Modified Cantor set with telescoping remaining lengths ($L_n=\frac{1}{3}+\frac{2}{3\cdot 2^n}$)

Starting from $[0,1]$, remove the central fraction $1/(2^n+2)$ of every stage-$n$ interval. The total length obeys $L_{n+1}/L_n=(2^n+1)/(2^n+2)$, which telescopes to the displayed value. The limiting intersection has Lebesgue outer measure at most $1/3$ by covering it with the stage intervals and slightly enlarging them to open intervals. This argument uses no integration.