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

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:

  1. 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 $(A_i,1)$.

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

  3. 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: 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 $\mathbb{R}^k$.