I am trying to understand the following fact:



> 
> Suppose $\{B\_i\}\_i$ are disjoint balls in $\mathbb R^n$, and $A\_i \subset 100 B\_i$ is a subset with $|A\_i| \geq c |B\_i|$. Then for any nonnegative $f$, we have $\sum\_i |B\_i| \inf\_{A\_i} f \lesssim \int\_{\cup\_i A\_i} f$, where the implied constant depends only on $c$ and the dimension $n$.
> 
> 
> 


(Here, $|\cdot|$ denotes Lebesgue measure, and $100B$ denotes the ball with the same center as $B$ and $100$ times the radius.)


Is there a way to prove this with (some combination of) covering lemmas, maximal function estimates, or weighted inequalities? I couldn't see an easy way to prove this.




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Some background (which isn't needed for my question): The statement above is taken from Chapter 13 of David and Semmes's *Singular integrals and rectifiable sets in $\mathbb R^n$*. (It appears in the middle of a proof. They do not state this as a separate lemma.)


Here is a sketch of the proof in the book:


Let $p \in (1, \infty)$ and for each $i$, let $w\_i$ be a function on $A\_i$ (all TBD). By Holder,


\begin{align\*}
\inf\_{A\_i} f 
\leq
\left(\frac{1}{|A\_i|}\int\_{A\_i} f^{1/p} \right)^{p}
\leq
\left(\frac{1}{|A\_i|}\int\_{A\_i} f w\_i \right) 
\left(\frac{1}{|A\_i|}\int\_{A\_i} w\_i^{-p'/p} \right)^{p/p'}
\end{align\*}


so


\begin{align}
\sum\_i |B\_i| \inf\_{A\_i} f 
&\lesssim
\sum\_i
\left(\int\_{A\_i} f w\_i \right) 
\left(\frac{1}{|A\_i|}\int\_{A\_i} w\_i^{-p'/p} \right)^{p/p'}
\\
&\leq
\left(\int f \textstyle\sum\_i 1\_{A\_i} w\_i \right) 
\left(\sup\_i\frac{1}{|A\_i|}\int\_{A\_i} w\_i^{-p'/p} \right)^{p/p'}
\end{align}


To complete the proof, we just need to choose $p$ and $w\_i$ so that (i) $\sum\_i 1\_{A\_i} w\_i \lesssim 1$ and (ii) $\sup\_i\frac{1}{|A\_i|}\int\_{A\_i} w\_i^{-p'/p} \lesssim 1$. This can be accomplished as follows:


Let $p = 3$. Introduce an ordering on the indices so that $i \prec j$ if $|B\_i| < |B\_j|$ (and break ties arbitarily). Set $w\_i(x)^{-1/2} = \sum\_{j \preceq i} 1\_{A\_j}(x) = \# \{ j : x \in A\_j \text{ and } j \preceq i\}$.


Note that if $j \preceq i$ and $A\_j \cap A\_i \neq \emptyset$, then $B\_j \subset 300B\_i$. This, with the disjointness of the $B\_j$, implies $$\int\_{A\_i} w\_i^{-1/2} \leq \sum\_{j \preceq i, A\_j \cap A\_i \neq \emptyset} |A\_j| \approx \sum\_{j \preceq i, A\_j \cap A\_i \neq \emptyset} |B\_j| \leq |300B\_i| \approx |A\_i|.$$


This proves (ii). (Also, this implies $w\_i(x) > 0$ for almost every $x \in A\_i$.)


Finally, for any fixed $x$, if $w\_i(x) = w\_j(x) \neq 0$, then $i=j$. Since $w\_i$ takes values in $\{m^{-2} : m \in \mathbb N\} \cup \{0\}$, we have the pointwise bound $\sum\_i 1\_{A\_i} w\_i \leq \frac{\pi^2}{6}$, which shows (i) holds and completes the proof.


I don't really have a good intuition for this proof, especially how to motivate the choice of $p$ and $w\_i$ (other than "because it works"). In particular, I am mystified (and amazed) at how the authors use $\sum\_{m=1}^\infty m^{-2} < \infty$ to control the overlap of the $\{A\_i\}\_i$. This is why I'd be interested to see if there was another proof.



Can this inequality be proved using weighted maximal function estimates?
It suffices to show that
$$ \sum\_i |B\_i| 1\_{\inf\_{A\_i} f > t} \lesssim \int\_{\bigcup A\_i} 1\_{f>t}$$
for any $t>0$, since the claim follows by integrating in $t$ and using the Fubini-Tonelli theorem (i.e., use the [layer cake decomposition](https://en.wikipedia.org/wiki/Layer_cake_representation)). (Equivalently: to prove the claim, it suffices to do so in the special case when $f$ is an indicator function.) But one has
$$ M (1\_{\bigcup A\_i} 1\_{f>t})(x) \gtrsim 1$$
whenever $x \in B\_i$ and $\inf\_{A\_i} f>t$, so the claim follows from the Hardy-Littlewood maximal inequality.


