Definition
Let $A$ be a subset of $\Bbb R^n$. We say $A$ has measure zero in $\Bbb R^n$ if for every $\epsilon>0$ there is a covering $Q_1,Q_2,...$ of $A$ by countably many rectangles such that $$ \sum_{i=1}^\infty v(Q_i)<\epsilon $$ If this inequality holds, we often say that the total volume of the rectangles $Q_1,Q_2,...$ is less than $\epsilon$.
Theorem
If $S$ is a subset of $\Bbb R^n$ that has measure zero then for any $\epsilon,\delta>0$ it can be covered by countably many closed cubes, each of width less than $\delta$, having total volume less than $\epsilon$.
Proof. See page 152 of Analysis on Manifolds by James Munkres.
Definition
Let $S$ be a bounded set in $\Bbb R^n$. If the constant function $1$ is integrable over $S$, se say that $S$ is rectifiable, and we define $(n-\text{dimensional})$ volume of $S$ by the equaction $$ v(S):=\int_S 1 $$ Note that this definition agrees with the definition of volume when $S$ is a rectangle.
Theorem
A subset $S$ of $\Bbb R^n$ is rectifiable if and only if $S$ is bounded and $\partial S$ has measure zero.
Proof. See page $112$ of Analysis on Manifolds by James Munkres.
Now studing the proof of change of variables theorem in the text "Analysis on Manifols" by James Munkres it seems that here it is used implicitly the following statement.
Statement
If $S$ is a rectifiable subset of $\Bbb R^{n-1}$ then the projection on $S':=S\cap(\Bbb R^{n-1}\times\{t\})$ on $\Bbb R^{n-1}$ is rectifiable too in $\Bbb R^{n-1}$.
So first of all we observe that $S'$ is rectifiable since $\Bbb R^{n-1}\times\{t\}$ is closed and so $\partial S'\subseteq\Bbb R^{n-1}\times\{t\}$ so that if the last set has measure zero (see here) then $\partial S'$ has measure zero. Then the restriction $\phi$ to $\Bbb R^{n-1}\times\{t\}$ of the projection map $\pi_{\Bbb R^{n-1}}$ of $\Bbb R^{n-1}\times\{t\}$ is a homeomorphism so that $\phi[\partial S']=\partial\big(\phi[S']\big)$ thus to prove the statement we have to prove that $\phi[\partial S']$ has measure zero but unfortunately I don't be able to do this. Indeed to prove this last thing I imaged that if $S$ is a subset of $\Bbb R^n$ that has measure zero then $S':=\pi_{_{\Bbb R^m}}[S]$ has zero measure in $\Bbb R^m$ for any $m<n$ but unfortunately this seems uncorrect because for example $\Bbb R^{n-1}\times\{t\}$ has measure zero but $\pi_{\Bbb R^{n-1}}\big[\Bbb R^{n-1}\times\{t\}\big]=\Bbb R^{n-1}$ has not measure zero. To follow for sake of completeness I show the passage of the proof that I don't understand.
So if $S_t$ is rectifiable then by previous lemma $F$ is here integrable and by the additivity property of integrale the last equality holds. So how prove that $S_t$ is rectifiable? So could someone help me, please?
