Let $A \subseteq \mathbb{R}^d$. Let $B_1,B_2,\ldots \subseteq \mathbb{R}^d$ be a countable collection of sets $A = \bigcup_i B_i$ with finite overlap $N$. This means that each $x \in A$ belongs to at most $N$ of the sets $B_i$. Let $f$ be a function defined on $A$. Assume the sets and functions are measurable with respect to a measure $\mu$ on $\mathbb{R}^d$.
Question 1:
If $f \geq 0$, do we have $$ \sum_{i} \int_{B_i} f(x) d\mu(x) \leq C_N \int_A f(x) d\mu(x) $$ for some constant $C_N>0$ depending on $N$? If so, how can it be proved and can $C_N$ be written explicitly?
Motivation:
If $A$ is countable and $\mu$ is counting measure, the left-hand side is $$ \sum_{i} \sum_{x \in B_i} f(x). $$ For each $x \in A$, the summand $f(x)$ can appear at most $N$ times in this sum. So the sum is $$ \leq N \sum_{x \in A} f(x). $$ This argument feels right. But I feel like it can be made more clear using intersections to form a disjoint cover of $A$. And maybe that is the way to get the result for general measures $\mu$. The argument also relies on being able to re-arrange a sum of positive numbers. This suggests something like this should work if $f$ is is absolutely integrable with respect to $\mu$, but not necessarily positive.
Question 2:
If $f$ is absolutely integrable with respect to $\mu$, do we have $$ c_N \int_A f(x) d\mu(x) \leq \sum_{i} \int_{B_i} f(x) d\mu(x) \leq C_N \int_A f(x) d\mu(x)? $$ How to prove it? Can $c_N$ and $C_N$ be written explicitly.
Update: I figured out the answer to Question 1. The trick is to use characteristic functions. The assumptions give
$$1_A \leq \sum_{i} 1_{B_i} \leq N,$$ whence
$$
\sum_{i} \int_{B_i} f d\mu = \int_A \sum_{i} 1_{B_i} \, f \, d\mu \leq N \int_A f d\mu.
$$
I learned the trick from the Besicovitch covering theorem article https://en.wikipedia.org/wiki/Besicovitch_covering_theorem
If you add some absolute value here and there, then Question 2 can be proven like this: $$ \left |\sum_i \int_{B_i} f \right| \le \sum_i \int_{B_i} |f| \le N \int_A|f|. $$ I doubt that the inequality in your post can be proven for arbitrary (not non-negative) $f$.