2

I really need an help with the following exercise.

Suppose that $A\subseteq \mathbb R^d$ is Lebesgue measurable. Let $f\colon A \to \mathbb R^k $ be a Lipschitz function. Show that $f(A)$ is $H^d$ measurable in $\mathbb R^k$, where $H^d$ is the $d-$ dimensional Hausdorff measure.

Now the main problem is that I don't know how to deal with this kind of problems.

I know that I can suppose $d\leq k$, otherwise $H^d(f(A))=0,$ so $f(A)$ is $H^d$ measurable.

Moreover our professor suggested us to prove it before for a closed set $A$. But I don't understand this hint, I mean, I don't understand how to use it. I know that $H^d$ is a Borel measure, so if I show that $f(A)$ is borel, then $f(A)$ would be $H^d$ measurable. But is it true that $f(A)$ is a Borel set provided that $A$ is closed? I don't know. Maybe, if I assume that $A$ is also bounded (hence compact) then $f(A)$ is also compact, then closed, then $H^d$ measurable. But if $A$ is not bounded?

And also, if I prove the statement for $A$ closed, I would have no idea how to extend the result for a generic $A$.

Can someone help me?

batman
  • 2,165

1 Answers1

4

It is a standard fact that a Lebesgue measurable set can be written as the union of an $F_\sigma$ set and a set of measure zero. You can adjust this result rather easily to show that a measurable set $E$ has the form $$E = \left( \bigcup_j K_j \right) \cup N$$ where $\{K_j\}$ is a sequence of compact sets and $N$ has Lebesgue measure zero. A continuous function carries compact sets to closed sets, and a Lipschitz function carries a set of $d$-dimensional Lebesgue measure zero to a set of $d$-dimensional Hausdorff measure zero. Since $$ f(E) = \left( \bigcup_j f(K_j) \right) \cup f(N)$$ you get that $f(E)$ is a union of closed sets and a set of Hausdorff measure zero.

Umberto P.
  • 54,204