Homeomorphism between cantor sets one with measure 0 and one with positive measure I defined a function $f$ as follows:
Let $I_{n,k}=\left(x_1,x_2\right)$ the removed open interval in step $n$ at position $k$ of the $\frac{1}{3}$-cantor set and $J_{n,k}=\left(y_1,y_2\right)$ the removed open interval in step $n$ at position $k$ of a arbitrarily fat cantor set. Map the closure of $I_{n,k}$ onto the closure of $J_{n,k}$ like:
$$f:clo(I_{n,k}) \rightarrow clo(J_{n,k}): x\mapsto \begin{cases}f(x):=y_1&x=x_1\\f(x):=y_2&x=x_2\\f(x):=\frac{y_2-y_1}{x_2-x_1}\cdot x & x\in\left(x_1,x_2\right)\end{cases}$$
Due to the construction $f$ is bijective and continuous at $\left(x_1,x_2\right)$ for every $I_{n,k}$ but I struggle with the continouity at the endpoint of the intervals $I_{n,k}$. The fact that I want to map onto the fat cantor set it's not possible to argue with $\left(\frac{1}{2}\right)^n$. What am I missing?
For example:
$clo\left(I_{1,1}\right)=clo\left(\left(\frac{1}{3},\frac{2}{3}\right)\right)=\left[\frac{1}{3},\frac{2}{3}\right]$ maps to $clo\left(J_{1,1}\right)=\left[y_1,y_2\right]$ with the definition of $f$, repeating for every $n \in \mathbb{N}$ with the inherent intervals.
That's what I am thinking:

So $f\left( \frac{1}{3}\right)=y_1$, $f\left(\frac{2}{3}\right)=y_2$ and for all $x \in I_{1,1}$ is $f\left(x\right)=\frac{y_2-y_1}{\frac{2}{3}-\frac{1}{3}}=3(y_2-y_1) \cdot x$.
 A: Your idea is   good. We already know that a continuous strictly increasing function is a homeomorphism. If we can find such a function   $f:[0,1]\to [0,1]$ which carries the complement of one set onto the complement of another, this function will also carry one Cantor set onto the other. 
But the task of piecing an infinite collection of linear functions together, checking the continuity of the whole thing, is rather tedious. I suggest building $f$ as a uniform limit of continuous maps $f_n$,   following the process of creation of Cantor sets themselves. That is, let $C_n$ and $F_n$ be pre-Cantor sets of $n$th generation; with $C_0=F_0=[0,1]$. Let $f_0(x)=x$.  Having defined $f_n$,   proceed to  $f_{n+1}=f_n$ on $[0,1]\setminus C_n$ and $f_{n+1}$ being an appropriate three-chain broken line on each component of $C_n$. 
Alternatively, define $f_n$ as the unique increasing piecewise linear function on $[0,1]$ that maps the boundary points of $C_n$ bijectively onto the boundary points of $F_n$, and only changes the slope at these points.    
This is essentially what you do already with linear pieces, with the difference that each $f_n$ is defined on all $[0,1]$. 
By construction,   $\sup |f_{n+1}-f_{n}| \le \max_k |J_{n,k}|\le 2^{-n}$. (I use $|\cdot|$ to denote length, and assume $J_{n,k}|$ is the same for all $k$, as on the picture.) Hence $\sup |f_{n+1}-f_{n }|\le  2^{-n}$, which implies that the sequence $f_n$ converges uniformly, by the Weierstrass $M$-test applied to $\sum (f_{n+1}-f_n)$. Let $f$ be its limit. Since 
$f_n(C_m)\subset F_m$ for all $n\ge m$, we have $f(C_m)\subset F_m$. Hence $f(C)\subset F$. 
Everything in the preceding paragraph  applies to the inverse maps $f_{n}^{-1}$: namely,  $\sup |f_{n+1}^{-1}-f_{n}^{-1}| \le \max_k |I_{n,k}|$, and so on.  Let $g= \lim_{n\to\infty} f_n^{-1}$. Then both 
$g\circ f = \lim_n g_n\circ f_n$ and $f\circ g = \lim_n f_n\circ g_n$ are identity maps. So, $g$ is the inverse of $f$. 
