Proposition 6.5 from Lee's Introduction to Smooth Manifolds is the proposition that if $A \subseteq \mathbb{R}^n$ has measure zero and $F : A \rightarrow \mathbb{R}^n$ is a smooth function, then $F(A)$ also has measure zero.
In the proof, we use the following: "Since $\overline{U}$ is compact $C$ such that $|DF(x)| \leq C$ for all $ x\in \overline{U}$". Initially my intuition for this statement was that F is bounded on a compact set, so its partial derivatives are bounded, so the determinant of the jacobian |DF(a)| is bounded, but I'm not sure if that makes much sense.
I'd appreciate any hints on how I can get started proving that the determinant of the jacobian is bounded on a compact set.
Note that the expression $|DF(x)|$ denotes the Frobenius norm, not the determinant. See Example B.47 in my smooth manifolds book.
It certainly does not follow from the fact that $F$ is bounded that its partial derivatives are also bounded. But because $F$ is smooth, the matrix entries of $DF(x)$ (which are just the partial derivatives of the component functions of $F$) are continuous functions of $x$. Thus $|DF(x)|$ is a continuous real-valued function of $x$, so it achieves a maximum value on the compact set $\overline U$.