Let $((0,1], \mathcal{B},\mu)$ be a measure space. Let $\{A_i\}_{i\ge 1}$ be a sequence of Borel measurable sets so that $\mu(A_i)\ge c$ for all $i\ge 1$ and some universal constants $c\in (0,1)$. Show that for all $k=1,2,3,\dots$ and $\epsilon>0$, there exists $i_1<i_2<\dots<i_k$ so that $$ \mu(A_{i_1}\cap A_{i_2}\cap\dots \cap A_{i_k})>c^k-\epsilon $$

I am stuck on this question. I try to upper bound $$ (\sum_{i=1}^n \mu(A_i))^k $$ for some $n\ge 1$...

I also try to specific case $k=2$: to show that $$ \mu(A_{i_1}\cap A_{i_2})>c^2-\epsilon $$

We have $$ \mu((A_{i_1}\cap A_{i_2})^c)=\mu(A_{i_1}^c\cup A_{i_2}^c)\le \mu(A_{i_1}^c)+\mu(A_{i_2}^c)\le 2(1-c) $$ So $$ \mu(A_{i_1}\cap A_{i_2})\ge 1-2(1-c)=2c-1 $$

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    $\begingroup$ Are you perhaps missing some assumptions? What happens if $A_i\equiv A$ and $\mu(A)=c$? $\endgroup$ Apr 16 at 8:12
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    $\begingroup$ @Keen-ameteur $c^{k}-\epsilon < c^{k} \le c$. $\endgroup$ Apr 16 at 8:13
  • $\begingroup$ @geetha290krm Yeah, you're right. I misunderstood the question and jumped to comment. $\endgroup$ Apr 16 at 8:19
  • $\begingroup$ Is $\mu$ maybe a finite measure? Otherwise, what happens if the $A_i$ are all pairwise disjoint? $\endgroup$
    – PhoemueX
    Apr 17 at 19:04
  • $\begingroup$ @PhoemueX Yes, $\mu$ should be finite because $\mu([0,1])=1$ and any subset $\mu(A)\le 1$ for $A\subset [0,1]$. $\endgroup$
    – H.Y Duan
    Apr 18 at 2:25

1 Answer 1


The proof is very neat and rather elementary. I will only write the details for the special case $k=2$.

For this to work, we assume that $\mu(X)=1$, which, besides, is the OP's assumption.

Noticing that $\mu(A_i)=\int \mathbb{1}_{A_i} \ d\mu$ we have that for all $n\in \mathbb{N}$ $$ (n\cdot c)^2 \leq \left( \int \sum_{i=1}^n \mathbb{1}_{A_i} \ d\mu \right)^2 \leq \int \left( \sum_{i=1}^n \mathbb{1}_{A_i} \right)^2 d\mu, $$ by assumption and an application of Cauchy-Schwarz. Then, expanding the square we have that $$ (n\cdot c)^2 \leq \sum_{i=1}^n \mu(A_i) + 2\sum_{1\leq i< j \leq n} \mu(A_i \cap A_j).$$ Let $\alpha \in (0,1)$ and assume for contradiction that $\mu(A_i \cap A_j) \leq \alpha \cdot c^2$, for all $i<j$. (Note that I simply rewrote the condition in an equivalent form which is slightly more convenient to work with).

Then, we must have that $$ (n \cdot c)^2 \leq n + n(n-1) \cdot \alpha \cdot c^2, $$ which implies that $$ c^2 - \frac{1-c^2}{n-1}=\frac{(n \cdot c)^2 - n }{n^2-n} \leq \alpha \cdot c^2,$$ for all $n\in \mathbb{N}$, which yields a contradiction if $n$ is chosen large enough.

An analogous argument starting from $(n\cdot c)^k \leq \left( \int \sum_{i=1}^n \mathbb{1}_{A_i}\ d\mu \right)^k$ and using Cauchy-Schwarz once again works for the general case too.

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    $\begingroup$ Just to nickpick: you actually need to assume $\mu(X) \leq 1$ for this to work, not just that $\mu(X)$ is finite. $\endgroup$
    – abacaba
    Apr 20 at 23:26

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