Computing $n$-dimensional Lebesgue measure with Euclidean $n$-balls I am studying the coarea formula proof from Evans and Gariepy's Measure Theory and Fine Properties of Functions. At the start of lemma 3.5, the authors are assuming that we can compute the $n$-dimensional Lebesgue measure from coverings of closed balls, namely, that the Lebesgue measure of a measurable set $A\subseteq\mathbb R^n$ equals
$$\text{inf}\left\{\left.\sum_{i=1}^\infty\mathcal L^n(B_i)\,\right|A\subseteq\bigcup_{i=1}^\infty B_i,\,B_i\text{ closed ball in }\mathbb R^n\,\right\}$$
I don't think this assumption is obvious at all. I have searched for a while and I have not encountered yet a justification for this. In the only answer to this question, some corollary of the Vitali covering lemma is applied. However, no-one can guarantee that the disjoint closed balls that approximate the rectangles of any covering of $A$ are going to contain, in their union, the set $A$ we are interested in...
Thanks in advance for your answers.
 A: The formula
$$\mathcal L^n(A)=\text{inf}\left\{\left.\sum_{i=1}^\infty\mathcal L^n(B_i)\,\right|A\subseteq\bigcup_{i=1}^\infty B_i,\,B_i\text{ closed ball in }\mathbb R^n\,\right\}$$
can be proved as follows. Note only the inequality $\ge$ needs to be proved.
From the construction of product measure, we know that $A$ has a cover by boxes $R_i$ of total measure at most $\mathcal L^n(A)+\epsilon \,.$
As noted in Gerald Edgar's answer to a previous question, the Vitali covering lemma yields  that for each  box $R_i$, there exist disjoint balls $B_{i1}, B_{i2}, \dots$ and a Lebesgue-null set $N_i$ so that
$$
\bigcup_{k=1}^\infty B_{ik} \subseteq R_i \subseteq N_i \cup \bigcup_{k=1}^\infty B_{ik} .
$$
A box can be covered by cubes of  total measure at most twice the measure of the box. Therefore the null-set $N:=\cup_i N_i$ can be covered by cubes $Q_j$ of total measure at most $\epsilon$. for each cube $Q_j$, the smallest ball $B_j$ containing it has volume at most $C_n \mathcal L^n(Q_j),$ where $C_n$ just depends on the dimension $n$.
Combining $\{B_j\}_{j \ge 1}$ with the balls $\{B_{ik}\}$ discussed earlier gives a cover of $A$ by balls of total  volume at most  $\mathcal L^n(A)+(1+C_n)\epsilon \,.$
