FALSE: in the substitution theorem the continuity of $f$ may be weakened to integrability, $f\circ\varphi$ still being integrable

$d_S \ge 0$ everywhere, $d_S(u) = 0$ for $u \in S$, and $d_S(u) > 0$ for $u \notin S$: $S$ is closed by [L1], so some $\rho>0$ has $(u-\rho,u+\rho)\cap S = \varnothing$, whence $|u-s| \ge \rho$ for every $s \in S$ and $d_S(u) \ge \rho$.

$d_S$ is continuous by [L3], hence integrable on every $[0,x]$ with $x>0$ by [L4]; so $\varphi$ is defined on $[0,1]$, and by [L5] it is differentiable at every point of $[0,1]$ with $\varphi' = d_S$, which is integrable by [L4].

$\varphi$ is strictly increasing, hence injective. For $0 \le x < y \le 1$, $\varphi(y)-\varphi(x) = \int_x^y d_S \ge 0$ by [L5] and [L7]; if it were $0$ then [L6] would force $d_S \equiv 0$ on $[x,y]$, so $[x,y] \subseteq S$ by step 1.1, contradicting [L2]. Hence $\varphi(x)<\varphi(y)$. In particular $\varphi(0)=0<\varphi(1)$.

A quadratic contraction on $S$. Let $x<y$ both lie in $S$. For $u \in [x,y]$ one has $d_S(u) \le \min\{u-x,\ y-u\} \le (y-x)\cdot 2^{-1}$, since $x,y \in S$; so by [L5] and [L7], $0 \le \varphi(y)-\varphi(x) = \int_x^y d_S \le (y-x)^{2}\cdot 2^{-1}$.

$\varphi[S]$ has content zero. Fix $N \ge 1$ and for $i<N$ put $J_i := [\iota(i)/\iota(N),\ \iota(i+1)/\iota(N)]$, so the $J_i$ cover $[0,1]$ and each has length $1/\iota(N)$. If $S \cap J_i \ne \varnothing$ let $E_i := \varphi[S \cap J_i]$, a nonempty bounded set, and put $a_i := \inf E_i$, $b_i := \sup E_i$; otherwise put $a_i := b_i := 0$.

For $z,w \in E_i$ one has $|z-w| \le \bigl(1/\iota(N)\bigr)^{2}\cdot 2^{-1}$ by step 3.2, the two preimages lying in $S \cap J_i$; hence $b_i \le a_i + \bigl(1/\iota(N)\bigr)^{2}\cdot 2^{-1}$, since every $z \in E_i$ is at most $w + (1/\iota(N))^{2}2^{-1}$ for each fixed $w$, and then $w \ge b_i - (1/\iota(N))^{2}2^{-1}$ for every $w$.

Every point of $\varphi[S]$ lies in some $[a_i,b_i]$, because every point of $S$ lies in some $J_i$; and $\sum_{i<N}(b_i-a_i) \le \iota(N)\cdot \bigl(1/\iota(N)\bigr)^{2}\cdot 2^{-1} = 1/\bigl(2\,\iota(N)\bigr)$ by [L10].

Given $\varepsilon>0$, [L10] supplies $N \ge 1$ with $1/(2\iota(N)) \le \varepsilon$; so $\varphi[S]$ has content zero and therefore measure zero by [L9].

$f := \mathbf{1}_{\varphi[S]}$ is integrable on $J = [0,\varphi(1)]$. It is bounded, with values in $\{0,1\}$. $S$ is compact by [L1] and $\varphi$ is continuous by [L5] and [L3], so $\varphi[S]$ is compact, hence closed, by [L8]; therefore at every $z \in J \setminus \varphi[S]$ some neighbourhood misses $\varphi[S]$ and $f$ vanishes on it, so $f$ is continuous there. The discontinuity set of $f$ is thus contained in $\varphi[S]$, which is null by step 7.1, so $f$ is integrable by [L9].

$f\circ\varphi = \mathbf{1}_{S}$ on $[0,1]$. For $x \in [0,1]$: if $x \in S$ then $\varphi(x) \in \varphi[S]$ and $f(\varphi(x)) = 1$; if $x \notin S$ then $\varphi(x) \notin \varphi[S]$, since $\varphi$ is injective by step 3.1, and $f(\varphi(x)) = 0$. Also $\varphi[\,[0,1]\,] \subseteq J$ by step 3.1.

$\mathbf{1}_{S}$ is discontinuous at every point of $S$. Let $x \in S$ and $\rho>0$; the set $(x-\rho,x+\rho)\cap(0,1)$ contains a nonempty open interval, which by [L2] is not contained in $S$, so some $y$ in it has $\mathbf{1}_S(y) = 0$ while $\mathbf{1}_S(x)=1$; no $\delta$ works for $\varepsilon = 2^{-1}$. At $x \notin S$ the function vanishes on a neighbourhood, $S$ being closed, so it is continuous there.

The discontinuity set of $f\circ\varphi$ on $[0,1]$ is therefore exactly $S$, which is not null by [L1]; so $f\circ\varphi$ is bounded and not Riemann integrable, by [L9].

So $\varphi$ is differentiable on $[0,1]$ with $\varphi'$ integrable, $f$ is integrable on an interval containing $\varphi[\,[0,1]\,]$, and $f\circ\varphi$ is not integrable: the claim is false, and the continuity hypothesis on $f$ in Substitution: if $\varphi$ is differentiable on $[c,d]$ with $\varphi'$ integrable and $f$ is continuous on an interval containing $\varphi([c,d])$, then $\int_{\varphi(c)}^{\varphi(d)} f = \int_c^d (f\circ\varphi)\,\varphi'$ cannot be weakened to integrability.