# Lipschitz Condition on a Set of Measure 0

The question is as follows: Let $f:[0,1] \rightarrow R$ be a non-decreasing continuous function, and let $A \subset [0,1]$ be a set of Lebesgue measure $0$.

(a)Suppose that $f$ satisfies the following condition: (*) For some constant $C \geq 0$, $|f(s)-f(t)| \leq C|s-t|$ for all $s,t \in [0,1].$ Prove that $f(A)$ has measure zero.

This part is ok, I had no problem proving that. But part (b) has been giving me some issues:

(b) Suppose that $f$ does not satisfy condition (*). Does the conclusion in part (a) still hold?

Me and my office mate feel that if the lipschitz condition does not hold, then $f(A)$ will not have measure zero. We have tried to come up with counterexamples, but are having problems defining a continuous function on a set of measure 0.

-
google 'devil's staircase'. This will give you one counterexample for several questions you did not even ask ;-) – user20266 Aug 17 '12 at 15:35
See Cantor function. – Brian M. Scott Aug 17 '12 at 15:41
"if the lipschitz condition does not hold, then f(A) will not necessarily have measure zero". – user31373 Aug 17 '12 at 15:48

Consider $A$ – the standard embedding of the Cantor set $2^\omega$ into $[0,1]$ and the standard surjection $g:A\to [0,1]$, $g(\sum_n 3^{-n}a_n)=\sum 2^{-n}a_n/2$. Extend it to a continuous nondecreasing function from $[0,1]$ to $[0,1]$ (in the only possible way, segmentwise constant outside $A$). Note that $A$ has zero measure.

-