4

Let $f: [a,b] \rightarrow \mathbb{R}$ be a Lipschitz function.

Let $\epsilon > 0$.

Let $E$ be a set of measure zero. There exists countable, bounded, open intervals with the form $I_n=(x_n, y_n)$ such that $E \subseteq \bigcup\limits_n I_n$ and for some $x_k, y_k \in I_k$, then $|f(x_k)-f(y_k)| \leq c|x_k -y_k|$. I believe this sufficiently incorporates the definition of Lipshitz.

Now I want to cover $E$ with the intervals such that $\sum\limits_n l(I_k) < \frac{\epsilon}{c}$.

So, for every n we pick $I_{n}^{'} \subset f(I_n)$ which are open, bounded, and countable such that $l(I_{n}^{'}) \leq c l(I_{n})$. The collection $\{I_{n}^{'}\}$ is a cover for $f(E)$ and $\sum\limits_n l(I_{n}^{'}) \leq \sum\limits cl(I_{n})< c\frac{\epsilon}{c} =\epsilon$.

emka
  • 6,646
  • 2
    I can't easily read the proof you've offered, but I can tell you that not every uniformly continuous function sends measure 0 sets to measure 0 sets. There may be simpler counterexamples, but the two I think of are the Peano curve and the continuous function from a fat Cantor set to the standard one. So if you haven't applied Lipschitz, the proof is wrong. – Kevin Carlson Oct 03 '12 at 01:27
  • I don't follow: -what your intent is in covering $E$ with intervals $(x_n,y_n)$ on which you place no conditions -what the function $l$ is, if not length, but if it is, how you intend to find a set of intervals ${\bar{I}_k}$ contained in the countable set $f(l({I_n}))$. – Kevin Carlson Oct 03 '12 at 01:30
  • 1
    This may be helpful: http://math.stackexchange.com/questions/139883/why-does-a-lipschitz-function-f-mathbbrd-to-mathbbrd-map-measure-zero-s – Nana Oct 03 '12 at 01:32
  • The reason this is true is that Lipschitz continuity implies absolute continuity, which is essentially "continuous with the property that sets of measure zero are mapped to sets of measure zero". Try chasing down the definition of absolute continuity to prove that absolutely continuous functions map sets of measure zero to sets of measure zero, then show that Lipschitz implies absolute continuity. – Chris Janjigian Oct 03 '12 at 01:32
  • @KevinCarlson I believed I fixed it and made it clearer. – emka Oct 03 '12 at 02:00
  • @Chris while it's true that Lipschitz implies absolutely continuous, I think the "chasing down the definition of absolute continuity" part of your program is the less trivial of the two proposed steps, and it's the one emka has accomplished here in all but name. – Kevin Carlson Oct 03 '12 at 03:27
  • @KevinCarlson I wrote it up just to be sure. I don't think the proof is meaningfully different between absolute continuity and Lipschitz continuity, but with absolute continuity it feels a bit more natural to me. Anyway fair enough if you disagree. – Chris Janjigian Oct 03 '12 at 04:21

2 Answers2

3

Here is the proof for absolutely continuous functions, which Lipschitz functions are a special case of.

Let $f$ be an absolutely continuous function on $\mathbb{[a,b]}$ and let $A$ be a set of measure zero. By outer regularity, for any $\delta > 0$ there exists a countable disjoint set of open intervals $(a_i,b_i)$ so that $A\subseteq \cup_i (a_i,b_i)$ and $\sum_i b_i - a_i < \delta$.

Now fix $\epsilon > 0$, by absolute continuity there is a $\delta > 0$ so that for any finite collection of disjoint intervals $(x_i, y_i)$ with $\sum y_i - x_i < \delta$ we have $\sum_i |f(y_i) - f(x_i)| \leq \epsilon$. It actually doesn't change anything by taking the collection to be countable (take limits of the finite sums) so long as the sum of the lengths of the partition remains less than $\delta$.

