Show a function assumes all values in $[0, 1]$ on any open interval $(a, b) \subset (0, 1)$

Question

Let $x \in (0, 1)$, let $(x_i)_{i \in \mathbb{N}}$ be the decimal expansion of $x$. If $x$ has several decimal expansions, use the one that terminates with $0$. Define:

$$f(x) = \limsup_{n \rightarrow \infty} \left(\frac{1}{n}\#(i = 1, ..., n: x_i = 3)\right)$$

Note that $\#$ here is the cardinality function.

I can show that $f(x)$ is a Borel measurable function by writing it as a sum of indicator functions, as shown in this link. Next, I want to show that $f(x)$ can take any values between $[0, 1]$ on any intervals $(a, b) \subset (0, 1)$. I see there is a similar construction here for binary expansion, but not sure how to generalize it to decimal expansion.

Answer 1

$\large\textbf{1.}$ Note that the value of $f(x)$ does not depend on the finitely many leading digits in the decimal expansion of $x$:

$$ f(x) = \limsup_{n\to\infty} \frac{\mathbf{1}[x_{1}=3] + \sum_{i=1}^{n-1} \mathbf{1}[x_{i+1}=3]}{n} = f(10x \text{ mod }1) $$

An important consequence is that we have

$$ f\left(\left(\frac{k}{10^m},\frac{k+1}{10^m}\right)\right) = f((0, 1)) $$

for any integers $m \geq 0$ and $0 \leq k < 10^m$. Since any non-empty openset of $(0, 1)$ contains an interval of the form $(\frac{k}{10^m},\frac{k+1}{10^m})$ for some $m$ and $k$, it suffices to show that $f((0, 1))=[0,1]$ holds.

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$\large\textbf{2.}$ Among the many different ways of showing $f((0, 1)) = [0, 1]$, let me demonstrate a probabilistic one:

For each $p \in [0, 1]$, let $(Y_i)_{i=1}^{\infty}$ be a sequence of independent Bernoulli random variables with parameter $p$. Then by the strong law of large numbers,

$$ \lim_{n\to\infty} \frac{Y_1 + \cdots + Y_n}{n} = p \qquad \text{with probability one.} $$

Now by setting $X = \sum_{i=1}^{\infty} \frac{3 Y_i + 2(1 - Y_i)}{10^i}$ and letting $(X_i)_{i=1}^{\infty}$ the deciman expansion of $X$, we get $\mathbf{1}[X_i = 3] = Y_i$, hence

$$ f(X) = p \qquad \text{with probability one.} $$

Moreover, we have $0.222\ldots \leq X \leq 0.333\ldots$, hence $0 < X < 1$. This proves that there exists a number $x \in (0, 1)$ such that $f(x) = p$, hence the desired claim follows.

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$\large\textbf{3.}$ (Optional) A less elegant proof of $f((0, 1)) = [0, 1]$ is as follows: Fix $p \in [0, 1]$. We will construct a real number $x \in (0, 1)$ such that $f(x) = p$.

• Fix an increasing sequence $\bbox[color:blue; padding:2px; background:#F2FAFF;]{(N_k)_{k=1}^{\infty}}$ of positive integers such that the following conditions hold: $$ \frac{k}{N_k} \to 0 \qquad\text{and}\qquad \frac{N_{k+1} - N_k}{N_k} \to 0. $$ For example, we may set $N_k = k^2$. These technical conditions will only be used in the very last step, and until then, you don't have to worry about it.

