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I saw a topic on the subject but I did not quite understand, and it was a bit old and I didn't want to resurrect it.

I am going in the right direction, I just need a little nudge.

let $f: \mathbb R^n \to \mathbb R^n$ be a smooth function $C^{\infty}$ (meaning $f$ is continuous, differentiable, and the differential of any order is also continuous).

let $E \subset \mathbb R^n$ be a set of measure zero.

Show that $f(E)$ is zero measure.

Hint: a $C^1$ function is lipchitz over a compact set.

What I did:

$E$ is of zero measure, so for all $\epsilon >0$ there are boxes $U_i$ such that $E \subseteq \cup_i U_i$ and $\sum_i V(U_i) \leq \epsilon$ (here V stands for volume in the sense of $\mathbb R^n$).

Let us define $m(A)$ as the measure of set $A$. It is clear to see that $f(E) \subseteq f(\cup_i U_i)$ and as such:

$m(f(E)) \leq m(f(\cup_i U_i))$

Now I assume that $f$ is lipchitz. and that means $m((f(\cup_i U_i)) \leq Km(\cup_i U_i)$

Because $m(\cup_i U_i) \leq \epsilon$ we get that $m(f(E)) \leq K\epsilon$ since $\epsilon$ can be as small as we like, I am tempted to say that $f(E)$ is of zero measure but i am not entirely sure.

But even if it is. I made the assumption that $f$ is lipchitz, but what if it isn't?

Oria Gruber
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1 Answers1

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Here's one solution to your problem: define $$ E_n = E \cap [-k,k]\times \cdots \times [-k,k] $$ Note that while $f$ might not be Lipschitz in general over $\Bbb R^n$, it is certainly Lipschitz over the compact "hyper-cube" $[-k,k]\times \cdots \times [-k,k]$, so that $f(E_k)$ is of measure zero. Now, note that $$ f(E) = \bigcup_{k \in \mathbb{N}} f(E_k) $$

Ben Grossmann
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