Let $\mu$ be the Lebesgue outer measure in $\mathbb{R}$. Find an example of measurable sets $A_n\subset[0,1]$ such that $\mu(A_n)>0$ for every $n\in\mathbb{N}$ and
- $\mu(A_n\bigtriangleup A_m)>0$ if $n\neq m$, where $A_n\bigtriangleup A_m=(A_n\setminus A_m)\cup(A_m\setminus A_n)$,
- $\mu(A_n\cap A_m)=\mu(A_n)\mu(A_m)$ if $n\neq m$.
Ok, what I thought: I know this sequence $(A_n)$ must satisfy $A_n\cap A_m\neq\emptyset$ if $n\neq m$; otherwise we would have $$ 0=\mu(\emptyset)=\mu(A_n\cap A_m)=\mu(A_n)\mu(A_m) \Longrightarrow \mu(A_n)=0 \quad\mbox{or}\quad\mu(A_m)=0, $$ which is contrary to our hypothesis: $\mu(A_n)>0$ for every $n\in\mathbb{N}$.
I built the Middle$-\beta$ Cantor sets as in this paper On Cantor-like sets and Cantor-Lebesgue Singular functions on page 7. So $$ \mu(C_\beta)=0,\mbox{ for every } 0<\beta<1 $$ On the other hand, denoting by $K_\beta = (C^\beta)^c\cap [0,1]\subset [0,1]$ the complement of the Middle$-\beta$ Cantor set in $ [0,1] $, it follows that $$ 0=\mu(C^\beta)=\mu([0,1]\setminus K_\beta)=\mu([0,1])-\mu(K_\beta). $$ Thus, $\mu(K_\beta) = 1 $ for all $ 0 <\beta <1 $.
Given $ 0 <\epsilon<\delta <1 $, by the construction of $ C^\epsilon $ and $ C^\delta$ follows that $C^\delta \subset C^\epsilon $ since $ C_k^\delta \subset C_k^\epsilon $ for all $ k\in\mathbb{N} $. Thus, \begin{align*} C^\delta\subset C^\epsilon \Longrightarrow (C^\epsilon)^c\subset (C^\delta)^c \Longrightarrow (C^\epsilon)^c\cap[0,1]\subset (C^\delta)^c\cap[0,1] &\Longrightarrow K_\epsilon\subset K_\delta\\ &\Longrightarrow K_\epsilon\setminus K_\delta=\emptyset. \tag{$\star$} \end{align*} Now let's prove that $ \mu (K_\delta \setminus K_\epsilon)> 0 $. We have \begin{align*} K_\delta\setminus K_\epsilon &= \left(\bigcup_{k=1}^\infty (C_k^\delta)^c\cap[0,1]\right)\setminus\left(\bigcup_{k=1}^\infty (C_k^\epsilon)^c\cap[0,1]\right)\\ &=\left(\bigcup_{k=1}^\infty (C_k^\delta)^c\cap[0,1]\right)\cap\left(\bigcup_{k=1}^\infty (C_k^\epsilon)^c\cap[0,1]\right)^c\\ &=\left(\bigcup_{k=1}^\infty (C_k^\delta)^c\cap[0,1]\right)\cap\left(\bigcap_{k=1}^\infty C_k^\epsilon\cup[0,1]^c\right)\\ &=\left(\bigcup_{k=1}^\infty (C_k^\delta)^c\cap[0,1]\right)\cap\left(C^\epsilon\cup[0,1]^c\right)\\ &=\left(\bigcup_{k=1}^\infty (C_k^\delta)^c\cap[0,1]\right)\cap C^\epsilon =\bigcup_{k=1}^\infty (C_k^\delta)^c\cap[0,1]\cap C^\epsilon \end{align*} Note that $$ (C_1^\delta)^c\cap[0,1]\cap C^\epsilon = \left(\frac{1}{2}(1-\delta),\frac{1}{2}(1-\epsilon)\right]\cup\left[\frac{1}{2}(1+\epsilon),\frac{1}{2}(1+\delta)\right)\subset \bigcup_{k=1}^\infty (C_k^\delta)^c\cap[0,1]\cap C^\epsilon, $$ whose measure is $\mu((C_1^\delta)^c\cap[0,1]\cap C^\epsilon)=2\cdot\frac{1}{2}(\delta-\epsilon)=\delta-\epsilon>0$. By the monotonicity of $ \mu $, it follows that $$ \mu(K_\delta\setminus K_\epsilon)=\mu\left(\bigcup_{k=1}^\infty (C_k^\delta)^c\cap[0,1]\cap C^\epsilon\right)\geq \mu((C_1^\delta)^c\cap[0,1]\cap C^\epsilon)>0. \tag{$\star\star$} $$ Now for every $ n\in\mathbb{N} $ define the sequence $ A_n = K_{1/n} \subset [0,1] $, we have $ \mu(A_n) = 1> 0 $ for all $ n\in\mathbb{N} $. Assuming without loss of generality that $ n>m $, it follows by $ (\star) $ that $ A_m\subset A_n$ and hence $$ \mu(A_n\cap A_m)=\mu(A_m)=1=1\cdot 1=\mu(A_n)\mu(A_m). $$ Also, for $ (\star\star) $ we have \begin{align*} \mu(A_n\bigtriangleup A_m)=\mu((A_n\setminus A_m)\cup(A_m\setminus A_n))&=\mu(A_n\setminus A_m)+\mu(A_m\setminus A_n)\\ &\overset{(\star)}{=}\mu(\emptyset)+\mu(A_m\setminus A_n)\\ &\overset{(\star\star)}{=}\mu(A_m\setminus A_n)>0. \end{align*}
I want to know if my example is correct and if you guys have other examples please show them.