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 $[e^{(n)}_j, e^{(n)}_j + \lambda_n]$, $j < N_n$, 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 $i \ne j$ below $N_n$ one has $|e^{(n)}_i - e^{(n)}_j| > \lambda_n$, by induction on $n$ ([L9]). At $n = 0$ there is nothing to prove, since $N_0 = 1$. Assume it at $n$ and let $i \ne j$ below $N_{n+1} = N_n + N_n$. If both indices are $< N_n$, or both are $\ge N_n$, the two entries are $e^{(n)}_{i'}$ and $e^{(n)}_{j'}$ with $i' \ne j'$, 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 $e^{(n)}_{i'}$ and $e^{(n)}_{j'} + g_n$; if $i' = j'$ the difference is $g_n > \lambda_{n+1}$ by [L1]; if $e^{(n)}_{i'} - e^{(n)}_{j'} > \lambda_n$ then $e^{(n)}_{i'} - e^{(n)}_{j'} - g_n > \lambda_n - g_n = \lambda_{n+1}$, and if $e^{(n)}_{j'} - e^{(n)}_{i'} > \lambda_n$ then $e^{(n)}_{j'} + g_n - e^{(n)}_{i'} > \lambda_n > \lambda_{n+1}$, in each case by [L1] and [L10]. Consequently the pieces $[e^{(n)}_j, e^{(n)}_j + \lambda_n]$, $j < N_n$, are pairwise disjoint.

Every endpoint lies in $S$. Fix $n$ and $j < N_n$. For $m \le n$ one has $e^{(n)}_j$ and $e^{(n)}_j + \lambda_n$ in $S_n \subseteq S_m$ by [L1]. For $m \ge n$, an induction on $m$ ([L9]) gives indices $j', j'' < N_m$ with $e^{(m)}_{j'} = e^{(n)}_j$ and $e^{(m)}_{j''} + \lambda_m = e^{(n)}_j + \lambda_n$: at $m = n$ take $j' = j'' = j$; and if they exist at $m$, then $e^{(m+1)}_{j'} = e^{(m)}_{j'}$ works for the left endpoint, while $e^{(m+1)}_{N_m + j''} + \lambda_{m+1} = e^{(m)}_{j''} + g_m + \lambda_{m+1} = e^{(m)}_{j''} + \lambda_m$ 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 $x \notin S_m$ 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 $x \notin S_{n+1}$.

The removed pieces. Fix $n$ and $j < N_n$. By [L1] the pieces $[e^{(n)}_j, e^{(n)}_j + \lambda_{n+1}]$ and $[e^{(n)}_j + g_n,\ e^{(n)}_j + \lambda_n]$ both occur among the pieces of $S_{n+1}$, so a point $x$ of $[e^{(n)}_j, e^{(n)}_j + \lambda_n]$ outside $S_{n+1}$ satisfies $\lambda_{n+1} < x - e^{(n)}_j < g_n$, that is $x \in M^{(n)}_j$; hence $S_n \setminus S_{n+1} \subseteq \bigcup_{j<N_n} M^{(n)}_j$. Conversely $M^{(n)}_j \cap S_{n+1} = \varnothing$: a piece of $S_{n+1}$ coming from $i \ne j$ lies in $[e^{(n)}_i, e^{(n)}_i + \lambda_n]$, which is disjoint from $[e^{(n)}_j, e^{(n)}_j + \lambda_n] \supseteq M^{(n)}_j$ by step 1.3, while the two pieces coming from $j$ itself are disjoint from the open interval $M^{(n)}_j$ by [L10]. Finally each $M^{(n)}_j$ 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 $\varepsilon > 0$ be given; by [L1] and [L8] fix $n$ with $\lambda_n \le 2^{-n} < \varepsilon$. Since $x \in S_n$ there is $j < N_n$ with $x \in [e^{(n)}_j, e^{(n)}_j + \lambda_n]$; the two endpoints of that piece lie in $S$ by step 1.4, are distinct because $\lambda_n > 0$, and each is within $\lambda_n < \varepsilon$ of $x$ by [L10]. So at least one of them is a point of $S \cap N_\varepsilon(x)$ 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 $N_\varepsilon(x) \subseteq S$ for some $x$ and some real $\varepsilon > 0$; fix $n$ with $\lambda_n \le 2^{-n} < \varepsilon$ by [L1] and [L8], and $j < N_n$ with $x \in [e^{(n)}_j, e^{(n)}_j + \lambda_n]$. The point $w := e^{(n)}_j + (\lambda_{n+1} + g_n) \cdot 2^{-1}$ lies in $M^{(n)}_j$, since $\lambda_{n+1} < g_n$, and hence in $[e^{(n)}_j, e^{(n)}_j + \lambda_n]$, so $|w - x| \le \lambda_n < \varepsilon$ and $w \in N_\varepsilon(x) \subseteq S \subseteq S_{n+1}$; but $M^{(n)}_j \cap S_{n+1} = \varnothing$ 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 $(c_i,d_i) := \big(e^{(m-1)}_t + \lambda_{m}, \ e^{(m-1)}_t + g_{m-1}\big)$; 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 $\varepsilon := 4^{-1}$, 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.