One way to argue is this: We have $\nu(\Omega)<1$. (For example a $\sigma$-finite Borel measure would suffice.) If you fix the subdivisions $\beta_s$ of $I_0$, then for $\mathcal{H}^{2n}$-almost every $t \in C^n$ we have $\nu(\partial I)=0$ (I+t))=0$for every$I \in \bigcup_s \beta_s$. Therefore you can move the grid slightly to have the claim satisfied. Notice that I have not yet even read the whole paper up to that point (nor do I intend at this point), so I might be missing something. 1 First of all, it would have been nice if you would have written the backgrounds for your questions, at least the assumptions and the claim which you were wondering. However, as I am currently reading things somewhat related to the paper you are referring to I looked it up. One way to argue is this: We have$\nu(\Omega)<1$. (For example a$\sigma$-finite Borel measure would suffice.) If you fix the subdivisions$\beta_s$of$I_0$, then for$\mathcal{H}^{2n}$-almost every$t \in C^n$we have$\nu(\partial I)=0$for every$I \in \bigcup_s \beta_s\$. Therefore you can move the grid slightly to have the claim satisfied.