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



Injective and Integrable Mapping from $\mathbb R^3$ to $\mathbb R$
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 $x\_i = \left(\sum\_{j=0}^{i-1}2^{i-j}x\_j\right) - \lfloor x 2^i \rfloor$, 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 $\text{osc}\_f(Q) \le 8^{-i}$.


