Showing that there is a point belonging to $2000$ sets.

Question

Question: Let $E_j\subset[0,1]$ be a sequence of measurable sets satisfying $$m(E_i\cap E_j)\geq\frac{1}{i^2+j^2}$$ for all $i,j\geq1$. Prove that there is an $x\in[0,1]$ belonging to at least $2000$ sets $E_j$. Does there exist an $x\in[0,1]$ belonging to infinitely many $E_j$'s?

My thoughts: If we consider the function $f=\sum_j \chi_{E_j}$, then $f$ is a measurable function from $[0,1]$ to $[0,\infty]$. So, $f^{-1}([2000,\infty])$ is measurable, and $f^{-1}([2000,\infty])$ is the set of points $x\in[0,1]$ that belong in at least $2000$ of the $E_j$'s.

I'm not sure if what I have above is totally accurate, and for the second question, I am not sure if that would just, then, be obvious (well, I suppose, evidently not) or if there is a bit more to consider. Moreover, I suppose I never really used the "$m(E_i\cap E_j)\geq\frac{1}{i^2+j^2}$ for all $i,j\geq1$" part, so I feel like I have something incorrect here....

Any thoughts, suggestions, etc. are appreciated! Thank you!

Answer 1

As pointed out by @mathmandan, you have never shown that $f^{-1}([2000,\infty])$ is non-empty, which is equivalent to what the question is asking about.

• For the proof of the first part, let $f = \sum_{i\geq 1} \chi_{E_i}$ and note that $$ \int_{[0,1]} f^2 \, \mathrm{d}m = \sum_{i,j \geq 1} \int_{[0,1]} \chi_{E_i}\chi_{E_j} \, \mathrm{d}m = \sum_{i,j \geq 1} m(E_i \cap E_j) \geq \sum_{i,j \geq 1} \frac{1}{i^2+j^2} = \infty. $$ In particular, $f$ cannot be bounded above and hence the desired claim follows.

• On the other hand, if we set $E_i = (0,i^{-2})$, then $$ m(E_i \cap E_j) = \frac{1}{\max\{i^2, j^2\}} \geq \frac{1}{i^2+j^2}, $$ whereas each $x \in [0, 1]$ lies in only finitely many of $E_i$'s. (In fact, for this choice, we have $\sum_{i\geq 1} \chi_{E_i}(x) = \lfloor\frac{1}{\sqrt{x}}\rfloor$ for any $x \in (0, 1]$.)

Answer 2

I think that what you have said so far is fine...but I don't think you have shown that $f^{-1}([2000, \infty])$ is nonempty.

Consider the following sequence: $$ E_i = \left(\frac{1}{2^{i}}, \frac{1}{2^{i-1}}\right) $$

So $E_1 = \left(\frac{1}{2}, 1\right)$, and $E_2 = \left(\frac{1}{4}, \frac{1}{2}\right)$, etc. No point belongs to more than one of these sets; they are pairwise disjoint.

So there is no point which belongs to $2000$ or more of these sets. In other words, $f^{-1}([2000, \infty]) = \emptyset$. (This example shows that you'll need to use the given condition somehow.)