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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.

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  • $\begingroup$ Does the reader at this point know that Lebesgue measure and Haudsorff measure coincide? $\endgroup$
    – Behnam Esmayli
    Apr 6, 2021 at 17:40
  • $\begingroup$ Yes, and also Vitali covering theorem is available... $\endgroup$
    – Akerbeltz
    Apr 6, 2021 at 17:48
  • $\begingroup$ A priori I wanted an elementary proof of the fact, but propbably the Hausdorff measure approach might be smarter... $\endgroup$
    – Akerbeltz
    Apr 6, 2021 at 17:49

1 Answer 1

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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 $\mathcal L^n(A)+(1+C_n)\epsilon \,.$

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  • $\begingroup$ Sorry, $d$ was a typo, I usually use $d$ for dimension rather than $n$. I have now corrected this, let me know if it is clear. $\endgroup$
    – Yuval Peres
    Apr 6, 2021 at 23:25
  • $\begingroup$ Allright, your answer seems to be completely satisfactory. Thanks. $\endgroup$
    – Akerbeltz
    Apr 6, 2021 at 23:25
  • $\begingroup$ I think one explaination is required. You say that only the $\geq$ part needs to be proved. However, the covering by closed balls has a total volume of $\mathcal L^n(A)+(1+C_n)\varepsilon>\mathcal L^n(A)+\varepsilon=\sum_{i=1}^\infty\mathcal L^n(R_i)$. It seems that you are proving quite the opposite inequality... $\endgroup$
    – Akerbeltz
    Apr 6, 2021 at 23:51
  • $\begingroup$ Well, I guess what you really are doing is to construct a cover by closed balls whose total measure is arbitrarily close to the measure of the original set, and that suffices to prove the equality $\endgroup$
    – Akerbeltz
    Apr 7, 2021 at 18:36

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