# Find Lebesgue measurable sets such that the measure of intersection is product of measures.

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 $$A_n$$?

• 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 $1$). Sep 12, 2023 at 19:57
• So I was so close! @TobyBartels Sep 12, 2023 at 20:00

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 $$n \neq k$$, $$m(A_n \cap A_k) = \frac{1}{4}$$.

• Those trivial examples don't satisfy $m(A_n) \in (0,1)$.
– M W
Sep 12, 2023 at 19:41
• @MW are you referring to my examples or the one he put in the solution? Sep 12, 2023 at 19:45
• @MW : They all have $\displaystyle m(A_n)=\sum_{0\leq k\leq2^n-1}\bigg(\frac{2k+1}{2^{n+1}}-\frac{2k}{2^{n+1}}\bigg)=\sum_{0\leq k<2^n}\frac1{2^{n+1}}=\frac{2^n}{2^{n+1}}=\frac12\in(0,1)$. Sep 12, 2023 at 19:46
• ok thats what i thought @TobyBartels Sep 12, 2023 at 19:46
• @TobyBartels I was referring to the first statement that $A_n=\emptyset$ or $A_n=(0,1)$ works.
– M W
Sep 12, 2023 at 19:48

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.