})+\mathrm{d}\tilde F(\{1\})$ . From this, it immediately follows that $\mathrm{d} \tilde{F}(\{j/q\})=0$ for $j=1,\ldots, q-1$ , which leads to $\mathrm{d}\tilde F(\{0\})=\mathrm{d} \tilde G(\{0\})$ and $\mathrm{d}\tilde F(\{1\})=\mathrm{d} \tilde G(\{1\})$ . Second, we extend (3.14) to all continuous functions on [0, 1] in the following way. If $\psi\,:\, [0, 1]\to \mathbb{C}$ is continuous, define, for $n=1,2,\dots$ ,

\begin{equation*}\varphi_n(x)\,:\!=\, \begin{cases} \psi(x)& \textrm{ for } x\in \big[0,1-\frac 1 n\big], \\[7pt] n\big[\psi(0)-\psi\big(1-\frac{1}{n}\big)\big]\big(x-\big(1-\frac 1 n\big)\big)+\psi\big(1-\frac{1}{n}\big)& \textrm{ for } x\in \big(1-\frac{1}{n},1\big]. \end{cases}\end{equation*}

This is a uniformly bounded sequence of continuous and $\mathbb{Z}$ -periodic functions converging pointwise to $\psi$ on [0, 1). Since $\mathrm{d} \tilde F(\{1\})=\mathrm{d} \tilde G(\{1\})$ , it follows from Lebesgue’s dominated convergence theorem that

\begin{align*}\int_{[0, 1]} \psi(x) \,\mathrm{d} \tilde F(x)&= \mathrm{d} \tilde F(\{1\})\psi(1)+\lim_{n\to \infty} \int_{[0, 1)}\varphi_n(x) \,\mathrm{d} \tilde F(x)\\&=\mathrm{d} \tilde G(\{1\})\psi(1)+\lim_{n\to \infty} \int_{[0, 1)}\varphi_n(x) \,\mathrm{d} \tilde G(x)\\&=\int_{[0, 1]} \psi(x) \,\mathrm{d} \tilde G(x).\end{align*}

The maps $C_0([0, 1])\ni\psi \mapsto \int_{[0, 1]} \psi(x) \, \mathrm{d}\tilde{F}(x) \in \mathbb{C}$ $C_0([0, 1])\ni\psi \mapsto \int_{[0, 1]} \psi(x) \, \mathrm{d}\tilde{G}(x) \in \mathbb{C}$ are positive linear functionals and can be represented by a unique measure according to the Riesz–Markov theorem (see [Reference Riesz19]), thus $\mathrm{d}\tilde F=\mathrm{d} \tilde G$ on [0, 1]. Equivalently, $\tilde F$ satisfies (2.1) and the proof of Theorem 2.2(I) is complete.

Next, we prove Theorem 2.2(II). As the ‘if’ part of the proof follows from a direct calculation, we only prove that if $c\,:\!=\,\lim_{t\to \infty}f(t)\in \mathbb{C}$ exists, then $0\leq c\leq 1$ and, for every $x\in[0, 1]$ ,

(3.15) \begin{equation} F(x)=(1-c)x+cH(x) ,\end{equation}

where H is the Heaviside function. Since $\lim_{t \to \infty} f(t) = c$ , for any $a\in \mathbb{R}$ a straightforward calculation gives

\begin{equation*} \lim_{T \to \infty} T^{-1}\int_0^T f(t)\mathrm{e}^{-\mathrm{i}ta}\,\mathrm{d} t =\left\{\begin{matrix}c\quad {\rm if}\; a=0, \\[4pt]0\quad {\rm if}\; a\neq 0.\end{matrix}\right .\end{equation*}

Define

\begin{align*} f_T (x) \,:\!=\, \frac{1}{T}\int_0^T \mathrm e^{\mathrm{i}t(x-a)} \, \mathrm{d}t = \begin{cases} 1 & \textrm{ if } x=a, \\[4pt] (\mathrm{i}(x-a)T)^{-1}(\mathrm e^{\mathrm{i}T(x-a)}-1) & \textrm{ if } x \neq a.\end{cases}\end{align*}

Using Fubini’s theorem we obtain $\frac{1}{T}\int_0^T f(t)\mathrm e^{-\mathrm{i}ta} \, \mathrm{d} t = \int_{[0, 1]} f_T(x) \, \mathrm{d} F(x)$ . Since $\vert f_T(x)\vert \leq 1$ and $f_T(x)$ converges pointwise to the indicator function of the set $\{a\}$ , an application of Lebesgue’s dominated convergence theorem shows that the above limit equals $\mathrm{d} F(\{a\})$ . Consequently, $c=F(0)\in[0, 1]$ and F is continuous on (0, 1).

