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Is there a continuous, strictly increasing (real-valued) function on the interval $[0,1]$ that maps some set of positive (lebesgue) measure onto a set of measure zero?

Should I play with cantor function to find, if exist, such a function?

Cantor function + $x$ does not work.

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    $\begingroup$ If $f$ is continuous and strictly increasing, then $f((a,b)) = (f(a),f(b))$ by the intermediate value theorem, and since it's strictly increasming, $f(a) < f(b)$ if $a < b$, so the image is a non-empty interval, which obviously has a non-zero measure... $\endgroup$
    – fgp
    Commented Apr 2, 2014 at 13:58
  • $\begingroup$ @fgp Thanks for your comment. I think in my question, 'any set' must be replaced by 'a set' that is 'some set'. $\endgroup$
    – le4m
    Commented Apr 2, 2014 at 14:02
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    $\begingroup$ If you drop the continuity requirement, then you can set $f(x) = \sum_{q \in \mathbb{Q}, 0 \leq q \leq x} 2^{-n_q}$, where $n_q$ is the index of $q$ in some enumeration of the (countable!) set $\mathbb{Q} \cap [0,1]$. Then the image of any set contains only rationals, therefore is countable, and hence has measure zero. $\endgroup$
    – fgp
    Commented Apr 2, 2014 at 14:02
  • $\begingroup$ I think still not - wouldn't any such function induce an absolutely continuous measure, and wouldn't mapping a non-zero set to a zero set contradict that? Just a quick idea though - didn't think this through... $\endgroup$
    – fgp
    Commented Apr 2, 2014 at 14:05
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    $\begingroup$ What about the inverse mapping of Cantor function + $x$? Doesn't it happen to be the one for the requirement? $\endgroup$
    – le4m
    Commented Apr 2, 2014 at 14:16

1 Answer 1

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Fix a fat Cantor set $K$ of positive measure (constructed in the standard way, that is, obtained by starting with a closed interval, removing an open interval from it, then removing an open interval from each of the two remaining closed intervals, etc), and consider the unique order preserving bijection between it and the standard Cantor set. This bijection exists, since both sets are order isomorphic to $2^{\mathbb N}$ with the lexicographic ordering, and it is in fact a homeomorphism: The order preserving bijection with $2^{\mathbb N}$ has the property that two points $x,y$ in $K$ are close, iff their images have a long initial sequence of $0$s and $1$s in common. But this is precisely saying that the map is continuous (in both directions).

Now, any continuous function from $K$ to $\mathbb R$ can be extended to a continuous function from $[0,1]$ (or from $\mathbb R$, if you wish) to $\mathbb R$: Simply extend the function linearly on each interval contiguous to $K$.

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