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Could someone give me an example of a set $Y\subset \mathbb{R}$ that has zero Lebesgue measure and a continuous function $f:X\subset \mathbb{R}\to\mathbb{R}$ such that $Y\subset X$ and $f(Y)$ is not a set of zero Lebesgue measure?

Thanks.

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The devils staircase on the Cantor set. The Cantor set is a null set and its image is $[0,1]$

The Cantor set can be written as all $x$ of $\mathbb{R}$ such that $$x=\sum_{n=1}^\infty \frac{a_n}{3^n}$$ where $a_n\in\{0,2\}$ and your functions maps $x$ to $$f(x)=\sum_{n=1}^\infty \frac{a_n}{2^{n+1}}$$ when $x$ is in the Cantor set and the trivial extension on $[0,1]$ (It is constant outside the Cantor set).

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  • $\begingroup$ This seems to be "the one" answer, I've never seen a substantially different one. I'd be interested if someone knows a qualitatively different example. $\endgroup$ – user2566092 Aug 29 '13 at 17:46
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    $\begingroup$ @user2566092 Well a partial result could be that all lipschitz continuous functions maps null sets to nullsets $\endgroup$ – Dominic Michaelis Aug 29 '13 at 17:48
  • $\begingroup$ @user2566092 so we need some function which isn't local lipschitz anywhere, maybe the Weierstraßfunction ($f(x)=\sum_{n=0}^\infty \frac{\sin(2^k x)}{3^k}$) will work too $\endgroup$ – Dominic Michaelis Aug 30 '13 at 6:26
  • $\begingroup$ Sorry I mean $\sum_{k=0}^\infty \frac{2^k \sin(2^k x)}{3^k}$ for sure $\endgroup$ – Dominic Michaelis Aug 30 '13 at 7:59
  • $\begingroup$ @user2566092 Well, since every compact metrizable space is a continuous image of the Cantor set, the simplest and most natural example is a mapping of the Cantor set onto the closed unit interval. For the sake of variety, you could take a mapping of the (standard) Cantor set onto a fat Cantor set. $\endgroup$ – bof Nov 15 '17 at 5:01
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This is relates very closely to the Luzin N property, which informally states an absolutely continuous function maps measure zero sets to measure zero sets.

Formally, for $f:X\rightarrow Y$ (let's consider $X,Y\subset \mathbb{R}$ for simplicity), and $\mu(X)=0$, we have $\mu(f(X))=0$. Such a function is said to possess the Luzin N property. As it turns out, a function is absolutely continuous if and only if it possesses this property. Consider the cantor function $f:C\rightarrow [0,1]$, where $C$ denotes the standard ternary Cantor set. Note that $\mu(C)=0$ (measure of the cantor set is $0$) but $f(x)$ maps the Cantor set onto the interval $[0,1]$ and so $\mu(f(C))=1$. The Luzin N property fails, and as can be seen from many other proofs, Cantor's function is not absolutely continuous.

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  • $\begingroup$ Doesn't the said equivalence hold only under the assumption of bounded variation? (See this) $\endgroup$ – Pedro Nov 15 '17 at 5:03
  • $\begingroup$ absolute continuity implies bounded variation, so it is enough to say $f$ has Luzin N property if and only if absolutely continuous, because that will imply $f$ is of bounded variation $\endgroup$ – Kernel_Dirichlet Nov 15 '17 at 5:04

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