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I want to find a sequence of measurable sets $A_k$.

such that $A_k \subset [0,1]$, $\lim \lambda(A_k) =1$, but $\liminf A_k = \varnothing$.

There are some examples on function such as $\sin x \over x$ , but I can't apply on a set, $A_k$.

Please give me a simple example.

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up vote 5 down vote accepted

One technique you can use is a "running window". The sequence of intervals $I_1=[0,1/2),I_2=[1/2,1],I_3=[0,1/3),I_4=[1/3,2/3),I_5=[2/3, 1],I_6=[0,1/4),\ldots$ has the property that their measure goes to $0$ but every element of $[0,1]$ is in infinitely many of these intervals. So you can simply let $A_k=[0,1]\backslash I_k$.

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Thank you for your answer Michael, but I don't know in your example does there exist a limit of $I_k$? I think there exists not $lim$ but only sub limit of $A_k$. And I think that $liminf(A_k)$ is not $\varnothing $ but $[0,1]$.. because $lim(I_k)=\varnothing$. If I'm confusing something please notice me that. Thank you. – japee May 8 '12 at 13:49
I'm not sure what you mean. The sequence $(\lambda(A_k))$ is nondecreasing and bounded and certainly has a limit. The sequence $A_k$ does not have a limit in the sense that $\lim\inf A_k=\emptyset$ but $\lim\sup A_k=[0,1]$. This follows from the characterizations that $\lim\inf A_k$ is the set of points that are in all but finitely many of the $A_k$ and $\lim\sup A_k$ is the set of points that are in infitely many of the $A_k$. If any sequence $(A_k)$ of measurable sets has a limit $A$, then $\lim\lambda(A_k)=\lambda(A)$, so this is unavoidable. – Michael Greinecker May 8 '12 at 13:53
$\lim(I_k)\neq\emptyset$ since $\lim\sup(I_k)=[0,1]$. Every point in $[0,1]$ is in infinitely many thin intervals. I have to leave now, I can respond again in a few hours. – Michael Greinecker May 8 '12 at 14:04
Oh. I got it. Thank you very much Then I wanna prove next question above. Can I have some intution for proving that? – japee May 8 '12 at 14:37

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