It remains to verify (3.15). If $c=1$ , then $F=H$ and (3.15) follows. Assume $c<1$ . Then $G(x)\,:\!=\,(F(x) - cH(x))/(1-c)$ is a continuous CDF that also satisfies the stationarity condition (2.1). Thus, defining $g(t)\,:\!=\,\int \mathrm{e}^{\mathrm{i}tx} \, \mathrm{d} G(x)$ , we obtain the identity $g(2\pi kq)=g(2\pi k)$ for all $k\in \mathbb{Z}$ . A repeated use of this identity gives, for any $m\in\mathbb N$ and $k\in\mathbb{Z}$ ,

(3.16) \begin{equation} g(2\pi k q^{m})=g(2\pi k).\end{equation}

From the definition of G it follows that $\lim_{t \to \infty} g(t) = 0$ , and since c is real we also obtain $\lim_{t \to -\infty} g(t) = 0$ . Combining this with (3.16) where we take m to infinity, we conclude that $g(2\pi k)=0$ for all $k\in\mathbb{Z}\setminus\{ 0\}$ . If $h(t)\,:\!=\,\int_0^1 \mathrm{e}^{\mathrm{i}tx} \, \mathrm{d} x$ , then also $h(2\pi k)=0$ for all $k\in \mathbb{Z}\setminus\{0\}$ , and since $\mathrm{d} G(\{1\})=0$ it follows from the same arguments as in the proof of Theorem 2.2(I) that $\mathrm{d} G=\mathrm{d} x$ on [0, 1]. Consequently, $F(x)=(1-c)x+cH(x)$ for all $x\in[0, 1]$ .

3.5. Proof of Theorem 2.3

The Lebesgue–Radon–Nikodym theorem [Reference Billingsley3, Reference Folland7] leads to the decomposition $F(x) = \theta_1 F_1(x) + \theta_2 F_2(x) + \theta_3 F_3(x)$ for $x\in [0, 1]$ , where $\theta_1,\theta_2,\theta_3\geq 0$ and $\theta_1 + \theta_2 + \theta_3 = 1$ , $F_1$ is an absolutely continuous CDF on [0, 1], $F_2$ is a discrete CDF on [0, 1], and $F_3$ is singular continuous CDF on [0, 1]. Proposition 2.2 implies that $F^{\prime}_1 =1$ almost everywhere on [0, 1], and so since $F_1$ is absolutely continuous, $F_1(x)=x$ for all $x\in [0, 1]$ . Thus, $F_1$ satisfies (2.1) and it only remains to show that $F_2$ is as claimed in Theorem 2.3(II) and satisfies (2.1).

3.5.1. Proof of Theorem 2.3(II)

Lemma 3.3. Suppose that F satisfies (2.1) and $s\in[0, 1]$ is a discontinuity of F. Then there exists $n\in\mathbb{N}$ and a cycle $(s_1,\ldots,s_n)$ in the sense of (2.4) such that $s=s_1$ and $s_1,\ldots,s_n$ are discontinuities of F. Furthermore, the jumps of F at these n discontinuities are all equal.

Proof. We start by investigating what can happen at 0 and 1. Both $0=(0.\overline{0})_q$ and $1=(0.\overline{q-1})_q$ are purely repeating base-q numbers of order 1, and they can be discontinuity points because both H(x) and $H(x-1)$ satisfy (2.1). Therefore, in the following we will only consider possible discontinuities at $x\in (0, 1)$ . Our idea is then to show that each point of discontinuity belongs to a ‘cycle’ of finitely many points which are all discontinuities and the jump at each point is the same.

As in Section 2.3, define $S_j(x)\,:\!=\,(x+j)/q$ for $x\in (0, 1)$ and $j=0,\dots,q-1$ . Then, by Theorem 2.1, it follows that, for any $x\in (0, 1)$ , there exists a sufficiently small $\delta_0>0$ such that, for all $\delta\in (0,\delta_0)$ ,

(3.17) \begin{equation} F(x)-F(x-\delta)=\sum_{j=0}^{q-1}[F(S_j(x))-F(S_j(x-\delta))].\end{equation}

For $x\in (0, 1)$ , define $L_0(x)\,:\!=\,\{x\}$ and $L_n(x)\,:\!=\,\bigcup_{j=0}^{q-1}S_j(L_{n-1}(x))$ , $n=1,2,\ldots$ Furthermore, let $J_x\,:\!=\,\lim_{\delta\downarrow 0} [F(x)-F(x-\delta)]$ denote the jump of F at $x\in (0, 1)$ . Taking $\delta\downarrow 0$ in (3.17) shows that

