The Cantor function on $[0,1]$, defined on the Cantor set through ternary digits and extended constantly across each removed interval

Let $C$ be the Cantor set, $D$ the set of sequences with values in $\{0,2\}$ and $\Phi : D \to C$ the bijection $\Phi(a) = \sum_{k \ge 0} a_k 3^{-k-1}$ of The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$, and this gives a bijection with $\{0,1\}^{\mathbb{N}}$. Since $\Phi$ is a bijection it has a two-sided inverse $\Phi^{-1} : C \to D$, and that inverse is a single function, determined and not selected (Injection, surjection, bijection).

On the Cantor set. For $x \in C$ write $a := \Phi^{-1}(x)$ and put

$\gamma(x) \;:=\; \sum_{k=0}^{\infty} \big(a_k \cdot 2^{-1}\big)\, 2^{-k-1} .$

Each coefficient $a_k \cdot 2^{-1}$ is $0$ or $1$, so all the terms are nonnegative and every partial sum is at most $\sum_{k<n} 2^{-k-1} \le \sum_{k=0}^{\infty} 2^{-k-1} = 1$ (For $|r| < 1$, $\sum_{k \ge 0} r^k = 1/(1-r)$, and for $|r| \ge 1$ the series diverges, Integer powers $a^m$, Laws of integer exponents); hence the series converges and $\gamma(x) \in [0,1]$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). In words: $\gamma$ halves each ternary digit of $x$ and reads the result as a binary expansion.

On all of $[0,1]$. The Cantor function is $c : [0,1] \to \mathbb{R}$,

$c(x) \;:=\; \sup\{\, \gamma(t) : t \in C \text{ and } t \le x \,\} .$

The supremum exists and is a single real number. The set on the right is nonempty, because $0 \in C$ (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds) and $0 \le x$, and it is bounded above by $1$, because $\gamma$ takes values in $[0,1]$; so it has a least upper bound by completeness (Complete ordered field (least-upper-bound property), Lower bound, bounded below, bounded set), and that bound is unique (Suprema and infima are unique). Since $0 \le \gamma(0) \le c(x) \le 1$, the values of $c$ lie in $[0,1]$.

That $c$ really extends $\gamma$, that is, $c(t) = \gamma(t)$ for every $t \in C$, is not an observation but a small theorem: it needs $\gamma$ to be nondecreasing along $C$. It is claim 1 of The Cantor function is well defined, satisfies $c(x) \le c(y)$ whenever $x \le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set ↗, recorded in this item's justified_by, and until it is proved the two symbols are kept apart.

• Why the extension is a supremum and not a case distinction. Writing "$c$ is constant across each interval removed in the construction of $C$" presupposes a description of those intervals; the supremum formula presupposes nothing, is defined at every point of $[0,1]$ at once, and yields the constancy as a theorem (claim 4 of The Cantor function is well defined, satisfies $c(x) \le c(y)$ whenever $x \le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set ↗). It also makes the monotonicity of $c$ immediate, since the set whose supremum is taken grows with $x$.

• Nothing is claimed here about continuity. No definition of continuity for a real function of a real variable is available at this point in the reading order, so no statement about it is made, in either direction; the properties proved on this page are well-definedness, monotonicity in the sense $c(x) \le c(y)$ for $x \le y$, surjectivity onto $[0,1]$ and constancy across the gaps of $C$.

• The name. The function is also called the devil's staircase, because it climbs from $0$ to $1$ while being constant across every gap of $C$, and the gaps fill up all of $[0,1]$ except a set of measure zero (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points).

• The digits are halved, not truncated. $\gamma$ sends the ternary digit sequence $(a_k)$ with values in $\{0,2\}$ to the binary sequence $(a_k/2)$ with values in $\{0,1\}$, which is the bijection $D \to \{0,1\}^{\mathbb{N}}$ of claim 3 of The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$, and this gives a bijection with $\{0,1\}^{\mathbb{N}}$ read backwards. So $\gamma$ is the composition of $\Phi^{-1}$ with that bijection and with the binary summation, and its surjectivity onto $[0,1]$ is exactly the statement that every real of $[0,1]$ has a binary expansion, proved where it is used.