Any linear subspace has measure zero 
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 $\epsilon$.
Theorem
Let $A$ be open in $\Bbb R^n$; let $f:A\rightarrow\Bbb R^n$ be a function of class $C^1$. If the subset $E$ of $A$ has measure zero in $\Bbb R^n$, then the set $f[E]$ has also measure zero in $\Bbb R^n$.
Proof. See the lemma $18.1$ of the text "Analysis on Manifolds" by James Munkres.
Lemma
The subset $\Bbb R^m\times\{t_{m+1}\}\times...\times\{t_{m+(n-m)}\}$ of $\Bbb R^n$ has measure zero in $\Bbb R^n$.
Proof. See here.
Theorem
Any linear subspace $W$ of $\Bbb R^n$ that has dimension $m<n$ has measure zero.

Fortunately I arranged the following proof but I doubt there are some imperfections.

Proof. First of all if $W$ is a subspace of $\Bbb R^n$ of dimension $m<n$ then
$$
W\equiv\big<w_1,...,w_m\big>
$$
for some $w_1,...,w_m\in\Bbb R^m$ that are linearliy independent thus we have to show that the set of linear combination of these vectors has measure zero. Now if $\mathcal E:=\big\{e_1,...,e_n\big\}$ is the canonical base then we define the linear transformation $t:\Bbb R^n\rightarrow\Bbb R^n$ through the condition
$$
t(e_i):=\begin{cases}w_i,\,\,\,\text{if}\,\,\,i\le m\\0,\,\,\,\text{otherwise}\end{cases}
$$
for any $i=1,...,n$ so that $t\big[\Bbb R^n\big]=W$. So we extend the set $\big\{w_1,...,w_m\big\}$ to a basis $\mathcal W:=\big\{w_1,...,w_m,w_{m+1},...,w_n\big\}$ and then we consider the (linear) diffeomorphism $f$ of class $C^1$ defined trough the condition
$$
f(e_i):=w_i
$$
for all $i=1,...,n$. So if $f[W]$ has measure zero then $W$ has measure zero too. So since $f[W]=\Bbb R^m\times\{0\}^{n-m}$ the theorem holds.

So is my proof correct? Then unfortunately I don't be able to prove that $f[W]=\Bbb R^m\times\{0\}^{n-m}$. So could someone help me, please?
 A: Using the notation in your theorem, let $A = \mathbb{R}^n\subset \mathbb{R}^n$ so that $A$ is open and we search for a diffeomorphism on $A$ so that $\mathbb{R}^m\times\{0^{n-m}\}$ is mapped to $W$ where we assume without loss of generality that $\dim(W) = m$. Since $W$ is a subspace of $\mathbb{R}^n$ then we may find a basis for $W$ and label these vectors $\{w_1, \ldots w_m\}$. We may also find an additional $n-m$ vectors such that $\{w_1, \ldots w_m, w_{m+1}, \ldots w_{n}\}$ is a basis for $\mathbb{R}^n$. Let $\{e_1,\ldots e_n\}$ be the standard basis for $\mathbb{R}^n$. Consider the linear transformation defined by $$ f(e_i) = w_i$$
Then $f:\mathbb{R}^n\to\mathbb{R}^n$ is a linear bijection and thus is $C^1$. Notice that $E = span\{e_1\ldots e_m\} = \mathbb{R}^m\times\{0^{n-m}\}$ and that $$f(E) = span\{f(e_1),\ldots f(e_m)\} = span\{w_1,\ldots w_m\} = W $$
A: Not exactly an answer, but doesn't fit in an comment.
It is a consequence of a general result which is that if $p:\mathbb{R}^n \to \mathbb{R}$ is a polynomial then either $p=0$ or non zero almost everywhere.
There is a concise proof here.
If $W$ is a proper subspace of $\mathbb{R}^n$, then it is contained is some hyperplane $H$ and we can write $H= \{ x | \phi(x) = \alpha \}$ where $\phi$ is a non zero linear functional. Since the polynomial $p(x)=\phi(x)-\alpha$ is a non zero polynomial in $x_1,..,x_n$ we see that $H$ has measure zero.
