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Let $X\subset\mathbb{R}^n$ have zero measure in $\mathbb{R}^n$, and $f:X\to\mathbb{R}^n$ be locally Lipschitz (for every $x$ in domain there is neighbourhood of $x$ in which $f$ is Lipschitz). Then $f(X)$ also has zero measure in $\mathbb{R}^n$.

What I tried so far: let $\varepsilon>0$, then there is a countable family of (closed) rectangles $(Q_i)_{i\in\mathbb{N}}$ which covers $X$ and has total volume lesser than $\varepsilon$. Then the countable family $(f(Q_i))_{i\in\mathbb{N}}$ covers $f(X)$ but I don't know how to prove that it has total volume less than $\varepsilon$. I think I actually must consider a family of subsets of each $Q_i$ in which $f$ is Lipschitz. But I don't see how to connect the estimation of the volume of a rectangle with the Lipschitz condition.

Ain't No O
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2 Answers2

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Suppose first that $f$ is Lipschitz, say with constant $C>0$.

Claim: Given $x\in\mathbb{R}^n$ and $r>0$, $m(f(B_r(x))\le C^n m(B_r(0))$, where $m$ denotes the Lebesgue measure.

Proof: If $y\in B_r(x)$, then $|f(x)-f(y)|\le Cr$. So $f(B_r(x))\subset B_{Cr}(f(x))$. Now use that $m(B_{sr}(y))=s^nm(B_r(0))$ for any $s>0$ and $y\in\mathbb{R}^n$.

Now suppose $X\subset \bigcup_i B_i$ and $\sum_i m(B_i)<\varepsilon$, then $f(X)\subset \bigcup_i f(B_i)$. Say $B_i$ has radius $r_i$. Then $$ m(f(X))\le \sum_i m(f(B_i))\le C^n\sum_i m(B_i)<C^n\varepsilon $$ This completes the argument under the assumption that $f$ is Lipschitz.

Now, if $f$ is locally Lipschitz, it is Lipschitz on every compact set (see here). By the first part of the argument, the $f(X\cap B_n(0))$ has measure $0$ for every $n\in\mathbb{N}$. But then $f(X)=\bigcup_n f(X\cap B_n(0))$ has measure $0$, and we're done.

Reveillark
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  • My definition of a set $X$ having zero measure is just that they can be covered by countable families of rectangles of total volume as small as we want. It does not include this "measure function" as it appear in your claim. Can you explain more on that definition of measure? – AnalyticHarmony Jul 07 '20 at 16:31
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Isn't it true that, if $f: \Bbb R^n \rightarrow \Bbb R^n$ is Lipschitz with constant $\alpha$ and Q is a rectangle in $\Bbb R^n$, then $m(f(Q))\le \alpha^n m(Q)$? I think that must be true and that together the countable cover of measure less than $\epsilon$ should do it.

richard gayle
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