note: $m(A\triangle B)$ means $(A-B)\cup (B-A)$
Let $m$ be lebesgue measure. Find an example of Lebesgue measurable subsets $A_1,A_2,...$ of $[0,1]$ such that $m(A_n)>0$ for each $n$. $m(A_n\triangle A_m)>0$ if $n\neq m$ and $m(A_n\cap A_m) = m(A_n)m(A_m)$
What I tried to do for this one was to stick to sets of measure 0 or 1, but the condition that $m(A_n\triangle A_m)>0$ makes this unfeasible. Another idea I had was to make $A_n$ a set of measure $1/n$, but it is still unclear how I can satisfy the condition that $m(A_n\cap A_m) = m(A_n)m(A_m)$.
This is exercise 4.9 in http://bass.math.uconn.edu/3rd.pdf. Any hints or solutions would be appreciated.
Every $x\in[0,1)$ has a binary expansion, which is unique if we require it to contain infinitely many zeros. For $n\geq 1$, let $A_n$ be the set of $x\in[0,1)$ such that the $n$th digit in the binary expansion is zero. Thus $$ A_1=\Big[0,\frac{1}{2}\Big)\quad A_2=\Big[0,\frac{1}{4}\Big)\cup\Big[\frac{1}{2},\frac{3}{4}\Big)\quad A_3=\Big[0,\frac{1}{8}\Big)\cup\Big[\frac{1}{4},\frac{3}{8}\Big)\cup\Big[\frac{1}{2},\frac{5}{8}\Big)\cup\Big[\frac{3}{4},\frac{7}{8}\Big)$$ and so on. Since $$ m(A_n\cap A_m)=\frac{1}{4}=m(A_n)m(A_m) $$ for all $n\neq m$, it follows that $\{A_n\}$ has the desired properties.
This example may seem more intuitive if you think of the binary expansion of each $x\in[0,1)$ as recording the outcome of an infinite sequence of flips of a fair coin, with $0$ meaning tails and $1$ heads. $A_n$ is then the event that the $n$th flip is tails.