I have a countable collection of proper disjoint closed intervals $\mathcal{C} = \left\lbrace I_n\right\rbrace$ in the unit interval $[0,1]$. I want to construct a certain non-decreasing surjective continuous function $f:[0,1]\to [0,1]$ which is locally constant (only) on the above intervals.
More precisely, $f$ is surjective, $t_1 \leq t_2$ should imply $f(t_1)\leq f(t_2)$, and
$$
f(t_1) = f(t_2) \iff t_1, t_2 \in I_n \text{ for some } n.
$$
In general the union of these disjoint intervals can have full measure so I guess I would have to imitate the Cantor function construction? So letting $U_n = [0,1] \setminus \bigcup\limits_{i=1}^n I_n$ be written as a disjoint union of open (in $[0,1]$) intervals $U_n = \bigcup V_j$, my thought was to define
$$f_1(x) = \frac{1}{|V_1| + |V_2|}\int^x_0 \left(1_{V_1}+1_{V_2}\right)
$$
and iterate the process by splitting $V_i$ based on where the next $I_n$ showed up? I'm not sure this works since I'm not convinced there will be some uniform convergence.
Can someone please help me out?
Best Answer
Here's one way of doing this that imitates a sequential approach to defining the Cantor function. Note that we'll write $I_n = [a_n, b_n]$.
We'll also assume for convenience that among the $I_n$ there's no interval of the form $[0, b]$ or $[a, 1]$ in our collection. If one or both is in play, the process below can be adjusted slightly. (We'd want to take all the functions below to be $0$ on $[0, b]$, and $1$ on $[a, 1]$.)
Step 0: Define $f_0(x) = x$ for all $x \in [0, 1]$.
Step 1: Define $f_1$ as follows. On $I_1 = [a_1, b_1]$, set $f_1$ equal to the average value of $f_0$ on $I_1$; call this value $y_1$. (Equivalently, this is the value of $f_0$ at the interval midpoint $(a_1 + b_1) / 2$.) Then linearly interpolate between $f_1(0) = f_0(0) = 0$ and $f_1(a_1) = y_1$; and between $f_1(b_1) = y_1$ and $f_1(1) = f_0(1) = 1$.
Step 2: For $j \geq 2$, define $f_j$ as follows. First locate $I_j$ with respect to $I_1, I_2, \ldots I_{j-1}$: suppose that $I_j$ is immediately to the right of $I_{j_1} = [a_{j_1}, b_{j_1}]$, and immediately to the left of $I_{j_2} = [a_{j_2}, b_{j_2}]$. (If $I_j$ doesn't have a neighbor to one side, we can adjust what follows slightly.) For $x \leq b_{j_1}$ or $x \geq a_{j_2}$, set $f_j(x) = f_{j-1}(x)$. For $x \in I_j$, set $f_j(x)$ equal to the average value of $f_{j-1}$ on $I_j$; call this value $y_j$. (Equivalently, this is the value of $f_{j-1}$ at the interval midpoint $(a_j + b_j) / 2$.) Finally, linearly interpolate between $f_j(b_{j_1}) = f_{j-1}(b_{j_1})$ and $f_j(a_j) = y_j$; and between $f_j(b_j) = y_j$ and $f_j(a_{j_2}) = f_{j-1}(a_{j_2})$.
Step 3: Define $f(x) = \lim_{j \to \infty} f_j(x)$. I haven't been able to figure out how to show Cauchy convergence of the $f_j$ in the uniform norm, so as to simultaneously establish the existence of the limit and its continuity. But we can definitely prove the pointwise convergence of the $f_j$.
The most complicated part of the pointwise convergence, to my mind, is for a point $x \notin \bigcup I_n$ that is arbitrarily close to these intervals to both the left and right sides. To see how convergence works in this case, we first write $(0, 1) \setminus \left( \bigcup_{j=1}^{k} I_j \right)$ as the union of disjoint open intervals $V_{k,1}, V_{k,2}, \ldots, V_{k, k+1}$; and let $V_{k, \ell_k} = (c_k, d_k)$ be the interval containing $x$. Without loss of generality we can assume that $I_{k+1}$ is contained in $V_{k, \ell_k}$ for every $k$, i.e. that at each step the interval we're "removing" is in the remaining piece where $x$ lives.
Our assumption is now that $s_k = d_k - c_k \to 0$, and it's sufficient to show that $f_k(d_k) - f_k(c_k) \to 0$ as well. Recursively, we have $$f_k(d_k) - f_k(c_k) = [ f_{k-1}(d_{k-1}) - f_{k-1}(c_{k-1}) ] \frac{d_k - c_k + (b_k - a_k) / 2}{d_{k-1} - c_{k-1}},$$ and tracing this back gives the product expansion $$f_k(d_k) - f_k(c_k) = \prod_{j=1}^{k} \frac{d_j - c_j + (b_j - a_j) / 2}{d_{j-1} - c_{j-1}}.$$ Since $b_j - a_j < (d_{j-1} - c_{j-1}) - (d_j - c_j) = s_{j-1} - s_j$, we can estimate $$f_k(d_k) - f_k(c_k) \leq \prod_{j=1}^{k} \frac{s_{j-1} + s_j}{2 s_{j-1}} = \prod_{j=1}^{k} \left( 1 - \frac{1}{2} \left( 1 - \frac{s_j}{s_{j-1}} \right) \right).$$
Meanwhile, the similar-in-form products $$\prod_{j=1}^{k} \left( 1 - \left( 1 - \frac{s_j}{s_{j-1}} \right) \right)$$ simplify to $s_k$, which we know go to $0$.
Letting $r_j = 1 - s_j / s_{j-1} \in (0,1)$, we can reduce things to the following result for series: if $\{ r_j \}$ has values in $(0,1)$, and $\sum \log(1 - r_j)$ diverges, then $\sum \log(1 - r_j/2)$ diverges as well. This is true, for example, because the divergence of the first series implies the divergence of $\sum r_j$, which implies the divergence of $\sum r_j/2$, which implies the divergence of $\sum \log(1 - r_j/2)$.
After all of this, there's still the question of $f$ being continuous and locally constant only on the $I_n$.
Once we know the limit exists, it's evidently a monotonic function since each $f_j$ is, and this helps with the continuity. We can then look at pointwise continuity by breaking things into cases, for example based on whether there are $I_n$ arbitrarily close to the point to one or both sides (i.e. left and right). In some cases, continuity follows from a very similar argument to the one above establishing the existence of the limit.
Being locally constant only on the $I_n$ can be handled by a similar casewise approach. It helps here to recognize that the value of $f$ on each $I_n$ is distinct. (This can be seen inductively.) You can then argue, for example, that at points which have some $I_n$ between them, $f$ takes on distinct values.