Is $c( \mathrm{inf} \{ x : c(x) \ge y \}) = y$, for Cantor function $c : [0,1] \to [0,1]$? Let $c : [0,1] \to [0,1]$ the Cantor function, which is constructed by the following procedure. 


*

*Convert the input into base 3.

*If the converted input contains 1, replace every digit after the first 1 by 0.

*Replace all 2s with 1s, and thus the output in base 2.


It is well known that the Cantor function is uniformly continuous, but not absolutely continuous.  But my main question is,

Is $c( \mathrm{inf} \{ x : c(x) \ge y \}) = y$, for Cantor function $c : [0,1] \to [0,1]$?

My attempt
If $y$ is a dyadic rational, it holds directly, because I can explicitly find $x$ corresponds to $\mathrm{inf} \{ x : c(x) \ge y \}$. 
But if $y$ is an irrational number, then things get complicated, since I'm lost in finding one explicitly. Any help?
 A: Let $f:[0,1]\to [0,1]$ be a continuous increasing function. Consider
$$x_*=\inf\{x\in [0,1]| f(x)\ge y\}.$$ If $x_*\not\in \{x\in [0,1]| f(x)\ge y\}$ then it is $f(x_*)<y.$ Now, by definition of infimum, for all $n\in\mathbb{N}$ there exists $x_n\in \{x\in [0,1]| f(x)\ge y\}$ such that $x_n\le x_*+1/n.$ So using that $f$ is increasing we get
$$y\le f(x_n)\le f\left(x_*+\frac 1n\right).$$ Since $f$ is continuous taking limits we get
$$y\le \lim_{n\to \infty}f\left(x_*+\frac 1n\right)=f(x_*).$$ Thus we get a contradiction. So it is 
$$x_*=\inf\{x\in [0,1]| f(x)\ge y\}\in \{x\in [0,1]| f(x)\ge y\},$$ from where,
$$f(x_*)\ge y.$$
Edit
We will show that $f(x_*)=y$ under the additional assumption that $f(0)=0, f(1)=1.$
First of all, if $y=0$ then $x_*=0$ and we are done. 
So we will suppose $0<y<1$ (and thus $x_*>0$). Assume that $f(x_*)>y.$ Since $f$ is continuous there exists $z\in (0,x_*)$ such that $f(z)=y.$ This contradicts the definition of $x_*.$ Thus, we have shown that $f(x_*)=y.$
Finally, consider $y=1.$ Since we have shown that $f(x_*)\ge y=1$ and $0\le f\le 1$ it is $f(x_*)=y=1.$
