= Solution
The <Lebesgue outer measure> can be defined by
$$
m^*(E)=\inf\left\{\sum_{j=1}^{\infty}|I_j|:E\subseteq\bigcup_{j=1}^{\infty}I_j,\ I_j\text{ open intervals}\right\}.
$$
A <null set> has $m^*(E)=0$, equivalently it admits such a cover of total length below every positive $\varepsilon$.
The usual <Cantor set> is the <intersection> of the sets obtained from $[0,1]$ by repeatedly removing every open middle third. At stage $n$ there are $2^n$ closed intervals of length $3^{-n}$. Enlarge them slightly to open intervals whose total added length is below $\varepsilon$. Thus $m^*(C)\leq(2/3)^n+\varepsilon$ for every $n$ and $\varepsilon>0$, proving $m^*(C)=0$.
For the different deletion rule in this problem, let $L_n$ be the total length of the stage-$n$ intervals. Removing the stated fraction from each component gives
$$
L_{n+1}=L_n\frac{2^n+1}{2^n+2},\qquad L_0=1.
$$
For $n\geq1$ the <telescoping product> simplifies to
$$
L_n=\prod_{j=0}^{n-1}\frac{2^j+1}{2^j+2}
=\frac23\prod_{j=1}^{n-1}\frac{2^j+1}{2(2^{j-1}+1)}
=\frac{2^{n-1}+1}{3\,2^{n-1}}.
$$
Because $K\subseteq K_n$, the same finite <interval>-enlargement argument gives $m^*(K)\leq L_n$. Taking the limit proves
$$
\boxed{m^*(K)\leq\frac13}.
$$
Only <interval> covers and their lengths were used; no <continuity> theorem for integrals is needed.
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