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The Lebesgue outer measure can be defined byA null set has $m^{\ast }(E)=0$, equivalently it admits such a cover of total length below every positive $\epsilon$.

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 $\epsilon$. Thus $m^{\ast }(C)\le (2/3{)}^n+\epsilon$ for every $n$ and $\epsilon >0$, proving $m^{\ast }(C)=0$.

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 givesFor $n\ge 1$ the telescoping product simplifies toBecause $K\subseteq K_n$, the same finite interval-enlargement argument gives $m^{\ast }(K)\le L_n$. Taking the limit provesOnly interval covers and their lengths were used; no continuity theorem for integrals is needed.

For the base case $n=2$, the product is $1-1/4=3/4=(2+1)/(2\cdot 2)$. Assume the identity holds for some $n\ge 2$. Multiplying by the next factor givesThis is the desired expression with $n$ replaced by $n+1$, so mathematical induction proves the identity for every $n\ge 2$. The factorization $1-i^{-2}=(i-1)(i+1)/i^2$ also explains the cancellation as a telescoping product.

At $n=2$, the product is $1-1/4=3/4$, as required. Suppose the formula holds at $n\ge 2$. Appending the next factor giveswhich is exactly the formula at $n+1$. Thus mathematical induction gives $(n+1)/(2n)$ for every $n\ge 2$. The factorization $1-i^{-2}=(i-1)(i+1)/i^2$ also explains the cancellation in this telescoping product.