Definition
Let $A$ be a subset of $\Bbb R^n$. We say $A$ has measure zero in $\Bbb R^n$ iffor 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 hte rectangles $Q_1,Q_2,...$ is less than $\epsilon$.
Lemma
Let be $\mathfrak{X}:=\{X_i:i\in I\}$ a collection of topological spaces. So if $J\subseteq I$ then $X_J:=\Pi_{j\in J}X_j$ is embeddable in $X_I:=\Pi_{i\in I}X_i$.
Proof. See here.
Corollary
The subset $\Bbb R^m\times\{t_{m+1}\}\times...\times\{t_{m+(n-m)}\}$ of $\Bbb R^n$ is homeomorphic to $\Bbb R^m$ for any $m<n$.
Lemma
The subset $\Bbb R^m\times\{t_{m+1}\}\times...\times\{t_{m+(n-m)}\}$ of $\Bbb R^n$ has measure zero in $\Bbb R^n$.
Well the user Doctor Who showed to me here that this lemma is true in the case where $ m: = (n-1) $ so I tried to generalize the proof he gave.
Proof. So for any $\epsilon>0$ the collection $$ \mathcal C:=\{[-i,i]^m\times[t_{m+1}-\epsilon,t_{m+1}+\epsilon]\times...\times[t_n-\epsilon,t_n+\epsilon]:i=1,2,...\} $$ is clearly a rectangular and countable cover of $\Bbb R^m\times\{t_{m+1}\}\times...\times\{t_n\}$ that has total volume $(2i)^m\cdot (2\cdot \epsilon)^{n-m}$ so that if we find $\xi\in\Bbb R$ such that $$ (2i)^m\cdot\biggr(\frac{2\epsilon}{\xi}\biggr)^{n-m}=\frac{\epsilon}{2^{n-m}} $$ then clearly the collection $$ \mathcal Q:=\Biggr\{\big[-i,i\big]^m\times\biggr[t_{m+1}-\frac{\epsilon}\xi,t_{m+1}+\frac{\epsilon}\xi\biggr]\times...\times\biggr[t_n-\frac{\epsilon}\xi,t_n+\frac{\epsilon}\xi\biggr]:i=1,2,...\Biggr\} $$ is a rectangular and countable cover of $\Bbb R^m\times\{t_{m+1}\}\times...\times\{t_n\}$ of total volume less than $\epsilon$. So we pick $\xi=(2i)^m\cdot\epsilon^{n-m-1}$ and the lemma follows.
So I ask if my proof is correct and if not I ask to prove the last lemma. In particular I doubt that my proof is uncorrect because in the answer of the user Doctor Who the $\epsilon$ quantity was only at the numerator of $(n-m)$ last factors of each rectangule. So could someone help me, please?
