Let $\underline{\alpha}=(\alpha_1,\dots,\alpha_d) \in \mathbb{R}^d$. Show that the transformation $R_{\underline{\alpha}}=\mathbb{T}^d \rightarrow \mathbb{T}^d$ defined by
$R_{\underline{\alpha}}=(x_1 + \alpha_1 \mod \ 1, \dots, x_d + \alpha_d \mod \ 1 ) $
preserves the $d$-dimensional Lebesgue measure $\lambda$ on Borel sets $\mathcal{B}$.
Not really sure how to approach this.
I don't know how Lebesgue measure was defined on ${\mathbb T}^d$ in your case. I shall adopt a pedestrian version in the following. It will be sufficient to consider the case $d=1$. Any subset $X\subset{\mathbb T}$ has a representant $\hat X$ in the fundamental domain $[0,1[\ $, and one puts $$\lambda(X):=\mu(\hat X)\ ,$$ where $\mu$ is the usual Lebesgue measure on ${\mathbb R}$.
For given $\sigma\in{\mathbb R}\ $ denote by $T_\sigma$ the shift $x\mapsto x+\sigma$ $\>(x\in{\mathbb R})$. Note that $\mu\bigl(T_\sigma(A)\bigr)=\mu(A)$ for any measurable set $A$.
Assume that an $\alpha\in[0,1[\ $ is given. Then the sets $$X':=\hat X\cap[0,1-\alpha[\ ,\quad X'':=\hat X\cap[1-\alpha,1[\ $$ form a partition of $\hat X$. Let $Y:=R_\alpha(X)$. Then $$\hat Y=T_\alpha(X')\cup T_{\alpha-1}(X'')\ .$$ The two sets on the right hand side are easily seen to be disjoint. It follows that $$\eqalign{\mu(\hat Y)&=\mu\bigl(T_\alpha(X')\bigr)+\mu\bigl(T_{\alpha-1}(X'')\bigr)\cr&=\mu(X')+\mu(X'')=\mu(X'\cup X'')\cr &=\mu(\hat X)\ .\cr}$$