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I am looking at an old qualifying exam to study for my finals; it asks the following true/false question:

Let $f$ be a continuous, non-decreasing function defined on $[0,1]$, and let $E$ be a set of Lebesgue measure zero. Then, $f(E)$ is a set of Lebesgue measure zero.

I suspect this is false, but am not sure. Can anyone think of a solution without using the notion of absolute continuity? The reason for this is because AC won't be covered on the final.

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up vote 7 down vote accepted

The Cantor-Lebesgue function $f$ — also known “the devil's staircase” — is continuous and monotonically increasing. It maps the unit interval onto the unit interval. Since the complement $[0,1] \smallsetminus C$ of the Cantor set $C$ has countable image, $f(C)$ must have measure one.

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@Thomas: I meant what I wrote. $[0,1] = f(C) \cup f(C^c)$ and I argued that $f(C^c)$ is a null-set. – t.b. Dec 10 '11 at 18:35
Ah, okay, thank you. I can't imagine coming up with a function like that on a qual unless one has seen it before! – Sid Raval Dec 10 '11 at 18:39
@Sid: I think that you could do worse than study that function for your qual -- no one would expect you to come up with it on your own, but they might expect you to know it. It is a standard counterexample to many things. For instance, it is differentiable almost everywhere, and wherever it exists the derivative is zero. Hence $f(x) \neq \int_{0}^x f'(t)\,dt$. Keep it in mind, spend an hour or two of looking at it, it's definitely worth it! – t.b. Dec 10 '11 at 18:43
+1 since it's the same thing I would have said. – Michael Hardy Dec 10 '11 at 19:32
This function is also the basis for many other counterexamples, for example the one I described today in this answer: a continuous function such that the image of some Lebesgue measurable set is not Lebesgue measurable - – Carl Mummert Dec 16 '11 at 20:55

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