# Does bounded integral over sequence of subsets of $X$ whose union is $X$ imply bounded integral over X?

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 $$a$$.

• The condition $\mu(A_i) = p$ might make some proof involving a bound that depends on $1/p$, but presumably this is also not desirable?
– T_M
Nov 14, 2019 at 17:09
• Hi T_M, thank you for your comment. I indeed need the bound to be related to $a$, since in my case, $a$ is some number that I can move to zero. (the choice of $A_i$ also depends on $a$) And yes, as you might have guessed, the final goal is to show the integral over $\Omega$ can be arbitrarily small, and if the statement indeed holds for all $a$, I can show the integral over $\Omega$ actually vinishs. Nov 14, 2019 at 17:33
• I think the best you could say is that the integral is at most $a/p$ (consider $f$ with constant value $a/p$), but to show that $a/p$ is indeed a bound seems tricky. Nov 14, 2019 at 18:11
• @RamirodelaVega Hi Ramiro, upper bounding is enough for my current purpose, but having $p$ in the bound is problematic. Nov 14, 2019 at 18:37
• Well, as I said, that is the best you can expect. You certainly cannot give a bound for the integral just in terms of $a$. Given any two positive numbers $a$ and $N$, you can find $f$ and $A_i$´s satisfying all the hypothesis but with $\int_{\Omega}f\ d\mu=N$ (e.g. working on the unit interval with Lebesgue measure, take $f$ with constant value $N$ and the $A_i$´s any collection of sets of measure $\leq a/N$ covering the interval). Nov 14, 2019 at 19:00

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 $$\tfrac{a N}{4}$$.
• Hi Mateusz, Thank you very much for your answer! But do you intend to provide a counterexample? In my case, $f$ is in fact bounded by a fixed number, say 5. Can you try with this assumption? Nov 14, 2019 at 20:48
• Then set $N$ so that $a N / 2$ is nearly equal to $5$. Then $f$ is bounded by $5$ and the integral is roughly $5/2$, no matter how small $a$ is. Nov 14, 2019 at 21:08