Let the function $f:[a,b] \to \mathbb R$ be Lipschitz. Show that $f$ maps a set of measure zero onto a set of measure zero

Let the function $f:[a,b] \to \mathbb R$ be Lipschitz, that is, there is a constant $c \geq 0$ such that for all $u,v \in [a,b]$, $|f(u)-f(v)| \leq c|u-v|$. Show that $f$ maps a set of measure zero onto a set of measure zero. Show that $f$ maps an $F_\sigma$ set onto an $F_\sigma$ set. Conclude that $f$ maps a measurable set to a measurable set.

I have a question about this. We know that a set of measure zero must either be of the form [0], or it must be a set of countable elements. But for this function, we know that the domain is an interval...and we cannot have an interval with only countable elements, because an interval contains all reals, right? So for the first part of this question, we need to prove that [0] is mapped to [0], right?

Thanks in advance

Solution 1:

If $F \subset [a,b]$ is an $F_{\sigma}$-set then $F = \bigcup_{n=1}^{\infty} K_n$ with $K_n$ compact (since closed subsets of $[a,b]$ are compact). By continuity of $f$ the set $f(F) = \bigcup_{n=1}^{\infty} f(K_n)$ is the countable union of the compact sets $f(K_n)$ and hence $f(F)$ an $F_\sigma$-set, so $f(F)$ is measurable.

If $N \subset [a,b]$ is a null set then for every $\varepsilon \gt 0$ there are intervals $I_n$ such that $N \subset \bigcup_{n=1}^\infty I_n$ and $\sum_{n=1}^\infty \lvert I_n\rvert \lt \varepsilon$. Now $f(I_n)$ is an interval of length at most $c|I_n|$ since $f$ is Lipschitz continuous, so $f(N) \subset \bigcup_{n=1}^\infty f(I_n)$ has outer measure at most $\sum_{n=1}^\infty c|I_n| \lt c\varepsilon$. Therefore $f(N)$ is a null set.

If $L$ is an arbitrary Lebesgue measurable set, then $L = F \cup N$ with $F$ an $F_\sigma$-set and $N$ a null set. Then $f(L) = f(F) \cup f(N)$ is the union of the $F_\sigma$-set $f(F)$ and the null set $f(N)$, so $f(L)$ is measurable.