• For each $k$ and with $\bbox[color:blue; padding:2px; background:#F2FAFF;]{M_k} = N_k - N_{k-1}$, let $\bbox[color:blue; padding:2px; background:#F2FAFF;]{(x_{k,i})_{i=1}^{M_k}} $ be a finite sequence consisting of $2$'s and $3$'s so that $\sum_{i=1}^{M_k} \mathbf{1}[x_{k,i} = 3] = \lfloor M_k p\rfloor$ holds. Note that we have $$ \left| \frac{\sum_{i=1}^{M_k} \mathbf{1}[x_{k,i} = 3]}{M_k} - p \right| = \left| \frac{\lfloor M_k p \rfloor}{M_k} - p \right| \leq \frac{1}{M_k}. $$

• We concatenate the sequences $(x_{k,i})_{i=1}^{M_k}$ for $k = 1, 2, \ldots$ to obtain a 1D sequence $\bbox[color:blue; padding:2px; background:#F2FAFF;]{(x_i)_{i=1}^{\infty}}$: \begin{align*} (x_i)_{i=1}^{\infty} &= (x_{1,i})_{i=1}^{M_1} \oplus (x_{2,i})_{i=1}^{M_2} \oplus \cdots \oplus (x_{k,i})_{i=1}^{M_k} \oplus \cdots \\ &= (x_{1,1}, \ldots, x_{1,M_1}, x_{2, 1}, \ldots, x_{2, M_2}, \ldots, x_{k, 1}, \ldots, x_{k, M_k}, \ldots). \end{align*} Then we define $\bbox[color:blue; padding:2px; background:#F2FAFF;]{x}$ as the real number having $(x_i)_{i=1}^{\infty}$ as its decimal expansion: $$ x = \sum_{i=1}^{\infty} \frac{x_i}{10^i} $$

Then for any $n$ and with $k$ such that $N_k < n \leq N_{k+1}$,

\begin{align*} &\left| \frac{\sum_{i=1}^{n} \mathbf{1}[x_i = 3]}{n} - p \right| \\ &\leq \frac{1}{n} \Biggl( \sum_{l=1}^{k} \left| \sum_{i=1}^{M_l} \mathbf{1}[x_{l,i} = 3] - M_k p \right| + \left| \sum_{i=1}^{n - N_k} \mathbf{1}[x_{k+1,i} = 3] - (n - N_k) p \right| \Biggr) \\ &\leq \sum_{l=1}^{k} \frac{M_l}{n} \left| \frac{\sum_{i=1}^{M_l} \mathbf{1}[x_{l,i} = 3]}{M_l} - p \right| + \frac{n - N_k}{n} \\ &\leq \frac{k}{N_k} + \frac{N_{k+1} - N_k}{N_k} \to 0. \end{align*}

Therefore we get $f(x) = p$.

Answer 2

Pick integers $N_1<N_2<\ldots$ such that $\sum_{j=1}^{k-1}10^{N_j}/10^{N_k}<1/k$. Let $s_0=0$ and $s_k=\sum_{j=1}^k10^{N_j}$ for $k=1,2,\ldots$. Let $I_k=(s_{k-1},s_k]$, so $|I_k|=10^{N_k}$.

Fix $y\in (0,1)$ and let $y=\sum_{j=1}^\infty y_j 10^{-j}$ be the decimal expansion of $y$.

Let $m_k=10^{N_k}\sum_{j=1}^{N_k}y_j10^{-j}$, which is an integer in $0$ and $10^{N_k}$. Let $S_k$ be the last $m_k$ members of $I_k$ and let $R_k=I_k\setminus S_k$. Let $x_i=3$ if $i\in \cup_{k=1}^\infty S_k$ and $x_i=0$ if $i\in \cup_{k=1}^\infty R_k$.

Note that $|\{i\leqslant n:x_i=3\}|/n$ will not increase as $n$ moves through $R_1$, then will not decrease as $n$ moves through $S_1$, will not increase as $n$ moves through $R_2$, then will not increase as $n$ moves through $S_2$, etc. So the $\lim\sup$ can be evaluated just at $n=s_k$, where it peaks. But at $n=s_k$, the first $s_{k-1}$ numbers are so few as to not matter ($s_{k-1}/s_k<1/k$ by choice). And we have $\frac{m_k}{s_k-s_{k-1}}\approx \frac{m_k}{s_k}=|\{i\leqslant s_k:x_i=3\}|/s_k$. But $m_k/{s_k-s_{k-1}}$ is the first $N_k$ digits of the decimal expansion of $y$, which will tend to $y$ as $k\to\infty$.