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Let $m$ denote the Lebesgue measure. Find Lebesgue measurable subsets $A_1,A_2,...$ of unit interval with $m(A_n) \in (0,1)$ for every $n$ such that $m(A_n \cap A_K)=m(A_n)m(A_k)$ whenever $n \neq k$.

So I first tried $A_n:=(0,\frac{1}{n+1})$ but these are nested so the criteria wont be satisfied. Then I thought up the following:

$$A_1=(0,1/2)$$ $$A_2=(0,1/4) \cup (3/4,1)$$ $$A_3:=(0,1/8) \cup (1/2,5/8).$$

Then we have $m(A_1 \cap A_2)=1/4=m(A_1)m(A_2)=(1/2)(1/2).$ But i dont think it holds for $A_1,A_3$

But I dont know how to recursively define the $A_n$. Any hints or tips greatly appreciated. Can I continue in this fashion by defining all the $A_n$?

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    $\begingroup$ If you use $A_2:=(0,1/4)\cup(1/2,3/4)$ (so the first half of each halves instead of the second half of the second half) and $A_3:=(0,1/8)\cup(1/4,3/8)\cup(1/2,5/8)\cup(3/4,7/8)$ (the first half of each fourth), then you get a pattern that works (and in fact is equal to the answer by @Cactus, except with the numbering off by $1$). $\endgroup$ Sep 12, 2023 at 19:57
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    $\begingroup$ So I was so close! @TobyBartels $\endgroup$
    – homosapien
    Sep 12, 2023 at 20:00

2 Answers 2

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First of all, $A_n = \emptyset$ or $(0,1)$ for all $n$ works but I guess you want a non trivial example.

The equality $m(A_n \cap A_k) = m(A_n)m(A_k)$ means, from the probabilitic point of view, that the $A_n$ are pairwise indépendant. Thus, you can try with something like $$ A_n = \bigcup_{0 \leqslant k \leqslant 2^n - 1} \left(\frac{2k}{2^{n + 1}},\frac{2k + 1}{2^{n + 1}}\right) $$ Then show that $m(A_n) = \frac{1}{2}$ for all $n$ and when $n \neq k$, $m(A_n \cap A_k) = \frac{1}{4}$.

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  • $\begingroup$ Those trivial examples don't satisfy $m(A_n) \in (0,1)$. $\endgroup$
    – M W
    Sep 12, 2023 at 19:41
  • $\begingroup$ @MW are you referring to my examples or the one he put in the solution? $\endgroup$
    – homosapien
    Sep 12, 2023 at 19:45
  • $\begingroup$ @MW : They all have $\displaystyle m(A_n)=\sum_{0\leq k\leq2^n-1}\bigg(\frac{2k+1}{2^{n+1}}-\frac{2k}{2^{n+1}}\bigg)=\sum_{0\leq k<2^n}\frac1{2^{n+1}}=\frac{2^n}{2^{n+1}}=\frac12\in(0,1)$. $\endgroup$ Sep 12, 2023 at 19:46
  • $\begingroup$ ok thats what i thought @TobyBartels $\endgroup$
    – homosapien
    Sep 12, 2023 at 19:46
  • $\begingroup$ @TobyBartels I was referring to the first statement that $A_n=\emptyset$ or $A_n=(0,1)$ works. $\endgroup$
    – M W
    Sep 12, 2023 at 19:48
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Denote the partition of $(0, 1)$ into $2^n$ equal intervals by $P_n$. Let $A_n$ be the union of the odd numbered intervals in $P_n$. The Lebesgue measure of $A_n$ is $1/2$. Now suppose $n > m$. Consider an interval $I$ of $A_m$. Clearly, $I$ has Lebesgue measure $1/2^m$ and contains $2^{n - m}$ intervals from $P_n$. Since only half the intervals in $P_n$ belong to $A_n$, $I \cap A_n$ has Lebesgue measure $$ \frac{2^{n-m}}{2} \frac{1}{2^n} = \frac{1}{2^{m+1}}. $$ Thus, $A_m \cap A_n$ has Lebesgue measure 1/4.

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