Let $(X,d)$ be an arbitrary metric space and $E \subset X$ also arbitrary. Fix $s \in (0,\infty)$.

Is it true that for any $ \delta > 0 $ and any collection of pairs $\{(A_i,a_i)\}_{i \in \mathbb{N}}$ where $A_i$ are subsets of $X$ and $a_i \in [0,\infty]$, if $$ \text{diam} \, A_i \leq \delta \quad for \ \ \ all \quad i\in \mathbb{N}, $$ and $$ \chi_E(x) \leq \sum_{i} a_i \, \chi_{A_i}(x) \quad for \ \ \ all \quad x \in X \ , $$ then $$ \mathcal{H}^s_{5\delta}(E) \leq C \sum_i a_i \, (\text{diam} \, A_i)^s \ , $$ with a $C$ that does not depend on $\delta$?

The infimum of all such $ \sum_i a_i \, (\text{diam} \, A_i)^s $ can be viewed as an "integral" of the characteristic function of $E$, or lternatively as the "weighted" Hausdorff measure of $E$.


1) $\chi_G$ stands for the characteristic function,

2) $\mathcal{H}^s_{5\delta}$ stands for the approximating Hausdorff measure. The ones that appear in the definition of the Hausdorff measure: $\mathcal{H}^s(G) = \lim_{\delta \to 0} \mathcal{H}^s_{\delta}(G).$


3) Of course the interesting case is when $ 0 < a_i \leq 1$ because if $a_i > 1$ then the pair $(A_i,a_i)$ can be replaced by $(A_i,1)$.

4) I am interested only in the asymptotic $\delta \to 0$, so feel free to assume $\delta$ is small.

5) I don't care if $5 \, \delta$ becomes $563 \, \delta$. As the joke goes, show it for some $5$ in the natural numbers!

Motivation: A proof of this will be a significant step in my attempt to give a proof of the coarea inequality, also known as the Eilenberg's inequality: Fix arbitrary metric spaces $X$ and $Y$, and pair of non-negative integers $\mu$ and $q$. Then for any lipschitz map $ g: X \to Y$ and any subset $E \subset X$, $$ \int^*_Y \mathcal{H}^\mu (g^{-1}(y) \cap E) d\mathcal{H}^q(y) \leq (\text{Lip g)}^q \ \frac{\omega _\mu \omega_q}{\omega_{\mu+q}}\mathcal{H}^{\mu + q}(E) \, . $$ Here $\omega_k$ is the volume of unit ball in $\mathbb{R}^k$.

  • $\begingroup$ Is the coarea inequality in your "motivation" not just Eilenberg's inequality? en.wikipedia.org/wiki/Eilenberg%27s_inequality Or am I missing a subtlety? $\endgroup$
    – user142382
    Mar 19, 2020 at 14:21
  • $\begingroup$ It is. Another name for it. Eilenberg had proved it for a special case. As far as I know Federer gave a proof of its general form, albeit, for boundedly compact metric spaces. The restriction was later removed thanks to a result of Davies. $\endgroup$ Mar 19, 2020 at 15:11

1 Answer 1


It is true and deserves to be known better (it is morally equivalent to the statement that the Frostman lemma is just an exercise in linear programming duality though there are some pesky details I don't want to go into now).

I'll assume that all $A_j$ are balls $B_j$. It may change 5 to 10 but I don't think you care.

Consider first the case when the cover is finite. Then it is possible to choose the non-negative coefficients $a_j$ so that $\sum_ja_j\chi(B_j)\ge \chi_E$ and $\sum_j a_jd(B_j)^s$ is the minimal possible under this restriction. Now, in this optimal choice, take any ball with $a_j>0$ and consider all other balls $B_i$ that intersect it.

Claim $a_jd(B_j)^s+\sum_i a_id(B_i)^s\ge\frac 12d(B_j)^s$

Proof If $a_j\ge \frac 12$, there is nothing to do. Otherwise we can replace $a_j$ by $a_j-h$ and all $a_i$ by $a_i(1+2h)$ and still have an admissible choice of coefficients. Since our choice was optimal, we have the inequality $$ -hd(B_j)^s+2h\sum_i a_id(B_i)^s\ge 0 $$ and we are done again.

Now just run the standard Vitali construction throwing out the biggest participating ball $B_j$ with all balls that intersect it and $3B_j$ from $E$ and reoptimizing the remaining picture after each throw (the last step is needed because otherwise the inequality in the claim may involve some balls that are already gone).

The reduction of the infinite case to the finite one is standard. Choose a finite part of the sum $\sum_j a_jd(B_j)^s$ that is at least $0.9$ of the whole sum and consider the set on which the corresponding linear combination of the characteristic functions is at least $1/3$. Then through these pieces away and triple the remaining coefficients. Repeat. The point is that if $\sum_ku_k\ge 1$ then there exists $u_k$ that is at least $3^{-k}$, so no point of $E$ will escape and $\mathcal H^s_{10\delta}$ is countably subadditive.

  • $\begingroup$ Thank a lot for the answer. i need time to writ out the details. I will get back to you. $\endgroup$ Mar 19, 2020 at 18:18
  • $\begingroup$ @BehnamEsmayli Sure. Feel free to ask questions :-) $\endgroup$
    – fedja
    Mar 19, 2020 at 20:27
  • $\begingroup$ This answer is awesome! So many beautiful arguments here and there. Thanks! Questions though: 1) When you say it is "standard" in the last paragraph, can you please reference some literature where it is done? Because the argument in Federe's proof of Coarea Inequality does not use this and that is the main difficulty of generalizing the formula. The solution came later when Davies proved that for increasing sequence of subsets, $\mathcal{H}^s_\delta(E_n) \to \mathcal{H}^s_\delta(\cup E_n)$, which is not an easy result (surprisingly?!) $\endgroup$ Mar 20, 2020 at 14:29
  • $\begingroup$ 2) In the last paragraph again, do I get a sum like $\sum_1^N a_id(B_i)^s + 3 \sum_{N+1}^M a_id(B_i)^s + 9 \sum_{M+1}^P a_id(B_i)^s+\cdots$ that is nevertheless comparable to $\sum_1^\infty a_id(B_i)^s$, and the former is at least $H^s_{10\delta}(E_1) + H^s_{10\delta}(E_2)+\cdots \geq H^s_{10\delta}(E)?$ Thanks in advance for continued attention. $\endgroup$ Mar 20, 2020 at 14:29
  • 1
    $\begingroup$ @BehnamEsmayli No, I haven't read Federer that far though I've gone through a noticeable portion of it when preparing to teach my measure theory class. The proof I posted was motivated by my attempt to understand the logic behind the Frostman lemma, as I said. $\endgroup$
    – fedja
    Mar 21, 2020 at 2:59

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