Approximation of an integral with balls

Question

Let's say I have nonnegative functions $f \in C_0^\infty(\mathbb R^d)^+$ smooth and compactly supported and I have a set $A \subset \mathbb R^d$ of infinite Lebesgue measure. I would like to estimate the integral over balls of fixed radius $\ell > 0$; since $\mathbb R^d$ is $\sigma$-compact I can cover $A$ with countably many balls, i.e. there is a countable set $\mathcal S$ for which
$$ \int_A f(x)dx \leq \sum_{x_i \in \mathcal S} \int_{B_\ell(x_i)} f(x)dx. $$
But I would like to continue working with an integral; is it possible to get
$$ \int_A f(x)dx \leq \sum_{x_i \in \mathcal S} \int_{B_\ell(x_i)} f(x)dx \leq \int_A \left(\int_{B_\ell(x)}f(y)dy\right)dx? $$
(Or is there an upper bound up to a constant $C$ independent of $f$ but possibly depending on $\ell$?) I have no idea how to start – part of me thinks the answer is yes because I am replacing a Riemann type sum with an integral and I know also for instance that the average integral of a ball converges to its centre as the radius gets small, but the other part thinks these arguments should yield a constant with dependence on $f$.

Answer 1

From knowing that "the covering of $A$ is countable and of fixed radius $l$" we can say nothing about how big $\sum_{x_i \in \mathcal S} \int_{B_\ell(x_i)} f(x)dx$ can get, since the overlapping of the covering balls can make this sum of integrals infinitely big.
The problem with thinking about it as a "Riemann sum" is that such sums are disjoint, an your sum allows for infinite overlaps.

Maybe you could try a similar thing with disjoint (f.i. half-open) cubes instead of balls.