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Let $Q$ be a closed, $n$-dimensional rectangle/cuboid and $N \subseteq Q$.

Let $ U \supseteq Q $ be an open subset.

Let $\varphi:U \to \mathbb{R}^n$ be a continuously differentiable function.

First claim Show that there exists a constant $L \geq 0$ such that the image $\varphi(Q)$ is contained within a cuboid $Q_{\varphi}$, whose maximal side length is at most $L$ times as long as the longest side of $Q$

Second claim We call $Q$ cube-like if $$\max_{k = 1,..,n}(b_k-a_k)\leq 2 \min_{k = 1,..,n}(b_k-a_k)$$ ($a_k$ and $b_k$ are the coordinates of the $k$th side of the cuboid.)

Show that Jordan-null-sets can also be defined with closed cube-like cuboids instead of open cuboids.

Third claim Show that $\varphi(N)$ is a null-set if $N$ is a null set.



Hints

The exercise gives me the hint to show in 1. that $\varphi$ is Lipschitz continuous.

My thoughts on how to solve this:

I have no idea how I can show that $\varphi$ on $Q$ is Lipschitz continuous (I'd need your help here).

  1. no idea how to prove.

  2. We know that $N$ is a (Lebesgue) null set iff $\forall \epsilon > 0 \ \exists \{U_n \}_n (N \subset \bigcup_1 ^\infty U_n \ \land \ \sum_1^\infty vol(U_n) < \epsilon )$ where $U_i$ is an open cuboid, I need to show that closed cuboids work as well. I would do this using defining the closed cubes $\overline{U _i} \supseteq U_i \implies N \subset \bigcup_1 ^\infty \overline{U _i} $. When then need to somehow show that $ \sum _1 ^\infty vol(\overline{U _i}) < \epsilon$

  3. Maybe we can use the fact that in 2. we have shown that open cuboids, as well as closed cube-like cuboids, can be used to define null sets.

Your help is very appreciated since I don't even really know where to start.

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First claim. Since the function $\varphi$ is differentiable, it is continuous. Since the set $Q$ is compact, the set $f(Q)$ is compact too, therefore it is a bounded subset of $\Bbb R^n$.

Second claim. Each cuboid $Q$ (resp. cube $C’$) is contained in an interior a cube $C$ (resp. cuboid $Q’$) with parallel sides whose volume is at most $3^n$ bigger than the volume of the cuboid $Q$ (resp. cube $C’$). Applying this to finite covers of sets, we obtain the claim.

Third claim. I guess that Lipschitz continuity of $f$ on $Q$ follows from the continuity of its derivatives on a compact set $Q$.

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  • 1
    $\begingroup$ Third claim is implied by the fact that hausdorff dimension is conserved by Lipchitz image. $\endgroup$ – Diesirae92 Jan 25 '18 at 20:55

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