# Show that $\mu(A_{i_1}\cap A_{i_2}\cap\dots \cap A_{i_k})>c^k-\epsilon$

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 $$i_1 so that $$\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 $$\mu(A_{i_1}\cap A_{i_2})\ge 1-2(1-c)=2c-1$$

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

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 $$i. (Note that I simply rewrote the condition in an equivalent form which is slightly more convenient to work with).
Then, we must have that $$(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.
• Just to nickpick: you actually need to assume $\mu(X) \leq 1$ for this to work, not just that $\mu(X)$ is finite. Apr 20 at 23:26