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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.

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    "if the lipschitz condition does not hold, then f(A) will not **necessarily** have measure zero".2012-08-17

1 Answers 1

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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.