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 $a$.

  • $\begingroup$ 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? $\endgroup$
    – T_M
    Nov 14, 2019 at 17:09
  • $\begingroup$ 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. $\endgroup$
    – Yi Huang
    Nov 14, 2019 at 17:33
  • $\begingroup$ 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. $\endgroup$ Nov 14, 2019 at 18:11
  • $\begingroup$ @RamirodelaVega Hi Ramiro, upper bounding is enough for my current purpose, but having $p$ in the bound is problematic. $\endgroup$
    – Yi Huang
    Nov 14, 2019 at 18:37
  • $\begingroup$ 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). $\endgroup$ Nov 14, 2019 at 19:00

1 Answer 1


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 $\tfrac{a N}{4}$.

  • $\begingroup$ 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? $\endgroup$
    – Yi Huang
    Nov 14, 2019 at 20:48
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 14, 2019 at 21:08
  • $\begingroup$ Got it! Thank you so much! $\endgroup$
    – Yi Huang
    Nov 14, 2019 at 21:30

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