I came across the following problem while doing a piece of research on automata theory.


Suppose we have a probability space $(\Omega, \mathcal{F}, \mu)$, where $\Omega$ is a set, $\mathcal{F}$ is a $\sigma$-algebra, and $\mu$ is a probability measure. Suppose we have **non-negative bounded** function $f:\Omega\rightarrow\mathbb{R}$. Suppose $A\_i$, $i=1, 2,\dots$, is a sequence of sets in $\mathcal{F}$ that satisfy $\int\_{A\_i}f\ d\mu \leq a \mu\left(A\_i\right)$ for each $i$ and $\mu\left(\bigcup\_{i=1}^{\infty}A\_i\right)=1$. My question is:


Does $\int\_{\Omega}f\ d\mu$ is upper bounded by something **related to $a$**?


In my case, I also have $\mu(A\_i)=p$ for all $i$, but I don't think the condition is essential to the question. However, if it is indeed essential or could simplify the proof, please feel free to use it. 


The integral is bounded because $f$ is bounded and $\mu$ is a probability measure, but I need the integral to be bounded by something related to $a$, like a scalar of $a$.



Does bounded integral over sequence of subsets of $X$ whose union is $X$ imply bounded integral over X?
Set $\Omega = [0,1)$, $\mu$ the Lebesgue measure, $f = \tfrac{a N}{2} \times \mathbb{1}\_{(0,1/2)}$ for an arbitrarily large integer $N$, and
$$ A\_k = \bigl([\tfrac{1}{2}, 1) \setminus [\tfrac{N+k-1}{2N}, \tfrac{N+k}{2N})\bigr) \cup [\tfrac{k-1}{2N}, \tfrac{k}{2N}) ,$$
where $k = 1, 2, \ldots N$. Then the union of $A\_k$ is $\Omega$, $\int\_{A\_k} f d\mu = \tfrac{a N}{2} \times \tfrac{1}{N} = a \mu(A\_k)$, but the integral of $f$ is an aribitrarily large number $\tfrac{a N}{4}$.


