Cantor construction is continuous I define a function $f:\mathbb{R}\to\mathbb{R}$ as follows:
$f(x)=0$ for $x\le 0$. $f(x)=1$ for $x\ge1$.
$f(x)=\dfrac12$ for $x\in\left[\dfrac13,\dfrac23\right]$.
$f(x)=\dfrac14$ for $x\in\left[\dfrac19,\dfrac29\right]$, $f(x)=\dfrac34$ for $x\in\left[\dfrac79,\dfrac89\right]$.
and so on.
So this function has been defined on $\mathbb{R}$, except for the Cantor set. How can we fill in the function on the Cantor set, so that we get a continuous function?
 A: Denote the funcion you constructed on $\mathbb{R}\setminus C$ by $f$. The question is how to define Cantor's function on $[0,1]$. You can use the formula
$$
F(x)=\sup_{t\in[0,x]\setminus C} f(t),\qquad x\in[0,1]
$$
The function $F$ is continuous thanks to the following lemma.

Lemma. Let $F:[a,b]\to\mathbb{R}$ is non decreasing and $F([a,b])$ is dense in $[F(a),F(b)]$, then $F$ is continuous.

Proof. Since $F$ is non decreasing there exist finite one sided limits at any $x\in[a,b]$, denote them $F(x+)$ and $F(x-)$. Assume $F$ is discontinuous at $c\in[a,b]$, then $f(c-)<f(c+)$. In this case $F([a,b])\cap[F(c-),F(c+)]=\{F(c)\}$, so $F([a,b])$ is not dense in $[F(a),F(b)]$. Contradiction.
A: Define the cantor function as defined in wikipedia http://en.wikipedia.org/wiki/Cantor_function , see section "iterative construction". Then prove that 
$$|f_{n+1}(x)-f_{n}(x)|\leq \frac {1}{2^n}$$
and use this to prove that $\sup_{x\in[0,1] }|f_m(x)-f_n(x)|\to 0$ as $m,n\to \infty$. Therefore $f_n$ converges uniformly, and the limit has to be continuous, since the functions $f_n$ are.
