I am reading "Lectures on the Lebesgue Integral" (in Japanese) by Hitoshi Arai.
The author used the following proposition without a proof in this book.
Let $I:=[0, 1)\times\dots\times [0,1)\subset\mathbb{R}^d.$
Let $A:=I\cap\mathbb{Q}^d.$
Let $a_{ij},b_{ij}$ be real numbers such that $a_{ij}<b_{ij}$ for any $i\in\{1,\dots, d\}$ and for any $j\in\{1,\dots,n\}.$
Let $I_j:=[a_{1j},b_{1j})\times\dots\times[a_{dj},b_{dj})$ for $j\in\{1,\dots, n\}.$
If $A\subset\bigcup_{j=1}^n I_j,$ then $I\subset\bigcup_{j=1}^n I_j.$
At first I thought it was very easy to prove this proposition and I thought I could prove it visually.
But I didn't have a visual proof.
I want to know a standard proof of this proposition.
My proof is the following:
Suppose there exists $x=(x_1,\dots,x_d)\in I$ such that $x\notin\bigcup_{j=1}^n I_j.$
Let $\epsilon':=\min\{1-x_1,\dots,1-x_d\}.$
For each $j\in\{1,\dots, n\}$ and for each $i\in\{1,\dots,d\}$, if $x_i<a_{ij}$, then let $\epsilon_{ij}:=\min\{a_{ij}-x_i,\epsilon'\}$ and if $a_{ij}\leq x_i$, then let $\epsilon_{ij}:=\epsilon'.$
For each $i\in\{1,\dots,d\}$, let $q_i$ be a rational number such that $x_i\leq q_i<x_i+\min\{\epsilon_{i1},\dots,\epsilon_{in}\}.$
Then, $(q_1,\dots,q_d)\in A.$
Since $x\notin\bigcup_{j=1}^n I_j,$ there exists $i_j\in\{1,\dots, d\}$ such that $x_{i_j}\notin [a_{i_jj},b_{i_jj})$ for each $j\in\{1,\dots,n\}.$
If $x_{i_j}<a_{i_jj}$, then $x_{i_j}\leq q_{i_j}<x_{i_j}+\min\{\epsilon_{i_j1},\dots,\epsilon_{i_jn}\}\leq x_{i_j}+\epsilon_{i_jj}\leq x_{i_j}+(a_{i_jj}-x_{i_j})=a_{i_jj}.$
So, $q_{i_j}\notin [a_{i_jj},b_{i_jj}).$
If $b_{i_jj}\leq x_{i_j}$, then $b_{i_jj}\leq x_{i_j}\leq q_{i_j}.$
So, $q_{i_j}\notin [a_{i_jj},b_{i_jj}).$
Therefore, $(q_1,\dots, q_d)\notin\bigcup_{j=1}^n I_j.$
But $A\subset\bigcup_{j=1}^n I_j.$
So, this is a contradiction.