Let $K$ a compact set. If $f:\mathbb{R}^n→\Bbb R^m$ is $C^1$ with $n<m$ then $f(K)$ has null measure.

Question

Let $n\lt m$. Let $K\subset\mathbb{R}^n$ compact and $U\subset\mathbb{R}^n$ open. If $f:U→ℝᵐ$ is $C^1$ then $f(K)$ has null measure. I stuck im this problem.

Answer 1

I assume that's supposed to be $n<m$.

One way to proceed is as follows. Define $\tilde f: \Bbb R^m \to \Bbb R^m$ by $\tilde f(x_1,\dots,x_m) = f(x_1,\dots,x_n)$. Now, note that $\mu(f(K) = \mu(\tilde f(K \times [0,1]^{m-n})) \leq \mu(K \times [0,1]^{m-n}) \max_{x \in K \times [0,1]^{m-n}} J(\tilde f(\mathbf x))$, where $J(\tilde f)$ denotes the absolute value of the Jacobian determinant.

Since $J(\tilde f(\mathbf x)) = 0$ for all $\mathbf x \in \Bbb R^m$, conclude that $\mu(f(K)) = 0$.

Answer 2

Hint: Special case: $f:[0,1]\to \mathbb R^2.$ Because $f\in C^1,$ $f$ is Lipschitz, so $|f(y)-f(x)|\le C|y-x|$ on $[0,1].$ It follows that $f([x,y])$ is contained in the disc with center $f(x)$ and radius $M|y-x|.$ The area measure of this disc is $\pi|y-x|^2.$ That will be much smaller than $|y-x|$ if $y$ is close to $x.$ So consider a partition of $[0,1]$ of small mesh size.