Is there an injective and Riemann integrable map $f:\mathbb R^3\rightarrow\mathbb R$? (Of course such a map cannot be continuous.)

Consider for example $f \colon [0,1]^3 \to \mathbb{R}$ defined as $$f(x,y,z) = \sum_{i=0}^\infty 8^{i}(x_i+2y_i+4z_i),$$ where $x_i = \left(\sum_{j=0}^{i1}2^{ij}x_j\right)  \lfloor x 2^i \rfloor$, and $y_i, z_i$ are defined the same way. In other words, $(x_i)_i$ is a specific dyadic representation of $x$. Continuity: The mapping $f$ is continuous outside the union of the boundaries of dyadic cubes $$Q_{i,k_1,k_2,k_3} = 2^{i}\left([k_1,k_1+1]\times [k_2,k_2+1] \times [k_3,k_3+1]\right),$$ for $i,k_1,k_2,k_3 \in \mathbb{Z}$. To see this, take $x \in [0,1]^3 \setminus \bigcup \partial Q_{i,k_1,k_2,k_3}$. Take $i \in \mathbb{N}$ and notice that for some $k_1,k_2,k_3 \in \mathbb{Z}$ (abbreviating $Q := Q_{i,k_1,k_2,k_3}$) we have $x \in \text{int}(Q)$. On the other hand $\text{osc}_f(Q) \le 8^{i}$. 


beginning Start with the unit cube $E$ in $\mathbb R^3$ and the unit interval $[0,1]$ in $\mathbb R$. Choose an injective map $\phi : E \to [0,1]$. This is the remaining question: do this so that $\phi$ is Riemann integrable. That is, the set of discontinuities has measure zero. Then cover $\mathbb R^3$ by disjoint unit cubes, make them disjoint by including boundary points in only one of the possible cubes. Call these $F_n, n \in \mathbb N$. Choose disjoint intervals $I_n$ going to zero fast enough. Define $f$ using $\phi$ with translation and dilation to map $F_n$ injectively into $I_n$. If the original $\phi$ has set of discontinuities of measure zero, then this piecedtogether function $f$ does too, since its set of discontinuities can be at most the discontinuities of a copies of $\phi$ in the interiors of the $F_n$, together with the boundary planes of the $F_n$. So $f$ is (improperly) Riemann integrable. 

