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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 $$m(\bigcap_{k=1}^{\infty}E_{nk})=0.$$

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

Andrés E. Caicedo
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Bozo Vulicevic
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  • could you go into more detail, if it's not a problem? I really do have troubles with these type of questions. – Bozo Vulicevic Oct 29 '15 at 07:28
  • Maybe this gives some idea: http://math.stackexchange.com/questions/519646/sequence-of-lebesgue-measurable-sets-a-k-such-that-a-k-subset0-1-lim-l?rq=1 – Arash Oct 29 '15 at 07:42
  • 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. – fweth Oct 29 '15 at 20:23
  • Thanks, I would love that. :) – Bozo Vulicevic Oct 30 '15 at 17:47

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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.

fweth
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