4
$\begingroup$

I self-study measure theory from Stein-Shakarchi. For practice, I tried to prove on my own that $L^1$ is complete (without having seen the proof). During my attempt, I got stuck on the following question. Beware that this question could be more difficult than the target theorem (which is why I am posting here), but still it will be very illuminating for me to learn the answer. Here is the question:

Is the following statement true? For every $\epsilon>0$, there exists a $\delta(\epsilon)>0$ such that

  1. $\delta(\epsilon)\to 0$ as $\epsilon\to 0$.
  2. For every sequence $E_1,E_2,\dots$ of measurable subsets of $[0,1]$, each of measure at most $\epsilon$, there exists a measurable $E\subseteq [0,1]$ of measure at most $\delta(\epsilon)$, that contains infinitely many $E_n$.

The case where $E_n$ are intervals is easy (pigeonhole principle). If the general question is too difficult, the simpler version where each $E_n$ is a union of finite number of intervals suffices for my purpose. Note that this number of intervals can grow with $n$ (otherwise it is easy).

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ This cannot work if the sets get more spread out as $n$ increases. For example, cover $[0,1]$ by intervals of length $\epsilon$, then let the sets of the next generation be unions of $1/\epsilon$ intervals of length $\epsilon^2$ each etc. Then we must cover some $E_n$, which already gives us measure $\ge\epsilon$, and now a set of a later generation will have overlap at most $\epsilon^2$ with the first set, so the measure increases by $(1-\epsilon)\epsilon$. Still later on, we will be forced to add measure $\ge (1-\epsilon)^2\epsilon$, so the overall measure will not be small. $\endgroup$ Commented Feb 1 at 5:08
  • 2
    $\begingroup$ Kolmogorov told that probability is measure theory plus independence. This is an instructive example: on measure theoretical language, the question is enigmatic. But replace the sets to events and measure to probability and you immediately see that for independent events of the same ptobability $\varepsilon$ the probability of every infinite union is 0, as in bof's example. $\endgroup$ Commented Feb 1 at 7:15
  • $\begingroup$ @FedorPetrov I think you typoed $0$ for $1$ or else union for intersection. :-) (And I guess you want $0\lt\epsilon\lt1$.) $\endgroup$ Commented Feb 1 at 7:21
  • $\begingroup$ @bof Ah yes, 0 for 1 $\endgroup$ Commented Feb 1 at 8:25
  • $\begingroup$ Thank you all, including @bof for your excellent replies! They are all the same construction approached from different perspectives: visually, probabilistically, and symbolically. Beautiful! Thanks again! $\endgroup$ Commented Feb 3 at 2:50

1 Answer 1

Reset to default
5
$\begingroup$

Suppose $\epsilon\gt0$; I will show that $\delta(\epsilon)\ge1$.

Choose an integer $r\ge\max\{\frac1\epsilon,2\}$. Let $E_n$ be the set of all numbers in $[0,1]$ whose expansion in base $r$ has $0$ in the $n^\text{th}$ place. Then $E_n$ is the union of $r^{n-1}$ intervals and has measure $\frac1r\le\epsilon$. The union of any infinite subsequence of $\{E_n\}$ has measure $1$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.