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.
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.
