11
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Your first statement is not true. Have you heard of the Cantor set? It's uncountable but has measure zero. Also, the approach you should take is: Let $A\subseteq[a,b]$ have measure zero. Now you want to show $f(A)\subseteq f[a,b]$ has measure zero $\endgroup$ – Clayton Feb 12 '13 at 11:40
  • 2
    $\begingroup$ I think you first need to understand the definition of a set of measure $0$ and see some examples. It is true that a countable set has measure zero. But there are also uncountable sets of measure zero, perhaps the nicest is cantor set en.wikipedia.org/wiki/Cantor_set $\endgroup$ – Lior B-S Feb 12 '13 at 11:41
  • 1
    $\begingroup$ @Artus: Every nondegenerate closed interval contains a Cantor set. Thus, you have to take into account all possible sets of measure zero, and show that the image has measure zero. $\endgroup$ – Clayton Feb 12 '13 at 11:54
  • 2
    $\begingroup$ See this for the first part of the assertion. $\endgroup$ – David Mitra Feb 12 '13 at 13:08
  • 2
    $\begingroup$ For the second part of the assertion, use (and prove, as appropriate in your situation) these facts: Every $F_{\sigma}$ set is a countable union of compact sets, the image of a compact set under a continuous function is a compact set (and hence closed), and functions preserve (arbitrary, and hence countable) unions. $\endgroup$ – Dave L. Renfro Feb 12 '13 at 20:04
13
+50
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ How to show that and Lebesgue measurable set can be expressed as union of an $F_\sigma$ and a Null set? $\endgroup$ – Ravi Upadhyay Mar 24 '13 at 14:23
  • $\begingroup$ Ok, I solved it. Thanks @Martin! $\endgroup$ – Ravi Upadhyay Mar 24 '13 at 14:26
  • $\begingroup$ You can show that a Lebesgue measurable set may be expressed this way from Theorem 11 part iv (page 40) in Royden's Real analysis text. (This problem is from Royden). $\endgroup$ – Eric Sep 16 '14 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy