2
$\begingroup$

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 $a$.

Improve this question
$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – Ramiro de la Vega
    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). $\endgroup$
    – Ramiro de la Vega
    Nov 14, 2019 at 19:00

1 Answer 1

Reset to default
3
$\begingroup$

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

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Mateusz Kwaśnicki
    Nov 14, 2019 at 21:08
  • $\begingroup$ Got it! Thank you so much! $\endgroup$
    – Yi Huang
    Nov 14, 2019 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.