 Solution

The Lebesgue outer measure of $E$ is the infimum of ${\sum }_j|I_j|$ over countable coverings by open intervals. A set is a null set when this infimum is zero, equivalently when it has such a cover of total length less than every prescribed positive number.

The Cantor set is obtained from $[0,1]$ by deleting the open middle third of every remaining interval at each stage. After $n$ stages it lies inside $2^n$ closed intervals of length $3^{-n}$. Slightly enlarging these finitely many intervals to open intervals shows its outer measure is at most $(2/3{)}^n$ for every $n$, hence zero.

For the modified construction, let $L_n$ be the total length of the $2^n$ intervals in $K_n$. The removed fraction at stage $n$ is exactly $1/(2^n+2)$, soThe product telescopes because $2^n+2=2(2^{n-1}+1)$, givingSince $K\subseteq K_n$, the same finite interval-cover argument gives $m^{\ast }(K)\le L_n$ for every $n$. Taking the infimum provesOnly interval lengths and the definition of outer measure were used; the three numbered clauses in the source describe the construction rather than separate requests to prove additional statements.