$f : B \to \mathbb R^n$ such that $\mathcal L^n(N)=0 \implies \mathcal L^n(f(N))=0$

Question

Let $B$ an open ball in $\mathbb R^n$. Let $f:B \to \mathbb R^n$ measurable and satisfy the property that $N \subset B, \mathcal L^n(N)=0 \implies \mathcal L^n(f(N))=0$, where $\mathcal L^n$ is the Lebesgue measure. We also know that

\[ \lim_{r \to 0} \frac{\operatorname{diam}(f(B_r(x)))}{r} =0 \]

for almost every $x \in B$, where $\operatorname{diam}$ gives the diameter of a set and $B_r(x)$ is the ball centered at $x$ with radius $r$.

Show that $\mathcal L^n(f(B))=0$.

I tried to use Vitali Covering Lemma with a fixed radius $r$, combined with the Isodiametric Inequality, but I didn't come to an end.

Thanks!

Answer 1

Let $m=\mathcal L^n$.

Let $E$ be the set of $x\in B$ such that $$\lim_{r \to 0} \frac{\operatorname{diam}(f(B_r(x)))}{r} =0.$$ There is a sequence of compact sets $K_n\subset E$ such that $$m\left(E\setminus\bigcup_nK_n\right)=0.$$Since $f$ maps null sets to null sets it's enough to show that $$m(f(K))=0$$for every compact $K\subset E$.

The hypothesis shows that $f$ is continuous on $K$. For $r>0$ define $\rho_r:K\to[0,\infty)$ by $$\rho_r(x)=\sup_{y\in K,0<|x-y|<r}\frac{|f(y)-f(x)|}{|x-y|}.$$The fact that $f$ is continuous on $K$ shows that $\rho_r$ is continuous. Since $\rho_r$ decreases to $0$ as $r\to0$ it follows that $\rho_r\to0$ uniformly on $K$ and we're done.