Defining Measure 0 Sets with Open or Closed Rectangles

Question

I have been reading through Calculus on Manifolds by Michael Spivak, and I am not understanding what he states after defining measure 0 sets. On page 50, he states

"A subset $A$ of $\mathbb{R}^n$ has (n-dimensional) measure 0 if for every $\varepsilon > 0$ there is a cover $\{U_1, U_2, U_3, \dotsc\}$ of $A$ by closed rectangles such that $\sum_{i = 1}^\infty v(U_i) < \varepsilon$."

And follows this by stating

"The reader may verify that open rectangles may be used instead of closed rectangles in the definition of measure 0."

I have been trying to prove the equivalence between the definition of measure zero with open rectangles and closed rectangles. I have been able to prove that if we can do this with open rectangles, then we can do it with closed rectangles. However, I have not been able to prove the other direction. I have written the question as the following statement:

Suppose that for $A \subset \mathbb{R}^n$ and $\varepsilon > 0$, there is a cover $\{F_1, F_2, F_3, \dotsc\}$ of $A$ by closed rectangles such that $\sum_{i = 1}^\infty v(F_i) < \varepsilon$. Then, for any $\varepsilon' > 0$, there is a cover $\{U_1, U_2, U_3, \dotsc\}$ of $A$ by open rectangles such that $\sum_{i = 1}^\infty v(U_i) < \varepsilon'$.

Spivak Page 50

I included the Lebesgue measure tag, since I understand Spivak's definition of volume of a rectangle to be connected, but if this is incorrect, feel free to remove that tag.

Answer 1

Suppose that for $A\subset\mathbb{R}^n$ and $\varepsilon>0$, there is a cover $\{F_1, F_2, F_3, \dots\}$ of $A$ by closed rectangles such that $\sum_{i = 1}^\infty v(F_i) < \varepsilon$. Take $\varepsilon' > 0$, and consider a cover $\{F_1, F_2, F_3, \dots\}$ of $A$ by closed rectangles such that $\sum_{i = 1}^\infty v(F_i) < \frac{\varepsilon'}2$. Replace each $F_i$ by a larger open rectangle $U_i$ such that $v(U_i)<v(F_i)+\frac{\varepsilon'}{2^{i+1}}$. Then $\{U_1,U_2,U_3,\dots\}$ is a cover of $A$, and $\sum_{i=1}^\infty v(F_i) <\frac{\varepsilon'}2+\frac{\varepsilon'}2=\varepsilon'$.

Answer 2

If $F_k=[a_{k1},b_{k1}]\times [a_{k2},b_{k2}] \times ... \times [a_{kn},b_{kn}]$ take $U_k=(a_{k1}-\frac {\epsilon} {2^{k}},b_{k1}+\frac {\epsilon} {2^{k}})\times ... \times (a_{kn}-\frac {\epsilon} {2^{k}},b_{kn}+\frac {\epsilon} {2^{k}})$.