As shown on title for a function continuous differentiable $f: \mathbb{R}^{n} \to \mathbb{R^n} $ holds: $A \subseteq \mathbb{R^n}$ is a Nullset then $f(A)$ is also a Nullset.
What can be idea to show this statement?
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This is not correct. There are functions from $\mathbb{R}$ to itself that map the Cantor set to a set with positive measure. – Hans Engler Nov 11 '15 at 21:48
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@HansEngler I've forgotten to write that f is continuous differentiable! – Melina Nov 11 '15 at 21:49
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Since $f(\bigcup_n E_n)=\bigcup_n f(E_n)$, you can localize the statement, namely, assume that the derivative is bounded. In this case, the map doesn't enlarge domains a lot (the ratio is related to the bound of the derivative). Could you elaborate a proof? – Yai0Phah Nov 11 '15 at 22:16
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@FrankScience No, because I haven't understood well the concept! :( – Melina Nov 11 '15 at 22:29
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1For the moment, you can do the following exercises: Assume first that $f$ is Lipschitz continuous, namely there exists $M>0$ such that for each $x,y$, we have $\lvert f(x)-f(y)\rvert\le M\lvert x-y\rvert$ (we take the Euclidean norm). 1. There's a constant $C$ that only depends on $M$, such that for each closed ball $B$, we have the outer measure of $f(B)$ is no larger than $C$ times the volume of $B$; 2. Do the same for cubes; 3. Conclude; 4. Now assume that $f$ is a general continuously differentiable function. Write $\mathbb R^n$ as a countable union, and conclude. – Yai0Phah Nov 12 '15 at 23:15
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The argument consists of three steps, as outlined by Frank Science.
- Restrict attention to the ball $B_n=\{x:\|x\|\le n\}$, and observe that $f$ is Lipschitz continuous on $B$ (since its gradient is bounded there).
- Apply the result that Lipschitz maps preserve sets of measure zero: see Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?
- Countable union of sets of measure zero also has measure zero, and $f(A)=\bigcup_n f(A\cap B_n)$.
