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

Comparison of two digit sequences. Let $a\mathrm{\ne }b$ in $D$ and let $k$ be the least index with $a_k\mathrm{\ne }{b}_k$, which exists by [L5]; suppose $a_k=0$ and $b_k=2$. Then $\mathrm{\Phi }(b)-\mathrm{\Phi }(a)={\sum }_{j\ge 0}(b_j-a_j)3^{-j-1}$ by [L2], the terms with $j<k$ vanish, and the tail $R:={\sum }_{j\ge k+1}(b_j-a_j)3^{-j-1}$ satisfies $\mathrm{\mid }R\mathrm{\mid }\le {\sum }_{j\ge k+1}2\cdot 3^{-j-1}=3^{-k-1}$ by [L2], since $\mathrm{\mid }{b}_j-a_j\mathrm{\mid }\le 2$; hence $\mathrm{\Phi }(b)-\mathrm{\Phi }(a)\ge 2\cdot 3^{-k-1}-3^{-k-1}=3^{-k-1}>0$. The same computation with the halved digits gives $\gamma (\mathrm{\Phi }(b))-\gamma (\mathrm{\Phi }(a))=2^{-k-1}+R^{\mathrm{\prime }}$ with $\mathrm{\mid }{R}^{\mathrm{\prime }}\mathrm{\mid }\le {\sum }_{j\ge k+1}{2}^{-j-1}=2^{-k-1}$, so $\gamma (\mathrm{\Phi }(b))\ge \gamma (\mathrm{\Phi }(a))$. Consequently, for $s,t\in C$ with $s\le t$ one has $\gamma (s)\le \gamma (t)$: this is trivial if $s=t$, and otherwise the least index $k$ at which the digit sequences differ must have the digit of $t$ equal to $2$, by the first computation applied both ways.

Values at the endpoints. The constant sequence $\bar{0}$ has $\mathrm{\Phi }(\bar{0})=0$ and $\gamma (0)=0$; the constant sequence $\bar{2}$ has $\mathrm{\Phi }(\bar{2})={\sum }_{k\ge 0}2\cdot 3^{-k-1}=1$ and $\gamma (1)={\sum }_{k\ge 0}{2}^{-k-1}=1$, by [L2]. Both $0$ and $1$ lie in $C$ by [L3].

Claims 1 and 2. For $x\in [0,1]$ the set $A_x:=\{\gamma (t):t\in C,t\le x\}$ is nonempty and bounded above by $1$ by [L1], so $c(x)=\sup A_x$ exists, is unique and lies in $[0,1]$ by [L1] and [L4]; that is claim 1 apart from the extension property. If $0\le x\le y\le 1$ then $A_x\subseteq A_y$, so $c(x)\le c(y)$ by [L4], which is claim 2. And for $t\in C$: $\gamma (t)\in A_t$, while $\gamma (t)$ is an upper bound of $A_t$ by step 1.1, so $\gamma (t)=\sup A_t=c(t)$ by [L4].

The two endpoints of a gap carry the same value of $\gamma$. Let $u<v$ with $u,v\in C$ and $(u,v)\cap C=\mathrm{\varnothing }$, and put $a:={\mathrm{\Phi }}^{-1}(u)$, $b:={\mathrm{\Phi }}^{-1}(v)$, with $k$ the least index where they differ; by step 1.1 and $u<v$ we have $a_k=0$ and $b_k=2$. If some $j>k$ had $a_j=0$, let $a^{\mathrm{\prime }}$ agree with $a$ except that $a_j^{\mathrm{\prime }}=2$; then $\mathrm{\Phi }(a^{\mathrm{\prime }})\in C$, $\mathrm{\Phi }(a^{\mathrm{\prime }})>u$ by step 1.1, and $a^{\mathrm{\prime }}$ still differs from $b$ first at $k$ with $a_k^{\mathrm{\prime }}=0<2=b_k$, so $\mathrm{\Phi }(a^{\mathrm{\prime }})<v$ by step 1.1, putting $\mathrm{\Phi }(a^{\mathrm{\prime }})$ in $(u,v)\cap C$, which is empty. Hence $a_j=2$ for every $j>k$. Symmetrically, if some $j>k$ had $b_j=2$, replacing it by $0$ gives $b^{\mathrm{\prime }}$ with $\mathrm{\Phi }(b^{\mathrm{\prime }})<v$ and $\mathrm{\Phi }(b^{\mathrm{\prime }})>u$, again impossible; hence $b_j=0$ for every $j>k$. Writing $P:={\sum }_{j<k}(a_j{2}^{-1})2^{-j-1}={\sum }_{j<k}(b_j{2}^{-1})2^{-j-1}$, [L2] now gives $\gamma (u)=P+0+{\sum }_{j\ge k+1}{2}^{-j-1}=P+2^{-k-1}$ and $\gamma (v)=P+2^{-k-1}+0=P+2^{-k-1}$, so $\gamma (u)=\gamma (v)$.

