Some doubts regarding determinant lower bound of linear discrepancy I am reading the book on Discrepancy Theory by Matousek. A famous theorem of Lovasz et. al. implies the following lower bound on linear discrepancy of a set system. (I am attaching a screenshot of the theorem from the aforementioned book)

As shown in the picture, if $[-1,1]^n \subseteq \cup_{a\in\{-1,-1\}^n}(U+a)$, then $\mathbb{R}^n \subseteq \cup_{a\in 2\mathbb{Z}^n}(U+a)$ because $\mathbb{R}^n = \cup_{a\in 2\mathbb{Z}^n}([-1,1]^n+a)$. But I fail to understand how this implies the volume of $U$ be at least the volume of the cube $[-1,1]^n$. This may be very obvious from the preceeding arguments but I do not understand this. Can someone help?
Thanks
 A: Intuition : If the volume of $U$ is strictly less than $2^n$, there is no way that $\bigcup_{a\in2\mathbb Z^n}(U+a)$ covers the whole space, indeed it will cover strictly less than $\bigcup_{a\in 2\mathbb Z^n} ([-1,1]^n+a)$ which essentially partition $\mathbb R^n$ (the intersection of any two disjoint such set is volume $0$).
Let $A\triangleq\{ a\in 2\mathbb Z^n : [-1,1]^n\cap (U+a)\neq \emptyset \}=\{ a\in 2\mathbb Z^n : ([-1,1]^n-a)\cap U\neq \emptyset \}$. Now
\begin{align*}
\mathrm{vol}(U)&=\mathrm{vol}\left(\bigcup_{a\in A} ([-1,1]^n-a)\cap U\right)\\
&=\sum_{a\in A}\mathrm{vol}\left( ([-1,1]^n-a)\cap U \right)\\
&=\sum_{a\in A}\mathrm{vol}\left( [-1,1]^n\cap (U+a) \right)\\
&\geq \mathrm{vol}\left( \bigcup_{a\in A}[-1,1]^n\cap (U+a) \right)\\
&= \mathrm{vol}\left( [-1,1]^n \right)
\end{align*}
The equality from the first line to the second is due to the fact that $[-1,1]^n+a$ are essentially disjoint. The last equality is due to the fact that $\bigcap_{a\in A}(U+a)$ covers $[-1,1]^n$, this is true because $\bigcap_{a\in2\mathbb Z^n}(U+a)$ covers it (since it covers $\mathbb R^n$) and for any $a\notin A$, $[-1,1]^n\cap(U+a)=\emptyset$.
