# sequence of Lebesgue measurable sets $A_k$ such that $A_k\subset[0,1]$, $\lim \lambda(A_k)=1$, but $\lim \inf A_k=\emptyset$.

Give an example in $\mathbb{R}$ of a sequence of Lebesgue measurable sets $A_k$ such that $A_k\subset[0,1]$, $\lim \lambda(A_k)=1$, but $\lim \inf A_k=\emptyset$.

My thoughts: By definition, $\lim\inf A_k=\cup_{n=1}^{\infty}\cap_{k\ge n}A_k$, and we want this to be empty. Maybe we can construct a sequence of sets such that $\lambda(A_k)=1-1/k$, but $\cap_{k\ge n}A_k=\emptyset$ for all $n$.

This was mentioned here but was unanswered: Fatou's lemma and measurable sets

First note that $\liminf A_{k}=\emptyset$, is equivalent to the statement that that given any $x\in[0,1]$ for any $n\in\mathbb{N}$ there is $n_{0}\in\mathbb{N}$ s.t. $n_{0}>n$ and $x\notin{A_{n_{0}}}$.

Using this we can come up with a sequence using the following "pattern" $A_{1}=[0,\frac{1}{2}]$, $A_{2}=[\frac{1}{2},1]$, $A_{3}=[0, \frac{1}{3}] \cup[\frac{1}{3},\frac{2}{3}]$, $A_{4}=[0,\frac{1}{3}] \cup[\frac{2}{3},1]$, $A_{5}=[\frac{1}{3},\frac{2}{3}] \cup[\frac{2}{3},1]$...... which has the desired properties.

• Wait, so what does $[2/3, 1/3]$ mean? – Christmas Bunny Oct 9 '13 at 4:39
• That was a typo, sorry. It has been corrected. – UserB1234 Oct 9 '13 at 15:37

Let $$A_1 = [0,1] \setminus [0,\frac12]$$ and $$A_2 = [0,1] \setminus [\frac12,1]$$.

Then let $$A_3 = [0,1] \setminus [0,\frac14]$$ and $$A_4 = [0,1] \setminus [\frac14,\frac12]$$ and $$A_5 = [0,1] \setminus [\frac12,\frac34]$$ and $$A_6 = [0,1] \setminus [\frac34,1]$$.

Then let $$A_7 = [0,1] \setminus [0,\frac18]$$ and ... and $$A_14 = [0,1] \setminus [\frac78,1]$$.

And so on.