Cover of a set of measure zero inside of an open set Let $U$ be a subset of an open set $A\subset \Bbb{R^n}$. Suppose also that $U$ has measure zero. Then for all $\epsilon>0$, there exists a countable collection of rectangles $\{R_i\}$ that covers $U$ and with $\sum_{i=1}^\infty v(R_i)<\epsilon$.
My question: how to prove that we can always choose these rectangles $\{R_i\}$, so that $R_i\subset A,\ \forall i\in \Bbb{N}$?
I've been thinking about it for quite a lot of time yet didn't manage to come up with any sort of decent idea to rigorously prove this. The most obvious reason why this should intuitively be valid is the fact that we can choose a rectangle $S$ containing any point of $A$ such that $S\subset A$, so the rectangles $\{R_i\}$, if made small enough, can be still inside $A$. But as we decrease the size of them, it is possible that they won't cover $U$ anymore.
It may be possible that this is not even true and we need some stronger conditions(however, I stumbled upon this question while trying to prove one even more plausible statement, so it is highly likely to be true). If the statement as it is turns out to be false and needs to be modified, than how exactly and how to prove it?

Case 1:  $A$ is a rectangle.
Just take your original rectangles and intercept them with $A$.
Case 2: General case.
Write $A = \bigcup_{n\in \mathbb N} A_n$, where each $A_n$ is an open rectangle and solve the problem for $U_n=U\cap A_n$ by case
(1), obtaining rectangles $\{R^n_k\}_{k\in \mathbb N}$ inside $A_n$, and  covering $U_n$, with total measure $\varepsilon /2^n$.  Then
$$
  \{R^n_k:n, k\in \mathbb N\}
  $$
is the desired covering.
