The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero

Suppose, for contradiction, that claim 4 fails, and fix $(a_k)$, $(b_k)$ and $M$ as in [A1], so that $M<2^{-1}$.

$S$ is compact, claim 1. Each $S_n$ is the union of the finite list of closed sets $[e_j^{(n)},e_j^{(n)}+{\lambda }_n]$, $j<N_n$, hence closed by [L2]; so $S={\bigcap }_n{S}_n$ is closed by [L2], and $S\subseteq [0,1]$ is bounded by [L1] and [L2]; by [L3] it is compact.

Separation. For every $n$ and all $i\mathrm{\ne }j$ below $N_n$ one has $\mathrm{\mid }{e}_i^{(n)}-e_j^{(n)}\mathrm{\mid }>{\lambda }_n$, by induction on $n$ ([L9]). At $n=0$ there is nothing to prove, since $N_0=1$. Assume it at $n$ and let $i\mathrm{\ne }j$ below $N_{n+1}=N_n+N_n$. If both indices are $<N_n$, or both are $\ge N_n$, the two entries are $e_{i^{\mathrm{\prime }}}^{(n)}$ and $e_{j^{\mathrm{\prime }}}^{(n)}$ with $i^{\mathrm{\prime }}\mathrm{\ne }{j}^{\mathrm{\prime }}$, possibly both shifted by the same $g_n$, so the difference has absolute value $>{\lambda }_n>{\lambda }_{n+1}$ by [L1]. Otherwise the entries are $e_{i^{\mathrm{\prime }}}^{(n)}$ and $e_{j^{\mathrm{\prime }}}^{(n)}+g_n$; if $i^{\mathrm{\prime }}=j^{\mathrm{\prime }}$ the difference is $g_n>{\lambda }_{n+1}$ by [L1]; if $e_{i^{\mathrm{\prime }}}^{(n)}-e_{j^{\mathrm{\prime }}}^{(n)}>{\lambda }_n$ then $e_{i^{\mathrm{\prime }}}^{(n)}-e_{j^{\mathrm{\prime }}}^{(n)}-g_n>{\lambda }_n-g_n={\lambda }_{n+1}$, and if $e_{j^{\mathrm{\prime }}}^{(n)}-e_{i^{\mathrm{\prime }}}^{(n)}>{\lambda }_n$ then $e_{j^{\mathrm{\prime }}}^{(n)}+g_n-e_{i^{\mathrm{\prime }}}^{(n)}>{\lambda }_n>{\lambda }_{n+1}$, in each case by [L1] and [L10]. Consequently the pieces $[e_j^{(n)},e_j^{(n)}+{\lambda }_n]$, $j<N_n$, are pairwise disjoint.

Every endpoint lies in $S$. Fix $n$ and $j<N_n$. For $m\le n$ one has $e_j^{(n)}$ and $e_j^{(n)}+{\lambda }_n$ in $S_n\subseteq S_m$ by [L1]. For $m\ge n$, an induction on $m$ ([L9]) gives indices $j^{\mathrm{\prime }},j^{\mathrm{\prime }\mathrm{\prime }}<N_m$ with $e_{j^{\mathrm{\prime }}}^{(m)}=e_j^{(n)}$ and $e_{j^{\mathrm{\prime }\mathrm{\prime }}}^{(m)}+{\lambda }_m=e_j^{(n)}+{\lambda }_n$: at $m=n$ take $j^{\mathrm{\prime }}=j^{\mathrm{\prime }\mathrm{\prime }}=j$; and if they exist at $m$, then $e_{j^{\mathrm{\prime }}}^{(m+1)}=e_{j^{\mathrm{\prime }}}^{(m)}$ works for the left endpoint, while $e_{N_m+j^{\mathrm{\prime }\mathrm{\prime }}}^{(m+1)}+{\lambda }_{m+1}=e_{j^{\mathrm{\prime }\mathrm{\prime }}}^{(m)}+g_m+{\lambda }_{m+1}=e_{j^{\mathrm{\prime }\mathrm{\prime }}}^{(m)}+{\lambda }_m$ works for the right one, by [L1]. So both points lie in every $S_m$, hence in $S$.

The complement decomposes over the stages. $[0,1]\setminus S={\bigcup }_n(S_n\setminus S_{n+1})$. The inclusion $\supseteq$ holds because $S_n\subseteq S_0=[0,1]$ and $S\subseteq S_{n+1}$ by [L1]. For $\subseteq$, let $x\in [0,1]\setminus S$; then $x\in S_0$ and, $S$ being ${\bigcap }_m{S}_m$, the set of $m$ with $x\mathrm{\notin }{S}_m$ is nonempty, so by [L9] it has a least element $m_0$, and $m_0\ge 1$ since $x\in S_0$. Put $n:=m_0-1$; then $x\in S_n$ by minimality and $x\mathrm{\notin }{S}_{n+1}$.

