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?

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 $$f(E) = span\{f(e_1),\ldots f(e_m)\} = span\{w_1,\ldots w_m\} = W $$

I am not sure if the below approach is correct. Rudin Real and Complex Analysis book states the result asked in the question within the proof of Theorem 2.20 regarding the Lebesgue measure on ${\boldsymbol{R^{k}}}$. Hence trying to use only results upto this portion of the book.

Part (a) of Theorem 2.20 states that $m\left(W\right)=\text{vol}\left(W\right)$ for every $k$-cell $W$. Using this, we can wrap or cover every compact (closed and bounded) set, $E$, in the lower subspace, $Y$, with dimension, $m<k$, with a suitable $k$-cell from the higher dimensional space. The additional parameters for this $k$-cell in the higher dimension can be $-\frac{\epsilon}{2},\frac{\epsilon}{2}$ with $\epsilon>0$. This gives $m\left(E\right)=\text{vol}\left(E\right)\leq\epsilon^{k-m}\text{volm}$. Here, volm is the volume of the wrapper of $E$, in the $m$-dimensional sub space, $Y$.

Since $\epsilon$ was arbitraty, $m\left(E\right)=\text{vol}\left(E\right)=0$. The entire space (and hence the subspace) is $\sigma-$compact so it can be written as the countable union of compact sets (which are closed and bounded and hence covered like the set $E$ above) all of which have measure zero, so the measure of the entire subspace is zero.

Related Questions: Sub-dimensional linear subspaces of $\mathbb{R}^{n}$ have measure zero.

Lebesgue measure of a subspace of lower dimension is 0

Lebesgue measure of a subspace of lower dimension

Every subset of a subspace of $\mathbb{R}^n$ of dim $<n$ has measure 0

proof that the lebesgue measure of a subspace of lower dimension is 0.