Sub-dimensional subspaces a null set 
Let $m<n$ and $f: \mathbb{R}^m \rightarrow \mathbb{R}^n$ be continuously differentiable. Show $$\lambda^n(f(\mathbb{R}^m))=0$$ and conclude from this, that every linear subspace $E$ of $\mathbb{R}^n$ with $\dim E < n$ is a null set as well.

The result seems rather obvious, yet I'm not able to prove it.
About the second part: If $m:=\dim E<n$, $E$ it's clearly isomorphic to $\mathbb{R}^m$. But is there some canonical isomorphism that satisfies $f(\mathbb{R}^m)=E$ and is continuously differentiable?
I tried to apply some isomorphism-argument for the first part, too, but didn't find it very helpful because of the generality of $f$...
 A: We can apply Sard's lemma, since $m<n$ and $f$ is $C^1$ we see that $\operatorname{rk} Df(x) < n$ for all $x$. Hence $f(\mathbb{R}^m)$ has measure zero.
If $E \subset \mathbb{R}^n$ is a subspace with dimension $m<n$, then let $b_1,...,b_m$ be a basis and define $f: \mathbb{R}^m \to \mathbb{R}^n$ by $f(x) = \sum_k x_k b_k$. $f$ is smooth, hence from the above we have $f(\mathbb{R}^m) = E$ has measure zero.
A: Since $\mathbb R^m$ is a countable union of cubes, it is enough to show that $\lambda^n(f(Q))=0$ for any closed cube $Q\subset \mathbb R^m$. 
Translating and dilating, we may assume that $Q$ is the unit cube $[0,1]^m$. For any $K\in\mathbb N$, the cube $Q$ can be decomposed into $K^m$ cubes $Q_i$ of size $\frac1K\cdot$ Now, $f$ is $\mathcal C^1$ and hence Lipschitz on $Q$; so each $f(Q_i)$ is contained in a cube of size $\leq C\times\frac1K$, for some absolute constant $C$. Hence, $f(Q)$ is contained in the union of $K^m$ cubes of size $\leq C\times\frac1K$, so that 
$$\lambda^n(f(Q))\leq K^m\times \left(\frac{C}K\right)^n:=\frac{C'}{K^{n-m}}\, .$$
Since $K\in\mathbb N$ is arbitrary, this shows that $\lambda^n(f(Q))=0$.
For the second part, just choose (as you did) any linear isomorphism $f:\mathbb R^m\to E$, which is of course $\mathcal C^1$ (being linear) when considered as a map from $\mathbb R^m$ into $\mathbb R^n$.
