= Solution
{c}
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 $1/(2^n+2)$, so
$$
L_{n+1}=L_n\frac{2^n+1}{2^n+2},\qquad L_0=1.
$$
The product telescopes because $2^n+2=2(2^{n-1}+1)$, giving
$$
L_n=\frac13+\frac{2}{3\cdot2^n}.
$$
Since $K\subseteq K_n$, the same finite interval-cover argument gives $m^*(K)\leq L_n$ for every $n$. Taking the infimum proves
$$
\boxed{m^*(K)\leq1/3.}
$$
Only 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.
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