3
$\begingroup$

The problem is:

Let $f$ be Lipschitz with constant $M$. Show that if $E \subseteq [0,1]$ is of measure zero, then $f(E)$ is of measure zero.

My attempt:

Let $\epsilon > 0$, and choose a countable covering of $E$ by open intervals $\{U_i\}_{i=1}^\infty$ such that $\sum_{i=1}^\infty |U_i| < \epsilon/M$.

By the intermediate value theorem, $V_i := f((a_i,b_i))$ is also an interval, and for any two points $f(x), f(y) \in V_i$, we have $|f(x) - f(y)| \leq M|x-y| < M|U_i|$. Hence $|V_i| < M|U_i|$, and $$ \sum_{i=1}^\infty |V_i| < M \sum_{i=1}^\infty |U_i| = M \cdot \frac{\epsilon}{M} = \epsilon. $$ Since the $V_i$ cover $E$, we conclude that $f(E)$ is of measure zero.

Comments: I have made two large assumptions here: 1) that the $V_i$ cover $f(E)$, but I think this is true since the $U_i$ cover $E$, and 2) the estimate on the measure of $V_i$. I actually don't think that is correct.

$\endgroup$

4 Answers 4

3
$\begingroup$

As $f$ is $M$-Lipschitz, we know that $m(f(E)) \leq M m(E)$, where $m$ denotes the Lebesgue measure on $\mathbb{R}$. If $E$ has measure $0$, then $0 \leq m(f(E)) \leq M \cdot 0 = 0$.

The proof of $m(f(E)) \leq M m(E)$ goes exactly as your proof, which is correct (note you can prove something more general in the same way).

$\endgroup$
1
  • $\begingroup$ For $x<y,$ for $J= (x,y)$ and $f(J)=\{f(z): z\in J\cap $ dom$(f)\}$ we have $[\sup f(J)]-[\inf f(J)]\leq M (y-x).$ So $m^o(f(J))\leq M\cdot m(J)$ where $ m^o$ is Lebesgue outer measure. For $ e>0$ let $F_e=\{J_{n,e}\}_{n\in N}$ be a family of open intervals with $\cup F_e\supset E$ and $m(\cup F_e)<e.$.....+1 $\endgroup$ Mar 7, 2016 at 19:47
2
$\begingroup$

Your proof looks good. And the two assumptions that you are unsure of can be justified the following way.

  • Assumption 1). Note that \begin{align*} f(E)\subseteq f(\cup_{i=1}^{\infty} U_{i})=\cup_{i=1}^{\infty}f(U_i)=\cup_{i=1}^{\infty}V_{i}. \end{align*} So $(V_{i})_{i=1}^{\infty}$ is a cover of $f(E)$. If you need these intervals to be open (as in the def of Lebesgue measure), add $\frac{\varepsilon}{2^{i+1}}$ length to the endpoints of $V_{i}$, this will add up at most $\varepsilon$ to the overall measure.

  • Assumption 2). Let $x_i,y_i\in [a_i,b_i]$ be such that $f(x_i)=\inf_{x\in [a_i,b_i]}f(x)$ and $f(y_i)=\sup_{y\in [a_i,b_i]}f(y)$. Since $f$ is continuous and $[a_i,b_i]$ is compact, both of these points exist. Then \begin{align*} |V_{i}|=|f(U_i)|\leq |[f(x_i),f(y_i)]|=|f(x_i)-f(y_i)|\leq M|x_i - y_i|\leq M|a_i-b_i|=M|U_i|. \end{align*} So the second assumption is also correct.

$\endgroup$
1
$\begingroup$

Just to fill in one missing detail that you expressed concern about, you can prove rigorously that $|V_i| \le M |U_i|$ in the following fashion. Let $V_i = [c_i,d_i]$. Choose $a'_i,b'_i \in U_i$ such that $f(a'_i)=c_i$ and $f(b'_i)=d_i$. Then $$|V_i| = |c_i-d_i| \le M |a'_i - b'_i| \le M |a_i - b_i| = M|U_i| $$ Note: your question seems to have an unstated assumption, which I am using, that the domain of $f$ is an interval. I am also feeling free to think of $U_i$ and $V_i$ as closed intervals, in order to make use of their endpoints.

$\endgroup$
1
$\begingroup$

I think you are pretty close. You can bound your $V_i$'s with $W_i=f([a_i,b_i])$, if $U_i=(a_i,b_i)$. Since Lipschitz implies continuity, you know that $f$ achieves its maximum and minimum on $[a_i,b_i]$ for each $i$. Say that the minimum is located at $c_i$ and the maximum at $d_i$. Then $m(V_i)\leq m(W_i)\leq m([f(c_i),f(d_i)])=|f(d_i)-f(c_i)|\leq M|d_i-c_i|\leq Mm(U_i)$ The rest should follow from that, then.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .