Bounding an "integral" from below by the Hausdorff measure of the domain

Question

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 $E$.

Notation:

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: $\mathcal{H}^s(G) = \lim_{\delta \to 0} \mathcal{H}^s_{\delta}(G).$

Remarks:

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 $\mathbb{R}^k$.

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.