The removed pieces. Fix $n$ and $j<N_n$. By [L1] the pieces $[e_j^{(n)},e_j^{(n)}+{\lambda }_{n+1}]$ and $[e_j^{(n)}+g_n,e_j^{(n)}+{\lambda }_n]$ both occur among the pieces of $S_{n+1}$, so a point $x$ of $[e_j^{(n)},e_j^{(n)}+{\lambda }_n]$ outside $S_{n+1}$ satisfies ${\lambda }_{n+1}<x-e_j^{(n)}<g_n$, that is $x\in M_j^{(n)}$; hence $S_n\setminus S_{n+1}\subseteq {\bigcup }_{j<N_n}{M}_j^{(n)}$. Conversely $M_j^{(n)}\cap S_{n+1}=\mathrm{\varnothing }$: a piece of $S_{n+1}$ coming from $i\mathrm{\ne }j$ lies in $[e_i^{(n)},e_i^{(n)}+{\lambda }_n]$, which is disjoint from $[e_j^{(n)},e_j^{(n)}+{\lambda }_n]\supseteq M_j^{(n)}$ by step 1.3, while the two pieces coming from $j$ itself are disjoint from the open interval $M_j^{(n)}$ by [L10]. Finally each $M_j^{(n)}$ has length $g_n-{\lambda }_{n+1}={\lambda }_n-2{\lambda }_{n+1}=4^{-n-1}$, so ${\sum }_{j<N_n}{4}^{-n-1}=2^n\cdot 4^{-n-1}=4^{-1}\cdot 2^{-n}$ by [L1].

$S$ is perfect, claim 2. $S$ is closed by step 1.2. Let $x\in S$ and let the real $\epsilon >0$ be given; by [L1] and [L8] fix $n$ with ${\lambda }_n\le 2^{-n}<\epsilon$. Since $x\in S_n$ there is $j<N_n$ with $x\in [e_j^{(n)},e_j^{(n)}+{\lambda }_n]$; the two endpoints of that piece lie in $S$ by step 1.4, are distinct because ${\lambda }_n>0$, and each is within ${\lambda }_n<\epsilon$ of $x$ by [L10]. So at least one of them is a point of $S\cap N_{\epsilon }(x)$ different from $x$, and $x$ is not isolated in $S$; by [L4], $S$ is perfect.

$S$ is nowhere dense, claim 3. $S$ is closed by step 1.2, so it equals its closure, and by [L4] it suffices that its interior be empty. Suppose $N_{\epsilon }(x)\subseteq S$ for some $x$ and some real $\epsilon >0$; fix $n$ with ${\lambda }_n\le 2^{-n}<\epsilon$ by [L1] and [L8], and $j<N_n$ with $x\in [e_j^{(n)},e_j^{(n)}+{\lambda }_n]$. The point $w:=e_j^{(n)}+({\lambda }_{n+1}+g_n)\cdot 2^{-1}$ lies in $M_j^{(n)}$, since ${\lambda }_{n+1}<g_n$, and hence in $[e_j^{(n)},e_j^{(n)}+{\lambda }_n]$, so $\mathrm{\mid }w-x\mathrm{\mid }\le {\lambda }_n<\epsilon$ and $w\in N_{\epsilon }(x)\subseteq S\subseteq S_{n+1}$; but $M_j^{(n)}\cap S_{n+1}=\mathrm{\varnothing }$ by step 2.1, which is impossible. So no neighbourhood is contained in $S$ and $S$ is nowhere dense.

A cover of $[0,1]$ built from [A1] and the removed pieces. By [L6] fix a bijection $J$ and define sequences $(c_i)$, $(d_i)$ as follows: for $i\in \mathbb{N}$ write $(m,t):=J^{-1}(i)$; if $m=0$ put $(c_i,d_i):=(a_t,b_t)$; if $m\ge 1$ and $t<N_{m-1}$ put $(c_i,d_i):=(e_t^{(m-1)}+{\lambda }_m,e_t^{(m-1)}+g_{m-1})$; and otherwise put $(c_i,d_i):=(0,0)$. Then $c_i\le d_i$ for every $i$ by [L1], and ${\bigcup }_i[c_i,d_i]$ contains $S$ by [A1] and contains $[0,1]\setminus S$ by steps 1.5 and 2.1, hence contains $[0,1]$. For a partial sum, fix $i_0$; the pairs $J^{-1}(i)$ with $i<i_0$ are distinct, so by [L9] there is $P$ bounding both of their coordinates, and since all the terms are nonnegative [L7] gives ${\sum }_{i<i_0}(d_i-c_i)\le {\sum }_{t\le P}(b_t-a_t)+{\sum }_{n\le P}{\sum }_{t<N_n}{4}^{-n-1}\le M+{\sum }_{n\le P}{4}^{-1}{2}^{-n}\le M+4^{-1}\cdot 2=M+2^{-1}$, using [A1], step 2.1, [L7] and [L8].

By [L5] applied to $[0,1]$ and the cover of step 3.2, $M+2^{-1}\ge 1-0=1$, so $M\ge 2^{-1}$, contradicting step 1.1. Claim 4 therefore holds; and $S$ is not null, since nullity would give, at $\epsilon :=4^{-1}$, a cover of $S$ with all partial total lengths $\le 4^{-1}<2^{-1}$, which claim 4 forbids. With steps 1.2, 2.2 and 3.1 all four claims are proved.