Construct a sequence of measurable sets $E_n\subset[0,1],$ such that $$\underline{\lim}_{n\to + \infty}m(E_n)>\frac{3}{4}$$ and for every sub sequence $(E_{nk})$ applies $$m(\bigcap_{k=1}^{\infty}E_{nk})=0.$$

$$(\underline\lim A_n=\bigcup_{n=1}^{+\infty}\bigcap_{m=n}^{+\infty}A_m)\text{ -limit inferior}$$

  • $\begingroup$ could you go into more detail, if it's not a problem? I really do have troubles with these type of questions. $\endgroup$ – Bozo Vulicevic Oct 29 '15 at 7:28
  • $\begingroup$ Maybe this gives some idea: math.stackexchange.com/questions/519646/… $\endgroup$ – Arash Oct 29 '15 at 7:42
  • $\begingroup$ Sorry, my idea was wrong, I deleted the comment! I think the right idea would be $A_n=\{x\in[0,1):\lfloor 4^{n+1}x\rfloor\neq0\mod 4\}$. Do you get along with Arash's comment? If not, I will try to give you a detailed answer tomorrow. $\endgroup$ – fweth Oct 29 '15 at 20:23
  • $\begingroup$ Thanks, I would love that. :) $\endgroup$ – Bozo Vulicevic Oct 30 '15 at 17:47

Sorry, I just typed the whole thing before I saw that you require the $\lim\inf$ to be larger than $\frac{3}{4}$, I just worked with the $\lim\inf$ being exactly $\frac{3}{4}$. But of course you could e.g. work with $\frac{4}{5}$ everywhere in my proof instead of $\frac{3}{4}$.

Take $A_n=\{x\in[0,1):\lfloor 4^{n+1}x\rfloor\neq0\mod 4\}$. So the sets look like this: $A_0$ is the intervall $[\frac{1}{4},1)$, $A_1$ consists of the union of the intervalls $[\frac{1}{16},\frac{4}{16}),[\frac{5}{16},\frac{8}{16}),[\frac{9}{16},\frac{12}{16}),[\frac{13}{16},1)$ and so on. Maybe try to visualize those sets, their structure is indeed not as complicated as it may look in the formal definition.

I would now sketch the proof as follows:

  • Our aim is to show that if $A$ equals the intersection of some $A_i$ with $i<n$, then $\lambda(A\cap A_n)=\frac{3}{4}\lambda(A)$, from this the result follows easily.

  • $\lambda([\frac{k}{4^n},\frac{k+1}{4^n})\cap A_n)=\frac{3}{4^{n+1}}=\frac{3}{4}\lambda([\frac{k}{4^n},\frac{k+1}{4^n}))$ for $0\le k<4^n$

  • Now observe that any $A$ desribed as above can be written as the disjunct union of some elements of the form $[\frac{k}{4^n},\frac{k+1}{4^n})$ with $0\le k<4^n$, especially $A\cap[\frac{k}{4^n},\frac{k+1}{4^n})$ equals either $\emptyset$ or $[\frac{k}{4^n},\frac{k+1}{4^n})$ for any $0\le k<4^n$.

  • If $A\cap[\frac{k}{4^n},\frac{k+1}{4^n})=\emptyset$ then $\lambda(A\cap[\frac{k}{4^n},\frac{k+1}{4^n})\cap A_n)=0=\frac{3}{4}0=\frac{3}{4}\lambda(A\cap[\frac{k}{4^n},\frac{k+1}{4^n}))$. On the other hand if $A\cap[\frac{k}{4^n},\frac{k+1}{4^n})=[\frac{k}{4^n},\frac{k+1}{4^n})$ then $\lambda(A\cap[\frac{k}{4^n},\frac{k+1}{4^n})\cap A_n)=\lambda([\frac{k}{4^n},\frac{k+1}{4^n})\cap A_n)=\frac{3}{4}\lambda([\frac{k}{4^n},\frac{k+1}{4^n}))=\frac{3}{4}\lambda(A\cap[\frac{k}{4^n},\frac{k+1}{4^n}))$.

  • $\lambda(A\cap A_n)=\sum_{0\le k<4^n}\lambda(A\cap[\frac{k}{4^n},\frac{k+1}{4^n})\cap A_n)$ holds because we just partition $A\cap A_n$ in $4^n$ disjoint sets and sum over the measures of those. (Same with $A$.)

  • So if we put this together then $\lambda(A\cap A_n)=\sum_{0\le k<4^n}\lambda(A\cap[\frac{k}{4^n},\frac{k+1}{4^n})\cap A_n)=\sum_{0\le k<4^n}\frac{3}{4}\lambda(A\cap[\frac{k}{4^n},\frac{k+1}{4^n}))=\frac{3}{4}\lambda(A)$.

I hope I didn't make anything harder than it needs to be (something I tend to do and try to avoid)! I think the proof works rather well when visualizing the whole thing in mind, it just becomes cumbersome to write everything down.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.