Let $n = p + 1$ for $G, g_0, p, b$ as given in the hypothesis. We construct our Cantor set by removing a diminishing proportion of sets available. The construction will be accomplished using two techniques. First, a ``conversational" approach, then one with more mathematical symbolism, or ``rigor". === Step 1 === By definition of the Cantor collection we express $g_0$ as the union (with possible exception of a set of measure zero) of $n$ pairwise disjoint sets from $G$. These are sets of type $1$. Remove one of them, leaving $n - 1$ sets remaining each with measure $\le b / n$. The proportion of sets removed is $1 / n$ and the measure of the set removed is $\le b/n$. === Step 2 === Express each of the $n - 1$ sets of type $1$ as the union of $n^2$ disjoint sets from $G$. We now have $\paren {n - 1} n^2$ sets. These are sets of type $2$. Remove from each of the $n - 1$ sets of type $1$ one of the type $2$ sets. The proportion of sets removed is: :$\dfrac {n - 1} {n^2 \paren {n - 1} } = \dfrac 1 {n^2}$ And the measure of the union of the removed sets is $\le \paren {n - 1} b / n^3 \le b / n^2$. These two steps suggest an induction argument for: === Lemma === Given the sets and conditions of the Theorem hypothesis. Let $m$ be a natural number. Then we can remove from $g_0$ a pair-wise disjoint collection of sets whose total measure is $\le b / n^m$. === Proof === The preceding  steps have established the first part of the inductive argument. ==== Step $k - 1$ ==== Assume that there are $N$ sets of type $k - 1$ each with measure $m_0$. Remove (from the type $k - 1$ sets) the proportion $\dfrac 1 {n^{k - 1} }$. The measure of the removed sets is: :$M_{\left({k - 1}\right)} = \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right) N m_0 \le \dfrac b {n^{\left({k - 1}\right)} }$ ==== Step $k$ ==== After step $\left({k - 1}\right)$ there are  $\left({1 - \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right)}\right) N$ sets of type $\left({k - 1}\right)$. Expressing each of them as the union of n$^k$ disjoint subsets of $G$ yields: :$n^k \left({1 - \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right)}\right) N$ sets of type $k$, each with measure $\dfrac {m_0} {n^k}$. Removing one type $k$ set from each type $\left({k - 1}\right)$ set results in the removal of: :$\left({1 - \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right)}\right) N$ type $k$ sets. Thus the proportion removed is $\dfrac 1 {n^k}$. Let $M_{\left({k - 1}\right)}$ be the measure of the union of the sets removed at step $\left({k - 1}\right)$. Let $M_k$ be the measure removed at step $k$. Since: {{begin-eqn}} {{eqn | l = M_k       | r = \left({1 - \left({\dfrac 1 {n^{\left({k - 1}\right)} } }\right)}\right) N \dfrac {m_0} {n^k}       | c =  }} {{eqn | o = \le       | r = N \dfrac {m_0} {n^k}       | c =  }} {{eqn | r = \dfrac 1 n \dfrac 1 {n^{\left({k - 1}\right)} } N m_0       | c =  }} {{eqn | r = \dfrac {M_{\left({k - 1}\right)} } n       | c =  }} {{eqn | r = \dfrac b {n^k}       | c =  }} {{end-eqn}} {{explain|Since ... what? The above is linguistically a subordinate clause with nothing for it to be subordinate to.}} By the Principle of Mathematical Induction, if $m$ is a natural number, there exists a pairwise disjoint subcollection of $G$ with total measure $\le \dfrac b {n^m}$ that can be removed from $g_0$. Thus the Lemma is proved. {{qed|lemma}} Let $P$ denote the union of all the removed sets. Let $N_p = g_0 - P$. The proof that $N_p$ and $P$ satisfy the theorem will be presented following the rigorous construction of them.