The homeomorphism $x\mapsto x+c(x)$ sends the Cantor set onto a compact set of Lebesgue measure $1$

The complement $(0,1)\setminus C$ is open, because $C$ is closed by the definition of the Cantor set. By [L4] write it as a countable disjoint union ${⨆}_{k\in \mathbb{N}}{I}_k$ of nonempty open intervals. For each $k$ there are endpoints $u_k<v_k$ in $C$ with $I_k=(u_k,v_k)$ and $(u_k,v_k)\cap C=\mathrm{\varnothing }$.

Fix $k$. By [L3] the Cantor function is constant on $[u_k,v_k]$; write that value as ${\alpha }_k$. Then $\psi (x)=x+{\alpha }_k$ for every $x\in I_k$, so $\psi [I_k]=(u_k+{\alpha }_k,v_k+{\alpha }_k)$ is an open interval and $\lambda (\psi [I_k])=\lambda (I_k)=v_k-u_k$ by [L5].

The intervals $\psi [I_k]$ are pairwise disjoint because $\psi$ is injective by [L1]. They cover $\psi [(0,1)\setminus C]$, so countable additivity on the disjoint family and step 2.1 give $\lambda (\psi [(0,1)\setminus C])={\sum }_{k=0}^{\mathrm{\infty }}\lambda (\psi [I_k])={\sum }_{k=0}^{\mathrm{\infty }}\lambda (I_k)=\lambda ((0,1)\setminus C)=1$, because [L2] says $\lambda (C)=0$ and [L5] says $\lambda ([0,1])=1$.

The set $K=\psi [C]$ is compact: $C$ is closed and bounded, hence compact by [L8], and the continuous map $\psi$ carries compact sets to compact sets by [L1]. Also $[0,2]=K\bigsqcup \psi [(0,1)\setminus C]$, with disjointness from injectivity of $\psi$, so [L5] and step 3.1 give $2=\lambda ([0,2])=\lambda (K)+1$. Therefore $\lambda (K)=1$.