Claim 4, first half. Let $u<v$ with $u,v\in C$ and $(u,v)\cap C=\mathrm{\varnothing }$, and let $x\in [u,v]$. Every $t\in C$ with $t\le x$ satisfies $t\le u$ or $t=v$: indeed if $t>u$ then $t\le x\le v$ and $t\mathrm{\notin }(u,v)$ force $t=v$. In the first case $\gamma (t)\le \gamma (u)$ by step 1.1, and in the second $\gamma (t)=\gamma (v)=\gamma (u)$ by step 2.2. So $\gamma (u)$ is an upper bound of $A_x$ and belongs to it, whence $c(x)=\gamma (u)$ by [L4]: $c$ is constant on $[u,v]$, with the value $c(u)$ given by step 2.1.

Claim 3. Let $s\in [0,1]$. Let $T:\mathbb{R}\to \mathbb{R}$ be $T(r):=2r$ for $r<2^{-1}$ and $T(r):=2r-1$ for $r\ge 2^{-1}$, a definition by cases on the total order, and by [L5] let $(r_n)$ satisfy $r_0=s$ and $r_{n+1}=T(r_n)$; put ${\beta }_n:=0$ when $r_n<2^{-1}$ and ${\beta }_n:=1$ otherwise, so $r_{n+1}=2r_n-{\beta }_n$. An induction ([L5]) gives $r_n\in [0,1]$ for every $n$, since $0\le r<2^{-1}$ gives $0\le 2r<1$ and $2^{-1}\le r\le 1$ gives $0\le 2r-1\le 1$ by [L9]; a second induction gives $s={\sum }_{k<n}{\beta }_k{2}^{-k-1}+2^{-n}{r}_n$ for every $n$, the step being ${\sum }_{k<n+1}{\beta }_k{2}^{-k-1}+2^{-n-1}{r}_{n+1}={\sum }_{k<n}{\beta }_k{2}^{-k-1}+{\beta }_n{2}^{-n-1}+2^{-n-1}(2r_n-{\beta }_n)={\sum }_{k<n}{\beta }_k{2}^{-k-1}+2^{-n}{r}_n$. Hence $0\le s-{\sum }_{k<n}{\beta }_k{2}^{-k-1}\le 2^{-n}$, so by [L6] the partial sums converge to $s$ and $s={\sum }_{k\ge 0}{\beta }_k{2}^{-k-1}$. Now $a:=(2{\beta }_k{)}_k$ lies in $D$, the point $x:=\mathrm{\Phi }(a)$ lies in $C$ by [L1], and $\gamma (x)={\sum }_k{\beta }_k{2}^{-k-1}=s$; by step 2.1, $c(x)=\gamma (x)=s$. With step 1.2 and step 2.1 this also gives $c(0)=\gamma (0)=0$ and $c(1)=\gamma (1)=1$.

Claim 4, second half. Let $x\in [0,1]\setminus C$. The set $A:=\{t\in C:t\le x\}$ is nonempty by [L3] and bounded above by $x$, so $u:=\sup A$ exists by [L4]; by [L4] every $N_{\epsilon }(u)$ meets $A\subseteq C$, so $u\in \bar{C}=C$ by [L3], and $u\le x$ with $u\mathrm{\ne }x$, so $u<x$. The set $B:=\{t\in C:t\ge x\}$ is nonempty by [L3], since $1\in C$ and $x\le 1$, and is bounded below by $x$, so $v:=\inf B$ exists by [L4]; likewise $v\in C$ and $v>x$. If $t\in C$ satisfied $u<t<v$, then $t\le x$ would put $t\in A$ and force $t\le u$, while $t\ge x$ would put $t\in B$ and force $t\ge v$, and one of the two holds by totality of the order ([L9]); so $(u,v)\cap C=\mathrm{\varnothing }$. By step 3.1 the function $c$ is constant on $[u,v]$, and $N_{\delta }(x)\subseteq (u,v)$ for $\delta :=\min \{x-u,v-x\}>0$ by [L7], [L8] and [L9].

Claims 1 and 2 are step 2.1, claim 3 is step 3.2, and claim 4 is steps 3.1 and 4.1 together; so all four hold.