A set $X\subseteq\mathbb{R}^n$ is said to have $n$-dimensional volume zero iff for any $\epsilon>0$, there exist finitely many closed $n$-boxes $R_1,\cdots,R_s$ such that $X\subseteq \bigcup_i R_i$ and $\sum_{i} \text{vol}(R_i)<\epsilon$.
Given a set $X$ with volume zero, I would like to show that for any $\epsilon>0$, there exists such a cover where each point in $X$ is an interior point of the cover and its volume is less than $\epsilon$. In other words, is it possible to cover $X$ with open boxes instead (with total volume less than $\epsilon$)?
I encountered this problem in the proof of Proposition 7.1.8 in Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds by T. Shifrin. The proposition asserts that if the set $X$ of discontinuities of a bounded function $f:R\to\mathbb{R}$ has volume zero, then $f$ is integrable on $R$. In the proof, $X$ is covered in the manner mentioned in the previous paragraph so that the closure of the complement of the covered portion does not contain any discontinuity point. However, the author did not provide a proof for why such a cover exists. This is intuitive to me, but I struggle to produce a rigorous proof.
Best Answer
Hint: A closed box can be covered by an open box with slightly larger volume.
Hint 2: Trying to prove your statement about open boxes for some value of $\epsilon$, you want to apply the other statement that has closed boxes. But you don't need to apply it with the same $\epsilon$. You could apply it with, say, $\epsilon/2$ or $2\epsilon$ or $\epsilon^2$. Which of these would be useful for proving your statement?