(3.18) \begin{equation} J_x=\sum_{y\in L_1(x)} J_y.\end{equation}

Suppose $s\in (0, 1)$ is a discontinuity of F with jump $J_s>0$ , and let $k>1/J_s$ be an integer. First, we show that the sets $L_0(s), \dots, L_k(s)$ are not pairwise disjoint. For the purpose of a contradiction assume that $L_0(s), \dots, L_k(s)$ are pairwise disjoint. By assumption $s\not \in L_1(s)$ , thus replacing $x=s$ in (3.18) shows that F has a total jump of at least $2J_s$ : one $J_s$ from s, and the other $J_s$ from the accumulated contribution of all the points of $L_1(s)$ . Using (3.18) for $x=S_j(s)$ with $j=0,\ldots,q-1$ , we see that the possible jump at each $S_j(s)$ equals the total accumulated jump at the points of $L_1(S_j(s))$ . Hence, by assumption, the total jump of F at the points of $L_2(s)$ is again $J_s$ . Continuing this way we obtain that $\sum_{x\in L_j(s)} J_x =J_s$ for $j=0,1,\dots, k$ . By the choice of k this contradicts $F\leq 1$ and hence the sets $L_0(s), \dots, L_k(s)$ are not pairwise disjoint.

Next, let n denote the smallest integer (not necessarily larger than $1/J_s$ ) such that $L_0(s),\dots,L_n(s)$ are not pairwise disjoint. We will show that $s\in L_n(s)$ . Suppose this is not the case. Then by the choice of n there exist an integer m with $1\leq m<n$ and $j_1,\dots,j_{m},j^{\prime}_1,\dots,j^{\prime}_{n}\in \{0,\dots,q-1\}$ such that

(3.19) \begin{equation} S_{j^{\prime}_1}\circ \dots \circ S_{j^{\prime}_{n}}(s)=S_{j_1}\circ \dots \circ S_{j_m}(s).\end{equation}

If $s=(0.t_1 t_2\dots)_q$ , then from (3.19) it follows that $(0.j_1\dots j_m t_1t_2\dots)_q=S_{j_1}\circ \dots \circ S_{j_m}(s)=S_{j^{\prime}_1}\circ \dots \circ S_{j^{\prime}_{n}}(s)=(0.j^{\prime}_1\dots j^{\prime}_{n} t_1t_2\dots)_q$ , and thus $S_{j^{\prime}_{m+1}}\circ\dots \circ S_{j^{\prime}_{n}}(s)=s$ , contradicting the minimality of n. Hence, $s\in L_n(s)$ , which implies that there exist $i_1,\dots,i_n\in\{0,\dots,q-1\}$ such that $S_{i_1}\circ \dots \circ S_{i_n}(s)=s$ . Note that $(0.\overline{i_1\dots i_n})_q$ is also a fixed point of $S_{i_1}\circ\dots S_{i_n}$ , but since $S_{i_1}\circ\dots S_{i_n}$ is a contraction on [0, 1] (with Lipschitz constant $q^{-n}$ ) it has a unique fixed point and we must have $s=(0.\overline{i_1\dots i_n})_q$ .

By definition $S_{i_n}(s)\in L_1(s)$ , and from (3.18) we deduce that $J_s\geq J_{S_{i_n}(s)}$ . Letting $x=S_{i_n}(s)$ in the left-hand side of (3.18) we have that $J_s\geq J_{S_{i_n}(s)}\geq J_{S_{i_{n-1}}\circ S_{i_n}(s)}$ . Continuing in this way we see that $J_s\geq J_{S_{i_n}(s)}\geq \dots\geq J_{S_{i_2}\circ \dots \circ S_{i_n}(s)}\geq J_{S_{i_1}\circ \dots \circ S_{i_n}(s)}=J_{s}$ . This shows that the numbers $s,S_{i_n}(s),\dots, S_{i_2}\circ \dots \circ S_{i_n}(s)$ are discontinuities of F with the same jump. By the minimality of n these points are distinct and hence constitute a cycle.

Now, Theorem 2.3(II) follows from Lemma 3.3 and the fact that F has countably many points of discontinuity.

3.5.2. Proof that $F_2$ satisfies (2.1)

Because of Theorem 2.3(II), in order to show that $F_2$ satisfies (2.1), without loss of generality we may assume that $F_2(x)=\frac{1}{n}\sum_{j=1}^n H(x-s_j)$ , where $(s_1,\ldots,s_n)$ is a cycle. For each $j \in \{1,\dots,n\}$ , let $s_{j}(1)$ denote the first digit in the base-q expansion of $s_j$ and note that $qs_j=s_{j-1}+s_{j}(1)$ , where we define $s_0\,:\!=\,s_n$ . Hence, for any $k\in \mathbb{Z}$ , the characteristic function $f_2$ of $F_2$ satisfies

\begin{equation*} f_2(2\pi k q)=\frac{1}{n}\sum_{j=1}^n \mathrm{e}^{2\pi \mathrm{i} k q s_j} =\frac{1}{n}\sum_{j=1}^n \mathrm{e}^{2\pi \mathrm{i} k s_{j-1}}= f_2(2\pi k).\end{equation*}

Then, by Theorem 2.2(I) it follows that $F_2$ satisfies (2.1).