If $\phi$ is Cantor function, is $f(x)= \sup \{t \in [0,1] : \phi(t) \leq x\}$ strictly increasing? Let $\phi:[0,1] \to [0,1]$ be the Cantor-Lebesgue function.
Define $f:[0,1] \to [0,1]$ by $f(x)= \sup \{t \in [0,1] : \phi(t) \leq x\}$.
I am interested in analyzing properties of the function $f$. (For example, is it continuous? Measurable? What is its range? Is it strictly monotone?) To avoid asking an overly broad question, however, I will restrict the scope of this SE question to the following:

Is $f$ strictly increasing on $[0,1]$?

I see already that $f$ is an increasing function (in the not necessarily strict sense). Why? Given $0\leq a<b\leq 1$, we have $\{t \in [0,1]: \phi(t) \leq a\} \subseteq \{t \in [0,1]: \phi(t) \leq b\}$ obviously, implying that $sup\{t \in [0,1]: \phi(t) \leq a\} \leq sup\{t \in [0,1]: \phi(t) \leq b\}$. So $f(a) \leq f(b)$.
But I do not see how to show that $f(a) \neq f(b)$. My intuition says that it is probably true and has something to do with finding some $\phi(t)$ sandwiched between $a$ and $b$, but I don't see the rigorous argument.
 A: The result that $f$ is strictly increasing holds for any non-decreasing and continuous function $\phi$ satisfying $\phi(0)=0$ and $\phi(1)=1$, not just for the Cantor-Lebesgue function. 
Here is an example that if $\phi$ is not continuous, then $f$ need not be strictly increasing. Let $\phi(t)=0$ for $t<\frac12\,,$ let $\phi(t)=t$ for $t>\frac12$ 
(and let $\phi(\frac12)$ be any value between $0$ and $\frac12$ inclusive). Then $f(0)=\frac12=f(\frac12)$, and $f$ is constant $f\equiv\frac12$ on the interval $[0,\frac12]$, so $f$ is not strictly increasing. 
Now assume that $\phi$ is any non-decreasing and continuous function on $[0,1]$ 
with $\phi(0)=0$ and $\phi(1)=1$. 
We will show that $\phi(f(x))=x$ for every $x\in[0,1]$. 
We have $\phi(f(x))\le x$ by continuity of $\phi$. 
Indeed, if $\phi(f(x))>x$ then there is some $\varepsilon>0$ such that 
$\phi(f(x)-\varepsilon)>x$. 
(Note that $f(x)>0$ since $\phi(f(x))>x\ge0=\phi(0)$.)
If we let $s=f(x)-\varepsilon$ then 
$\phi(s)>x$, hence $f(x)\le s$ by the definition of $f(x)$, 
obtaining a contradiction $f(x)\le f(x)-\varepsilon$.
Next we show that $\phi(f(x))\ge x$. If $\phi(f(x))<x$ then 
by continuity of $\phi$ there is some $\varepsilon>0$ such that 
$\phi(f(x)+t)<x$ for all $t\in[f(x),f(x)+\varepsilon)$, 
resulting in the contradiction $f(x)\ge f(x)+\varepsilon$ 
(again using the definition of $f(x)$). 
(Note that $f(x)<1$ since $\phi(f(x))<x\le1=\phi(1)$.)
It follows that if $0\le a<b\le1$ then $0\le f(a)<f(b)\le1$. 
Indeed $\phi(f(a))=a<b=\phi(f(b))$, hence $f(a)\neq f(b)$, and it was already proved by the OP that $f(a)\le f(b)$. 
The rest of this answer deals specifically with the 
Cantor-Lebesgue function and treats the remaining questions. 
$f$ is not continuous because of the horisontal line segments on the graph of $\phi$. For example, if $x<\frac12$ then $f(x)<\frac13$, and it is easily seen that 
$\lim\limits_{x\to\frac12^-}f(x)=\frac13\neq f(\frac12)=\frac23$. On the other hand, $f$ is continuous from the right (using that in the definition of $f(x)$ we work with the inequality $\phi(t)\le x$ rather than $\phi(t)<x$). 
The range of $f$ is a subset of the Cantor (middle-third) set $C$. Indeed, if $(p,q)$ is any of the "middle-third" intervals removed in the construction of $C$, then the range of $f$ misses $[p,q)$. We have that $p<q$ and $(p,q)\times\{c\}$ is a subset of the graph $G$ of $\phi$ (and $p$ and $q$ are some fractions with denominators $3^n$, and $c$ is a fraction with denominator $2^n$). Then $\lim\limits_{x\to c^-}f(x)=p\neq q=f(c)=\lim\limits_{x\to c^+}f(x)$. 
It follows that $f$ is not measurable. Indeed take any set $M\subset[0,1]$ that is not measurable. Then $f(M)\subset C$ hence it has measure $0$ (as a subset of the measure $0$ Cantor set $C$), in particular it is measurable. The function $f$ is $1-1$ (being strictly increasing), hence $M=f^{-1}(f(M))$ showing that $f$ is not measurable. 
One could also easily sketch the graph of $f$, at least as easy as one could sketch the graph $G$ of the Cantor-Lebesgue function $\phi$. The function $f$ is "almost" an inverse function of $\phi$. We obtain the graph of $f$ by the following steps. (a) reflect the graph $G$ of $\phi$ about the diagonal $y=x$, (b) remove all but the very top point of each of the vertical intervals that we obtain this way (vertical intervals that result from the reflection about $y=x$ of the horizontal intervals on the graph $G$ of $\phi$). 
Here is a picture (from Wikipedia) of the graph of the 
Cantor-Lebesgue function $\phi$. 

Following is a picture of the graph of the 
Cantor-Lebesgue function reflected about the line $y=x$. 

Finally, a picture of the graph of $f$. 