Here's where Lipschitz is nice. For general absolutely continuous functions, notice that we can without loss of generality assume that the maximum occurs at the right endpoint and the minimum at the left endpoint, otherwise there is a point $z_i$ inside $[x_i,y_i]$ so that $f(z_i)$ is minimal on that interval and a corresponding point for the maximum, so without loss of generality we can replace $[x_i,y_i]$ with the interval of those two points in the definition of absolute continuity and nothing changes. It follows then that $$ \mu(f(A)) \leq \mu(\cup f([x_i,y_i]))\leq \sum_i |f(y_i) - f(x_i)| \leq \epsilon$$

$\epsilon$ is arbitrary so the result follows.

Chris Janjigian
  • 9,678
  • 4
  • 32
  • 46
1

OK, now this almost works. You can't necessarily cover $f(E)$ with open intervals contained in $f(I_n)$: consider the constant functions, which are very much Lipschitz, but whose images contain no interval. But as you know you can approximate $f(I_k)$ from the outside as well as you please by open sets, so do that instead.

That's the substantive point, but here's one on exposition, if you're interested: the way you define the $I_n$ makes it seem like you've fixed them once and for all, when in fact later on you require the sum of their lengths to be less than $\frac{\epsilon}{c}$. Instead, you might say "Then there exists a countable set of open intervals $\{I_k\}$ with $\sum_{k=1}^\infty l(I_k)<\frac{\epsilon}{c}$," and go on from there.

Kevin Carlson
  • 54,515
  • Does this mean that my proof is not substantially incorrect, but incorrect because of my order of exposition? – emka Oct 03 '12 at 04:52
  • The exposition bit isn't an error, just a suggestion. The first paragraph is regarding an actual error. – Kevin Carlson Oct 03 '12 at 05:10
  • What do you mean? Since $f(I_k)$ is measurable I find an $O_k$ such that $f(I_k) \subset O_k$. This using the outer approximation idea. – emka Oct 03 '12 at 05:19
  • Yeah, but this $O_k$ seems to correspond to $I'_n$ in your question, but you wrote $I'_n\subset f(I_n)$ instead of $f(I_n)\subset I'_n$. Maybe that was just a typo. – Kevin Carlson Oct 03 '12 at 05:30
  • Let me make sure I understand, in general I can't take the second set of covers, the $I_{n}^{'}\subseteq f(I_n)$? I realize that my type was using $\subset$ instead of $\subseteq$. – emka Oct 03 '12 at 06:03
  • No, using $\subset$ instead of $\supset$. There's not necessarily any open $I'_n \subset f(I_n)$. – Kevin Carlson Oct 03 '12 at 06:06
  • Continuity preserves open sets. Do you have an example of this not being true? – emka Oct 03 '12 at 06:42
  • @emka the function 1 maps an open set (the real line) to a closed set (the point {1}) – Chris Janjigian Oct 03 '12 at 11:42
  • If I just switch $\subset$ to $\supset$ would that complete the proof?

    I have a theorem that allows me to approximate the measure of a set $E$ by open sets $O$ such that $m^*(O\setminus E) < \epsilon$.

     – emka Oct 03 '12 at 17:00
  • Essentially: since the $O_k$ will be a little bigger than the $f(I_k)$, you should start with $\sum l(I_k)<\frac{\epsilon}{2c}$, say, so that the $O_k$ can have lengths summing to $\frac{\epsilon}{2}+\epsilon'<\epsilon$ since $\epsilon'$ is arbitrarily small.

    You should think carefully about the examples both Chris and I offered of continuous functions which don't preserve open sets, and find more of your own, if you're planning to study much more math. It's of utmost importance to realize that continuous functions only preserve open sets in reverse, i.e. $f^{-1}(O)$ is always open.

     – Kevin Carlson Oct 03 '12 at 18:16
  • Asserting that continuous functions preserve openess was a gut reaction.
    Anyway, any two points in $f(I_n)\leq cl(I_n)$ due to the Lipschitz property. This is why I presumed that the cover of $f(E)$ would be $I_{k}^{'} \subset f(I_k)$.     – emka Oct 03 '12 at 19:39