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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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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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    @Carl: Thanks for pointing this out. Indeed it is the failure of $f$ to be absolutely continuous that makes the example in your post work -- (which is also the reason that it doesn't satisfy the fundamental theorem of calculus as I mentioned above). It is also one of the standard examples of a function of bounded variation which isn't absolutely continuous.2011-12-16